For the light beam
a) The index of refraction is 2.24.
b) The wavelength of the same light in air is 551.04 nm.
c) the angle is 17.14°.
Finda) The index of refractionb) The wavelengthc) The angle of refraction(a) The refraction index of the diamond at this wavelength can be found using the formula n=c/v, where c is the speed of light in a vacuum and v is the speed of light in a diamond.
n = c/v = 3.00 x 10^8 m/s / 1.34 x 10^8 m/s = 2.24
Therefore, the index of refraction of the diamond at this wavelength is 2.24.
(b) When light travels through air, its wavelength changes due to the change in the medium, but its frequency remains the same. The relationship between the speed, frequency, and wavelength of light is given by the formula c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency.
We can rearrange this formula to solve for the new wavelength:
λ_air = c/f = (c/v) λ_diamond = n λ_diamond
where n is the index of refraction of a diamond. Substituting the values given,
λ_air = 2.24 x 246 nm = 551.04 nm
Therefore, the wavelength of the same light in air is 551.04 nm.
(c) According to Snell's law, n1 sinθ1 = n2 sinθ2, where n1 and n2 are the indices of refraction of the initial and final mediums, and θ1 and θ2 are the angles of incidence and refraction, respectively, with respect to the normal.
We can rearrange this formula to solve for θ2:
sinθ2 = (n1 / n2) sinθ1
Substituting the values given, we get:
sinθ2 = (1 / 2.24) sin 15°
θ2 = sin⁻¹(0.295) = 17.14°
Therefore, the angle at which the light ray will be refracted into the air is 17.14°.
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A bicycle wheel is rotating at 47 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration of 0.50 rad/s2A) Part complete What is the wheel's angular velocity, in rpm, 10 s later?ANswer: 95rpm Having trouble with part b How many revolutions does the wheel make during this time?B). How many revolutions does the wheel make during this time?
(A) The Angular velocity, is 95 rpm. (B) During the 10-second period, the wheel makes approximately 11.83 revolutions.
To find the number of revolutions the wheel makes during this time, we'll first calculate the final angular velocity and then use the equations of motion to find the total angular displacement.
A) We've already calculated the final angular velocity, which is 95 rpm.
B) To find the number of revolutions during this time, follow these steps:
1. Convert the initial and final angular velocities from rpm to rad/s:
Initial angular velocity (ωi) = 47 rpm * (2π rad/1 rev) * (1 min/60 s) = 4.928 rad/s
Final angular velocity (ωf) = 95 rpm * (2π rad/1 rev) * (1 min/60 s) = 9.95 rad/s
2. Use the formula for angular displacement with constant angular acceleration:
Δθ = ωi * t + 0.5 * α * ²
where Δθ is the angular displacement, t is the time (10 s), and α is the angular acceleration (0.50 rad/s²).
3. Plug in the values:
Δθ = (4.928 rad/s) * (10 s) + 0.5 * (0.50 rad/s²) * (10 s)²
Δθ = 49.28 rad + 25 rad
Δθ = 74.28 rad
4. Convert the angular displacement from radians to revolutions:
Number of revolutions = Δθ * (1 rev/2π rad)
Number of revolutions = 74.28 rad * (1 rev/6.2832 rad) ≈ 11.83 revolutions
During the 10-second period, the wheel makes approximately 11.83 revolutions.
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when 1.0-µc point charge is 15 m from a second point charge, the force each one experiences a force of 1.0 µn. what is the magnitude of the second charge? (k = 1/4πε0 = 9.0 × 109 n • m2/c2)
Therefore, the magnitude of the second charge is 0.066 µC.
The electric force between two point charges is given by Coulomb's law: Here F is the force, k is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
In this problem, we have two charges with equal magnitudes of 1.0 µC, and they each experience a force of 1.0 µN at a distance of 15 m. Plugging in the given values, we get:
[tex]F = k * (q1 * q2) / r^2[/tex]
1.0 µN =[tex](9.0 * 10^9 N*m^2/C^2)[/tex] *[tex](1.0 uC)^2[/tex] / (15 m)^2
The unknown charge q2:
q2 = [tex]\sqrt{((1.0 uN * 15 m)^2 / (9.0 * 10^9 N*m^2/C^2))}[/tex]
q2 = 0.066 µC
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A full elevator has a mass of 1785.kg. You would like the elevator to go down at a constant speed of 0.650 m/s. What is the power rating of the motor that can handle this?
Power rating of the motor will be 11.38 kW
To calculate the power rating of the motor needed to move the elevator at a constant speed, we can use the formula for power:
Power = Force x Velocity
First, we need to determine the force acting on the elevator. Since it is moving at a constant speed, the force is equal to the gravitational force:
Force = Mass x Gravity
Force = 1785 kg x 9.81 m/s²
Force = 17505.85 N
Now, we can calculate the power:
Power = Force x Velocity
Power = 17505.85 N x 0.650 m/s
Power = 11378.8025 W
So, the power rating of the motor required to move the elevator downwards at a constant speed of 0.650 m/s is approximately 11,378.8 W or 11.38 kW.
