Answer:
y = y₀ / 4
Explanation:
In a double slit experiment the constructive interference lines are given by
d sin θ = m λ
if we use trigonometry
tan θ = y / L
as in these experiments the angles are very small
tan θ = sin θ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
let's use a subscript "o" for the initial values
I = m λ₀ L /d₀
They ask us to decrease the wavelength by 2
λ = λ₀ / 2
the distance in the slit is increased by a factor of two
d = 2 d₀
we substitute
2d₀ y/L = m λ₀/2
y = m λ₀ L/d₀ ¼
y = y₀ / 4
You drive past a potential parking space in center city. Your new car is travelling at 85% the speed of light. If your car is 6.0 m long (which you measured the day you bought it) and you observe the space to be 3.0 m, should you try to park? Why is your friend on the sidewalk (who hasn't studied relativity) so sure that you can park? How does the situation appear to him?
Answer:
We should not try to park the car because its rest length is greater than the space available.
The car seems to be approximately equal to the friend (L = 3.16 m). Due to this reason he is sure to park.
Explanation:
We should not try to park the car because its rest length is greater than the space available.
The friend is sure about parking because the car appears short in length to him. For this, we will solve Einstein's length contraction formula from theory of relativity:
[tex]L = L_o\sqrt{1-\frac{v^2}{c^2}}[/tex]
where,
L = Relative length observed by friend = ?
L₀ = rest length = 6 m
v = relative speed = 85% of speed of light = 0.85c
Therefore,
[tex]L = (6\ m)\sqrt{1-\frac{(0.85c)^2}{c^2}}[/tex]
L = 3.16 m
Hence, the car seems to be approximately equal to the friend. Due to this reason he is sure to park.
What does Coulomb's law applies to:
1. Like charges and unlike charges
2. Charge motion
3. gravitational fields
4. moving objects
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s
Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.62 10-6 W/m2 at a distance of 165 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity [tex]I[/tex] is proportional to 1/(distance)²
i.e
[tex]I[/tex] ∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2
Hence,
[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
consider a circular loop of wire carrying a counterclockwise current as shown. Indicate the direction of the magnetic field at points both inside and outside of the loop.
Answer:
in this case around the loop the field points downwards on the outside and upwards on the inside
Explanation:
To find the direction of the magnetic field in a wire you must use the right hand rule.
The thumb points in the direction of the current flow and the other created fingers point in the direction of the magnetic field.
Therefore in this case around the loop the field points downwards on the outside and upwards on the inside
How can spectroscopy and infrared technology be useful in space? (5 points)
a.)They can enhance speed by making spacecraft fuel more efficient.
b.) They can measure magnetic fields produced by astronomical bodies.
c.) They can provide an emergency escape to the astronaut from a space center.
d.) They can determine the elements that make up the surface of astronomical bodies.
Answer:
B
Explanation:
i took the test
A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid inside a control surface and determine the change in internal energy, in kJ, of this control mass.
Answer: [tex]3590\ kJ[/tex]
Explanation:
Given
Paddle wheel work is [tex]W=-5090\ kJ\quad \text{work is done on the system}[/tex]
Heat transfer from the tank is [tex]Q=-1500\ kJ\quad \text{heat taken from the system}[/tex]
From the first law of thermodynamics
Change in the internal energy of the system is equal to the difference of heat and work .
[tex]\Rightarrow \Delta U=Q-W\\\Rightarrow \Delta U=-1500-(-5090)\\\Rightarrow \Delta U=3590\ kJ[/tex]
Therefore, the change in internal energy is [tex]3590\ kJ[/tex]
How much power is delivered to a light bulb on a 120V, 0.5A
circuit?
Answer:
60 w
Explanation:
Given :
V= 120V
I=0.5 A
Now,
power can be calculated as :
P=VI
where,
V is voltage
I is current
Now,
P=(120)(0.5)
P=60 W
Therefore, 60w power is delivered to a light
If a second ball were dropped from rest from height ymax, how long would it take to reach the ground
Answer:
[tex](b)\ t_1 - t_0[/tex]
[tex](d)\ t_2 - t_1[/tex]
[tex](e)\ \frac{t_2 - t_0}{2}[/tex]
Explanation:
Given
See attachment for complete question
Required
How long to reach the ground from the maximum height
First, calculate the time of flight (T)
[tex]T =t_2 - t_0[/tex]
The time taken (t) from maximum height to the ground is:
[tex]t = \frac{1}{2}T[/tex]
So, we have:
[tex]t = \frac{t_2 - t_0}{2}[/tex]
Another representation is:
At ymax, the time is: t1
On the ground, the time is t2
The difference between these times is the time taken.
So;
[tex]t = t_2 - t_1[/tex]
Since air resistance is to be ignored, then
[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height
Two people that have identical weight are holding onto a massless pole while standing on horizontal frictionless ice. 1)If the guy on the left starts to pull on the pole, where do they meet
Answer:
Explanation:
From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.