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a proton with a kinetic energy of 4.8×10−16 jj moves perpendicular to a magnetic field of 0.37 t. What is the radius of its circular path in cm?
To determine the radius of the proton's circular path, we can use the formula for the centripetal force acting on a charged particle moving in a magnetic field:
F = qvb
where F is the centripetal force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength.
The centripetal force is provided by the magnetic force, which can be expressed as:
F = mv^2 / r
where m is the mass of the particle and r is the radius of the circular path.
Equating these two expressions for F, we get:
mv^2 / r = qvb
Solving for r, we get:
r = mv / qb
To find the velocity of the proton, we can use the formula for the kinetic energy of a particle:
KE = (1/2)mv^2
Solving for v, we get:
v = sqrt(2KE / m)
Substituting this expression for v into the equation for r, we get:
r = sqrt(2KE / m) * m / qb
Substituting the given values, we get:
r = sqrt(2 * 4.8×10^-16 / 1.67×10^-27) * 1.67×10^-27 / (1.6×10^-19 * 0.37)
r = 0.010 cm (rounded to three significant figures)
Therefore, the radius of the proton's circular path is 0.010 cm.
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if a 100 microfarad capacitor is charged off of a 12 volt battery, how many volts will be present between the terminals of the capacitor?
If a 100 microfarad capacitor is charged off of a 12-volt battery, the voltage between the terminals of the capacitor will eventually reach 12 volts when it is fully charged. Initially, when the capacitor is uncharged, the voltage across its terminals is zero.
However, when it is connected to the battery, current flows from the battery to the capacitor, charging it up. As the capacitor charges, the voltage across its terminals increases until it reaches the same voltage as the battery (in this case, 12 volts). The time it takes for the capacitor to fully charge depends on the resistance of the circuit in which it is connected.
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Suppose you were looking at two stars, both at the same distance, but while star A is a G5 I, star B is a G5 III. How would they look different to you in a telescope?A. Star A would be brighterB. Star B would be brighterC. Both the same brightness
If you were looking at two stars, both at the same distance, with star A being a G5 I (a supergiant) and star B being a G5 III (a giant), the difference in appearance through a telescope would be their brightness. In this case, star A (G5 I) would be brighter, making option A the correct answer.
Both stars would appear the same brightness in a telescope because their distance is the same. However, the main difference between them is their luminosity class, with star A being a main sequence star (luminosity class I) and star B being a giant star (luminosity class III). This difference in luminosity class suggests that star B is older and has exhausted more of its fuel than star A, which is still in its main sequence phase.
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If m=50 kg and a=2 m/s², what is force?
Answer:
Explanation:
by Newton's second law:
F = m*a
F = 50*2 = 100 Newton
The Previous and current values of a metre are 5,600 Kw and 6, 800kw respectively. If the monthly metre Charge is # 20 and one unit of the kw is #4, calculate the amount to be paid
Explanation:
6800 - 5600 = 1200 kw-hr used
total charge is meter charge + kw-hr charge
= 20 + 4 (1200) = # 4820
A drum rotates around its central axis at an angular velocity of 19.5 rad/s. If the drum then slows at a constant rate of 5.35 rad/s2, (a) how much time does it take and (b) through what angle does it rotate in coming to rest?
The drum rotates through an angle of approximately 70.98 radians in coming to rest. To find the time it takes for the drum to come to rest, we need to use the formula:
final angular velocity = initial angular velocity + (angular acceleration x time)
In this case, the final angular velocity is 0 (since the drum comes to rest), the initial angular velocity is 19.5 rad/s, and the angular acceleration is -5.35 rad/s^2 (since the drum is slowing down).
So we have:
0 = 19.5 - 5.35t
Solving for t, we get:
t = 3.64 seconds
Therefore, it takes 3.64 seconds for the drum to come to rest.
To find the angle through which the drum rotates in coming to rest, we need to use the formula:
angular displacement = (initial angular velocity x time) + (0.5 x angular acceleration x time^2)
In this case, the initial angular velocity is 19.5 rad/s, the time is 3.64 seconds (as we just calculated), and the angular acceleration is -5.35 rad/s^2. So we have:
angular displacement = (19.5 x 3.64) + (0.5 x -5.35 x 3.64^2)
angular displacement = 70.98 radians (rounded to two decimal places)
Therefore, the drum rotates through an angle of approximately 70.98 radians in coming to rest.
(a) To find the time it takes for the drum to come to rest, we can use the formula for angular acceleration: α = (ωf - ωi) / t, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time. Since the drum comes to rest, ωf = 0.
Given that the drum slows at a constant rate of 5.35 rad/s², we have α = -5.35 rad/s² (negative because it's slowing down) and ωi = 19.5 rad/s.
Plugging these values into the formula, we get:
-5.35 = (0 - 19.5) / t
Solving for t, we find that t ≈ 3.64 seconds.
(b) To find the angle through which the drum rotates, we can use the formula θ = ωi*t + 0.5*α*t².
Plugging in the values,
we get θ = 19.5 * 3.64 + 0.5 * (-5.35) * (3.64)².
Calculating this, we find that θ ≈ 35.53 radians.
So, the drum takes approximately 3.64 seconds to come to rest and rotates through an angle of about 35.53 radians in the process.
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An iron nail is driven into a block of ice by a single blow of a hammer. The hammerhead has a mass of 0.5 kg and an initial speed of 2 m/s. Nail and hammer are at rest after the blow. How much ice melts? Assume the tempera- ture of both the ice and the nail is 0°C before and after. The heat of fusion of ice is 80 cal/g. Answer in units of g. Answer in units of g.
The amount of ice that melts is approximately 0.596 g.
How to solve for the iceTo solve this problem, we need to use the conservation of momentum and the conservation of energy.
Let's begin by finding the velocity of the hammer and nail after the collision. We can use the conservation of momentum to do this:
(mass of hammer + mass of nail) x initial velocity of hammer = (mass of hammer + mass of nail) x final velocity of hammer and nail
(0.5 kg + m) x 2 m/s = (0.5 kg + m) x 0
where m is the mass of the nail.
Solving for m, we get:
m = 0.25 kg
So the mass of the nail is 0.25 kg.
Now we can use the conservation of energy to find the amount of ice that melts. The initial kinetic energy of the hammer and nail is:
KE = 0.5 x 0.5 kg x (2 m/s)^2 = 1 J
The final kinetic energy of the hammer and nail is zero, since they come to rest.
The energy required to melt the nail and the ice that it is in contact with is:
Q = (mass of nail + mass of melted ice) x heat of fusion of ice
We can assume that all the energy from the hammer's kinetic energy is used to melt the nail and the ice.
So we have:
1 J = (0.25 kg + m) x 80 cal/g x 4.184 J/cal
Solving for m, we get:
m = 0.596 g
Therefore, the amount of ice that melts is approximately 0.596 g.
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In order to jump off the floor, the floor must exert a force on you a. in the direction of and equal to your weight. b. opposite to and equal to your weight. c. in the direction of and less than your weight. d. opposite to and less than your weight. e. opposite to and greater than your weight.
The correct option is b. When you jump off the floor, the floor exerts a force on you that is opposite to and equal to your weight. This force is called the reaction force and it is a fundamental law of physics known as Newton's Third Law of Motion.
When you push against the floor, the floor pushes back with the same force in the opposite direction, allowing you to jump upwards. This force is equal to your weight because of gravity, which is pulling you down toward the ground.
If the force was less than your weight, you would not be able to jump off the floor as you would not be able to overcome gravity. If the force was greater than your weight, you would be pushed upwards with a greater force and jump higher than intended.
Therefore, option B is the correct answer as the floor exerts a force opposite to and equal to your weight when you jump off the floor.
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all waves have which four characteristics
Answer:it is a
Explanation:because No matter whether you are talking about vibrations or waves, all of them can be characterized by the following four characteristics: amplitude, wavelength, frequency, and speed.
Answer:
A
Explanation:
These 4 are the main characteristics which you should have learned in your class. Hope it helps!
c) What is the initial velocity?
d) What is the final velocity at t=6
e) What is the average acceleration? (Use the graph)
(a) The initial velocity is 0 m/s.
(b) The final velocity at t=6s, is 10 m/s.
(c) The average acceleration is 1.67 m/s².
What is instantaneous velocity?Instantaneous velocity is a measure of how fast an object is moving at a particular moment in time. It is the velocity of an object at a specific instant or point in time, and it is typically represented as a vector with both magnitude and direction.
The initial velocity = 0 m/s
The velocity of the particle at time, t = 6 seconds = displacement/time
velocity = 60 m/ 6s = 10 m/s
The average acceleration = (v₂ - v₁) / (t₂ - t₁)
average acceleration = (10 m/s - 0 m/s )/ (6 s - 0 s) = 10/6 = 1.67 m/s²
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A spaceship and its shuttle pod are traveling to the right in a straight line with speed v, as shown in the top figure above. The mass of the pod is m, and the mass of the spaceship is 6m. The pod is launched, and afterward the pod is moving to the right with speed vp and the spaceship is moving to the right with speed vf where vf > v as shown in the bottom figure. Which of the following is true of the speed vc of the center of mass of the system after the pod is launched?
A)vc=vf
B) v
C) vc
D) vc=v
(The correct answer is D. Can anyone explain why?)
The speed v(c) of the center of mass of the system after the pod is launched is equal to v(f).
Mass of the pod, m₁ = m
Mass of the spaceship, m₂ = 6m
The conservation of momentum principle states that, within a given domain, the amount of momentum is constant such that, momentum is never created nor destroyed, but only modified by the application of forces.
So, according to the conservation of momentum, the momentum before launch and before launch must be equal. Therefore, the speed of the center of mass of the system becomes equal to the speed with which the spaceship is moving towards the right.
Therefore,
v(c) = v(f)
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(1)
A wheel of radius R and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses m, M, and 2M, respectively, are mounted on the rim of the wheel, If the system is in static equilibrium, what is the value of m in terms of M?
To find the value of m in terms of M, we need to use the principle of torque equilibrium. Since the system is in static equilibrium, the net torque acting on it must be zero.
Let's consider the torque acting on each of the masses. The torque due to gravity on the mass m is m*g*R*sin(theta), where g is the acceleration due to gravity and theta is the angle between the radius vector and the vertical direction. Similarly, the torque on the mass M is M*g*R*sin(theta), and the torque on the mass 2M is 2M*g*R*sin(theta).
Now, since the wheel is in static equilibrium, the net torque acting on it must be zero. This means that the sum of the torques due to gravity on the masses must be equal to zero.
m*g*R*sin(theta) + M*g*R*sin(theta) + 2M*g*R*sin(theta) = 0
Simplifying this equation, we get:
(m + 3M)*g*R*sin(theta) = 0
Since sin(theta) cannot be zero, we have:
m + 3M = 0
Therefore, the value of m in terms of M is:
m = -3M
Note that the negative sign indicates that the mass m is located on the opposite side of the wheel compared to the masses M and 2M, which is necessary for the system to be in static equilibrium.
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Consider a cylindrical specimen of a steel alloy 10.0 mm (0.39 in.) in diameter and 75 mm (3.0 in.) long that is pulled in tension. Determine its elongation when a load of 20,000 N (4,500 lbf) is applied.
the elongation of the cylindrical steel alloy specimen with a diameter of 10.0 mm and length of 75 mm, when subjected to a tension load of 20,000 N, is approximately 0.095 mm.
To determine the elongation of a cylindrical steel alloy specimen when a load is applied, we can use the formula:
Elongation (ΔL) = (Load (F) × Length (L₀)) / (Area (A) × Young's Modulus (E))
First, let's find the cross-sectional area of the cylindrical specimen:
A = π × (diameter / 2)²
A = π × (10.0 mm / 2)²
A = 78.54 mm²
Next, we need the Young's Modulus (E) of the steel alloy. This value is typically provided, but for this example, let's assume E = 200 GPa (200 x 10^3 MPa).
Now we can calculate the elongation:
ΔL = (20,000 N × 75 mm) / (78.54 mm² × 200,000 MPa)
ΔL = (1,500,000 N·mm) / (15,708,000 N)
ΔL ≈ 0.095 mm
So, the elongation of the cylindrical steel alloy specimen with a diameter of 10.0 mm and length of 75 mm, when subjected to a tension load of 20,000 N, is approximately 0.095 mm.
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. identify the expressions 1) that defines change in population in unit change in time, and 2) the final population size.
it can be expressed as:
r = (Nt - N0) / (t - t0)
where:
- Nt is the population size at time t
- N0 is the population size at time t0
- t is the final time
- t0 is the initial timetime.
The population growth equation is given by:
Nt = N0 * e^(rt).
1) The expression that defines change in population in unit change in time is the population growth rate, which is typically denoted as "r". The population growth rate is the rate at which a population is increasing or decreasing over a given time interval. Mathematically, it can be expressed as:
r = (Nt - N0) / (t - t0)
where:
- Nt is the population size at time t
- N0 is the population size at time t0
- t is the final time
- t0 is the initial time
2) The expression that defines the final population size is simply Nt, which represents the population size at a given time t.
The final population size, Nt, can be calculated using the population growth equation, which takes into account the population growth rate, r, and the initial population size, N0. The population growth equation is given by:
Nt = N0 * e^(rt)
where:
- e is the mathematical constant e (approximately 2.71828)
- r is the population growth rate
- t is the time interval over which the population is growing or declining
The population growth equation assumes exponential growth or decline, which means that the rate of change of the population is proportional to the current population size. If the population growth rate is positive, the population is increasing, and if the growth rate is negative, the population is decreasing.
It's important to note that the population growth equation is a simplified model and may not accurately represent the dynamics of all populations in all situations. Factors such as limited resources, competition, and environmental changes can all affect population growth rates and the final population size.
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what is magnitude of gravitational force acting on the space junk by the satellite?
The magnitude of gravitational force acting on the space junk by the satellite is determined by the masses of the objects and the distance between them.
To calculate the magnitude of the gravitational force acting on the space junk by the satellite, you need to use the formula for gravitational force: F = G × (m1 × m2) / r² where: - F is the gravitational force, - G is the gravitational constant (approximately 6.674 × 10⁻¹¹ N m²/kg²), - m1 and m2 are the masses of the satellite and space junk, respectively, - r is the distance between the centers of mass of the satellite and space junk.
The larger the masses and the closer the objects are, the stronger the gravitational force. However, it is important to note that space junk is typically not in orbit around a satellite and therefore not subject to its gravitational force. Instead, space junk is affected by the gravitational force of the Earth and other celestial bodies, as well as other forces such as atmospheric drag and solar radiation pressure.
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In a mass spectrometer, germanium atoms have radii of curvature equal to 21.0, 21.6, 21.9, 22.2, and 22.8 cm. The largest radius corresponds to an atomic mass of 76 u.
What are the atomic masses of the other isotopes?
m21.0, m21.6, m21.9, m22.2 = ?
The relative atomic masses of the four germanium isotopes with curvature radii of 21.0, 21.6, 21.9, and 22.2 cm are roughly 1.64 u, 1.68 u, 1.70 u, and 1.72 u.
Why are isotopes' atomic masses different?Isotopes are the same atomic number but different mass number atoms of the same element.
To solve this problem, we can use the equation for the radius of curvature of an ion in a magnetic field:
r = (mv) / (qB)
where r = radius of curvature,
m = mass of the ion
v = velocity of the ion
q = charge of the ion
B = magnetic field strength. We can assume that the charge of the germanium ions is +1 (since they are singly charged ions), and we can use the mass of the isotope with the largest radius of curvature (corresponding to an atomic mass of 76 u) to find the velocity of the ions.
m = (qrB) / v
We can substitute values,
For r = 21.0 cm:
m = (1 x 1.602 x 10^-19 C x 0.25 T x 21.0 cm) / [(2 x 1.67 x 10^-27 kg) x v]
m = 68.4 u / v
For r = 21.6 cm:
m = (1 x 1.602 x 10^-19 C x 0.25 T x 21.6 cm) / [(2 x 1.67 x 10^-27 kg) x v]
m = 70.1 u / v
For r = 21.9 cm:
m = (1 x 1.602 x 10^-19 C x 0.25 T x 21.9 cm) / [(2 x 1.67 x 10^-27 kg) x v]
m = 71.0 u / v
For r = 22.2 cm:
m = (1 x 1.602 x 10^-19 C x 0.25 T x 22.2 cm) / [(2 x 1.67 x 10^-27 kg) x v]
m = 71.8 u / v
The velocity of the ions can then be calculated using the curve with the biggest radius:
r = (mv) / (qB)
v = (qrB) / m
v = (1 x 1.602 x 10^-19 C x 0.25 T x 22.8 cm) / [(2 x 1.67 x 10^-27 kg) x 76 u]
v = 4.17 x 10^4 m/s
The atomic masses of the other isotopes can be determined by substituting this velocity back into each equation:
m21.0 = 68.4 u / 4.17 x 10^4 m/s = 1.64 u
m21.6 = 70.1 u / 4.17 x 10^4 m/s = 1.68 u
m21.9 = 71.0 u / 4.17 x 10^4 m/s = 1.70 u
m22.2 = 71.8 u / 4.17 x 10^4 m/s = 1.72 u
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A rotating flywheel of a diameter 40.0 cm uniformly acceleratesfrom rest to 250 rad/s in 15.0 s. (a) Find its angularacceleration. (b) Find the linear velocity of a pointon the rim of the wheel after 15.0 s. (c) How manyrevolutions does the wheel make during the 15.0 s?
(a)the angular acceleration of the flywheel is 16.7 rad/s^2.
(b)the linear velocity of a point on the rim of the wheel after 15.0 s is 50 m/s.
(c)the wheel makes approximately 29.8 revolutions during the 15.0 s.
(a) The initial angular velocity of the flywheel, ω0 = 0. The final angular velocity, ω = 250 rad/s. The time, t = 15.0 s. Using the formula,
ω = ω0 + αt
where α is the angular acceleration, we can solve for α:
α = (ω - ω0)/t = 250 rad/s / 15.0 s = 16.7 rad/s^2
Therefore, the angular acceleration of the flywheel is 16.7 rad/s^2.
(b) The linear velocity, v, of a point on the rim of the wheel is given by:
v = rω. where r is the radius of the wheel. Substituting r = 0.2 m and ω = 250 rad/s, we get:
v = (0.2 m)(250 rad/s) = 50 m/s
Therefore, the linear velocity of a point on the rim of the wheel after 15.0 s is 50 m/s.
(c) The number of revolutions made by the wheel during the 15.0 s can be calculated using the formula: θ = ω0t + (1/2)αt^2
where θ is the angular displacement of the wheel. Since the wheel starts from rest, ω0 = 0. Also, the final angular velocity, ω, is given by:
ω^2 = ω0^2 + 2αθ
Solving for θ, we get: [tex]θ = (ω^2 - ω0^2) / 2α = (250^2 - 0^2) / (2 x 16.7) = 187.1 rad[/tex]
The number of revolutions, N, made by the wheel can be calculated as:
N = θ / 2π = 187.1 rad / (2π) = 29.8 revolutions (approx)
Therefore, the wheel makes approximately 29.8 revolutions during the 15.0 s.
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Find a) any critical values and b) any relative extrema. f(x)= x2 - 4x +9 a) Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The critical value(s) of the function is/are (Use a comma to separate answers as needed.) B. The function has no critical values. b) Select the correct choice below and, if necessary, fill in the answer box(es) within your choice, O A. The relative maximum point(e) is/are and there are no relative minimum points. (Simplify your answer. Type an ordered pair, using integers of fractions. Use a comma to separato answers as needed.) OB. The relative minimum point(e) in/are and there are no relative maximum points (Simplify your answer. Type an ordered poir, using integers or fractions. Use a comma to separato answers as needed.) OC. The relative minimum point(s) is/are and the relative maximum point(s) is/are (Simplify your answers. Type ordered pairs, using integers or fractions. Use a comma to separato answers as needed.) D. There are no relative minimum points and there are no relative maximum points.
a. The critical value of the function the function f(x) = x² - 4x + 9 is 2 (Option A).
b. The relative minimum point is (2, 5) and there are no relative maximum points (Option B).
To find the critical values and relative extrema of the function f(x) = x² - 4x + 9, we first need to find its derivative:
f'(x) = 2x - 4
Now, we find the critical values by setting the derivative equal to zero:
2x - 4 = 0
2x = 4
x = 2
So, the critical value of the function is 2.
Now, to find the relative extrema, we analyze the concavity of the function at the critical point:
f''(x) = 2
Since f''(x) > 0 for all x, the function is concave up, which means we have a relative minimum at the critical point. To find the value of the function at the critical point, we plug x = 2 back into the original function:
f(2) = (2)² - 4(2) + 9
= 4 - 8 + 9 = 5
So, the relative minimum point is (2, 5) and there are no relative maximum points.
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Suppose that we want to make bulb H dimmer than it was in circuit 9 (when 1 glow flowed through it). What will we need to do to the flow through H I1 In circuit 9, suppose that the obstacle presented by the rheostat and the obstacle presented by bulb Hare both of size L) For the sake of being definite, suppose that 1 glow is flowing. If the obstacle presented by the rheostat is doubled then the flow from the battery will decrease Will the flow decrease to less than half of what it was or to more than half of what it was? Explain. 121 What will happen to the pressure difference across the rheostat? Explain. 131 Hint: calculate the pressure difference across the rheostat before and after the rheostat changes. Be sure to use the correct obstacle and flow when doing this What will happen to the pressure difference across bulb H? Explain. 12l What will this do to the brightness of bulb H? Explain
To make bulb H dimmer than it was in circuit 9, you will need to decrease the flow (current) through it.
If the obstacle presented by the rheostat is doubled, the flow from the battery will decrease to less than half of what it was. This is because the overall resistance in the circuit has increased, leading to a reduced current.
The pressure difference (voltage) across the rheostat will increase due to the increased resistance. To calculate the pressure difference before and after the rheostat changes, use Ohm's Law (V=IR) with the correct obstacle and flow values.
The pressure difference across bulb H will decrease, as the overall current in the circuit has reduced due to the increased resistance of the rheostat. This decrease in pressure difference across bulb H will cause its brightness to diminish, making it dimmer than before.
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Two masses ma = 15kg and mb = 28kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
The acceleration of the masses is 4.19 m/s² and the tension in the string is 221.4 N when the masses are released.
How to find the acceleration of the masses and the tension in the string?Since the string is light and inextensible, the tension in the string will be the same throughout the length of the string.
Let's assume that the acceleration of both masses is a and that the direction of acceleration is downwards for the mass ma and upwards for the mass mb.
Using Newton's second law of motion for both masses, we can write the following equations:
ma * g - T = ma * a ...(1)
T - mb * g = mb * a ...(2)
where g is the acceleration due to gravity.
Adding both equations, we get:
ma * g - mb * g = (ma + mb) * a
Simplifying and solving for a, we get:
a = (ma - mb) * g / (ma + mb)
Substituting the given values, we get:
a = (15 kg - 28 kg) * 9.81 m/s² / (15 kg + 28 kg) = -4.19 m/s²
The negative sign indicates that the acceleration is in the opposite direction to the assumed direction of motion for mass ma.
Substituting the value of a in equation (1), we get:
T = ma * g - ma * a = ma * (g - a) = 15 kg * (9.81 m/s² + 4.19 m/s²) = 221.4 N
Therefore, the acceleration of the masses is 4.19 m/s² and the tension in the string is 221.4 N.
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True or false (The fracture toughness of a material increases with increasing temperature. )
2/ . Creep can be best described as:
A/fast elongation under high load at elevated temperature
B/ slow elongation under high load at elevated temperature
C/ fast elongation under low load at elevated temperature
D- slow elongation under low load at elevated temperature
e/ fast elongation under impact load at elevated temperature
The fracture toughness of a material increases with increasing temperature, the given statement is false because it is not dependent on temperature. Creep can be best described as B. slow elongation under high load at elevated temperature
Fracture toughness is a property of a material that describes its resistance to crack propagation. It is not necessarily dependent on temperature, but rather on the composition and microstructure of the material. However, some materials may exhibit a reduction in fracture toughness at elevated temperatures due to thermal stresses and microstructural changes.
Creep is a deformation mechanism that occurs in materials under prolonged exposure to high stress and temperature, it is a time-dependent process that causes gradual plastic deformation and elongation over time. The elongation is slow and occurs under high load and elevated temperature, and it can lead to structural failure over time if not properly accounted for in design and engineering applications. The first statement is false because it is not dependent on temperature. the second question creep can be best described as B. slow elongation under high load at elevated temperature
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To "observe" small objects, one measures the diffraction of particles whose de Broglie wavelength is approximately equal to the object's size. Find the kinetic energy (in electron volts) required for electrons to resolve a large organic molecule of size 10 nm.
Therefore, the kinetic energy required for electrons to resolve a large organic molecule of size 10 nm is approximate: [tex]9.18 * 10^{-3 }[/tex] eV.
The de Broglie wavelength of a particle is given by:
λ = h/p
For a non-relativistic particle, the momentum can be expressed as:
p = mv
where m is the mass of the particle and v is its velocity.
Equating these two expressions for p and solving for v, we get:
v = p/m = h/(mλ)
The kinetic energy of the particle can be expressed in terms of its velocity as:
For an organic molecule of size 10 nm, we can estimate its effective radius as half its size, or 5 nm. The de Broglie wavelength required to resolve this object is therefore:
λ = h/p = h/(mv) = h/(m√(2K/m)) = h/√(2mK)
where we have used the expression for velocity in terms of kinetic energy derived above.
Equating λ with the size of the object, we get:
λ = 2r = 10 nm
Substituting for λ and solving for K, we get:
[tex]K = (h^2/8mr^2) = (6.626 * 10^{-34} J s)^2/(8 * 9.109 * 10^{-31 }kg * (5 * 10^{-9 }m)^2)\\ = 9.18 * 10^{-3} eV[/tex]
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in an oscillating lc circuit with l = 39 mh and c = 4.6 μf, the current is initially a maximum. how long will it take before the capacitor is fully charged for the first time?
It will take approximately 0.00194 seconds before the capacitor is fully charged for the first time in this oscillating LC circuit.
In an oscillating LC circuit with L = 39 mH and C = 4.6 μF, we can determine the time it takes for the capacitor to be fully charged for the first time by first calculating the angular frequency (ω) and then finding the time period (T).
The angular frequency is given by the formula:
ω = 1 / √(LC)
Plugging in the values, we get:
ω = 1 / √(0.039 H * 4.6 × 10^(-6) F) ≈ 809.3 rad/s
The time period (T) is the reciprocal of the angular frequency:
T = 2π / ω
T ≈ 2π / 809.3 ≈ 0.00776 s
Since the capacitor is fully charged for the first time at a quarter of the time period, we divide the time period by 4:
t = T/4 ≈ 0.00776 s / 4 ≈ 0.00194 s
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An electron has a linear momentum of 4.0 x 10^-25 kg m^-2. What is the li order of magnitude of the kinetic energy of the electron? A. 10^-50j
B. 10^-34j C. 10^-19j D. 10^6j
Hi! To find the order of magnitude of the kinetic energy of the electron, we can use the relationship between linear momentum (p), mass (m), and kinetic energy (KE):
p = √(2m * KE)
We know the linear momentum (p) and the mass of an electron (m = 9.11 × 10^-31 kg). Let's solve for KE:
KE = (p^2) / (2m)
Plugging in the values:
KE ≈ ((4.0 × 10^-25 kg m/s)^2) / (2 × 9.11 × 10^-31 kg)
KE ≈ 8.79 × 10^-19 J
The order of magnitude of the kinetic energy of the electron is 10^-19 J. So, the correct answer is C. 10^-19 J.
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light is incident on an equilateral glass prism at a 45° angle to one face. calculate the angle at which light emerges from the opposite face. assume the index of refraction of the prism is 1.52.
The angle at which light emerges from the opposite face of the equilateral glass prism is 51.2°.
What is Refraction?Refraction is the bending of light when it passes from one medium to another. The angle of refraction is determined by the angle of incidence and the refractive index of the medium through which it is travelling.
The angle at which light emerges from the opposite face of the equilateral glass prism can be calculated using the concept of refraction.
In this case, the angle of incidence is 45° and the refractive index of the medium is 1.52.
Using Snell's law of refraction, the angle of refraction can be calculated as follows:
n₁ sinθ₁ = n₂ sinθ₂
Where n₁ is the refractive index of the incident medium, θ₁ is the angle of incidence, n₂ is the refractive index of the emergent medium, and θ₂ is the angle of refraction.
Therefore, substituting the values for n₁, θ₁ and n₂, the angle of refraction can be calculated as follows:
1.52 sin 45° = n₂ sinθ₂
n₂ sinθ₂ = 1.52 sin 45°
n₂ sinθ₂ = 1.08
θ₂ = sin-¹ (1.08/n₂)
θ₂ = sin-¹ (1.08/1.52)
θ₂ = 51.2°
Therefore, the angle at which light emerges from the opposite face of the equilateral glass prism is 51.2°.
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if a sprinter reaches his top speed of 10.5 m/s in 2.44 s , what will be his total time? express your answer in seconds.
The total time taken by the sprinter is 3.73 seconds.
Let's assume that the sprinter maintains a constant speed of 10.5 m/s after reaching it.
The time taken to reach the top speed is given as 2.44 seconds.
The distance covered during the time taken to reach the top speed can be calculated using the formula:
[tex]d = (1/2)*a*t^{2}[/tex]
Assuming that the sprinter starts from rest, the initial velocity is 0 m/s. The acceleration can be calculated as:
[tex]a = (v_f-v_i)/t = (10.5m/s-0m/s)/2.44s = 4.30m/s^{2}[/tex]
Substituting the values, we get:
[tex]d = (1/2)*4.30m/s^{2} * (2.44s)^{2} = 13.5[/tex]
The time taken to cover the remaining distance at a constant speed of 10.5 m/s can be calculated using the formula:
[tex]t = d/v = 13.5/10.5 m/s = 1.29s[/tex]
Therefore, the total time taken by the sprinter is:
total time = time taken to reach top speed + time taken to cover distance at top speed
= 2.44 s + 1.29 s
= 3.73 s
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A boy runs on a circular path of radius R = 28 m with a constant speed u = 4 m/s. Another boy starts from the centre of the path to catch the first boy. The second boy always remains on the radius connecting the centre of the circle and the first boy and maintains magnitude of his velocity constant V = 4 m/s. If the time of chase is (10 + x) sec then
Answer:
We can solve this problem by using the concept of relative motion. Let's assume that the first boy is running in the clockwise direction and the second boy is chasing him in the counterclockwise direction.
Since the second boy always remains on the radius connecting the center of the circle and the first boy, the distance between them is always equal to the radius of the circle, which is 28 m.
Let's denote the distance covered by the first boy as S1 and the distance covered by the second boy as S2. We know that the first boy is running with a constant speed of 4 m/s, so we can write:
S1 = u*t1
where t1 is the time taken by the first boy to complete the chase.
The second boy is moving with a constant velocity of 4 m/s towards the first boy, so we can write:
S2 = V*t2
where t2 is the time taken by the second boy to catch up with the first boy.
Since the second boy is always moving on the radius connecting the center of the circle and the first boy, the distance covered by him is equal to the distance on the circumference of the circle covered by the first boy, minus the distance covered by the first boy along the radius. We can write:
S2 = S1 - 2*pi*R
where pi is the mathematical constant pi (approximately equal to 3.14).
Substituting the values of S1 and S2, we get:
u*t1 = V*t2 + 2*pi*R
Since the time of chase is (10 + x) sec, we can also write:
t1 + t2 = 10 + x
We have two equations and two unknowns (t1 and t2), so we can solve for them. First, we can solve for t2:
t2 = (u*t1 - 2*pi*R) / V
Substituting this in the second equation, we get:
t1 + (u*t1 - 2*pi*R) / V = 10 + x
Simplifying this equation, we get:
t1*(1 + u/V) = 10 + x + 2*pi*R/V
Finally, we can solve for t1:
t1 = (10 + x + 2*pi*R/V) / (1 + u/V)
Substituting the given values of R, u, and V, we get:
t1 = (10 + x + 56*pi) / 20
Simplifying this expression, we get:
t1 = 2.8*pi + 0.5*x + 2.8
Therefore, the time taken by the first boy to complete the chase is 2.8*pi + 0.5*x + 2.8 seconds.
Explanation:
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