So, if there is no external force;
The center of mass of the two people is:
[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]
Thus, In the system, no movement occurs and all forces remain the same.
What kind of model is shown below?
O A. A mathematical model
B. An experimental model
O C. A computer model
D. A physical Model
Answer:
B is excellent answer..............
The model of the brain that is shown here is the experimental model that is present in Option B, as it is used to study the brain's parts and its function, which is helpful for a better understanding of the brain.
What is an experimental model of the brain?There are various experimental models of the brain that have been developed to better understand its functions and mechanisms, such as Animal models, such as mice, rats, and primates, have been widely used to study the brain due to their similarity to the human brain in terms of structure and function. Computer models can simulate brain function and behavior at various levels of detail, from individual neurons to large-scale brain networks. These models are useful for testing hypotheses and predicting outcomes, as well as for designing new experiments.
Hence, the model of the brain that is shown here is the experimental model that is present in Option B.
Learn more about the experimental model of the brain here.
https://brainly.com/question/23802617
#SPJ7
Activity 1
The equation for the combustion of butane gas is given below.
1.1
1.2
AH < 0
butane(g) + 1302(g) → 8CO2(g) + 10H2O(g)
Define the term activation energy.
Is the combustion reaction of butane exothermic or endothermic? Give
reason for the answer.
Draw a sketch graph of potential energy versus course of reaction for
reaction above.
3
Clearly indicate the following on the graph:
o
Activation energy
Heat of reaction (AH)
Reactants and products
Determine the empirical formula of butane gas if it consists of 82,76%
and 17,24% hydrogen.
Answer:
I don't know hhaha ammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm I wish I could help
effieiency of simple machine is always less than 100% why
Answer:
efficiency of a machine is less than 100% because some part is energy is utilized to overcome some opposing forces like friction which is wasted as heat ,sound energy etc
Explanation:
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the 120-lb woman jogs up the flight of stairsThe 120-lb woman jogs up the flight of stairs in 5 seconds. Determine her average power output.
Answer:
Power = 24.41Watts
Explanation:
Find the diagram attached
Power output = Force * distance/Time
Given
Force = 120lb
Distance = 9inches
Time = 5sec
Since
1lb = 4.4482216153 N
120lb = 120 * 4.4482216153
120lb = 533.787N
9in to meters
9in = 0.0254*9
9in = 0.2286N
Power = 533.787*0.2286/5
Power = 24.41Watts
An illustration with two positive spheres 0.1m apart. The one on the left is labeled q Subscript 1 baseline = 6 microcoulombs and the sphere on the right is labeled q Subscript 2 baseline = 2 microcoulombs.
Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.
What is the force applied between q1 and q2?
In which direction does particle q2 want to go?
Answer:
F = 10.78 N
Hence q₂ will move away from the charge q₁ towards right side.
Explanation:
The force between two charged particles can be found by using Colomb's Law:
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where,
F = Force = ?
k = Colomb Constant = 8.99 x 10⁹ N.m²/C²
q₁ = charge on first particle = 6 μC = 6 x 10⁻⁶ C
q₂ = charge on second particle = 2 μC = 2 x 10⁻⁶ C
r = distance between particles = 0.1 m
Therefore,
[tex]F = \frac{(8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(2\ x\ 10^{-6}\ C)}{(0.1\ m)^2}[/tex]
F = 10.78 N
Since both particles have a positive charge. Therefore this force will be the force of repulsion.
Hence q₂ will move away from the charge q₁ towards right side.
Answer:
Explanation:
E2020
What kind of model is shown below?
о
A. Experimental model
O B. Computer model
O C. Mathematical model
O D. Physical model
Answer:
.....where's the model-
tha motor pumps water from a well 10 m deep and projects it at a speed of 15 meters per second.the water pause ftom the pipe at a rate of 1200 kg per minute calculate the power of the motor
Answer:
4210 W
Explanation:
We'll begin by calculating the energy required in lifting the water from the depth. This can be obtained as follow:
Mass (m) = 1200 Kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E₁) =?
E₁ = mgh
E₁ = 1200 × 9.8 × 10
E₁ = 117600 J
Next, we shall determine the energy required to project the water at 15 m/s. This can be obtained as follow:
Mass (m) = 1200 Kg
Velocity (v) = 15 m/s
Energy (E₂) =?
E₂ = ½mv²
E₂ = ½ × 1200 × 15²
E₂ = 600 × 225
E₂ = 135000 J
Finally, we shall determine the power of the motor. This can be obtained as follow:
Total energy (E) = E₁ + E₂
= 117600 + 135000
= 252600 J
Time (t) = 1 min = 60 s
Power (P) =?
P = E/t
P = 252600 / 60
P = 4210 W
Therefore, the power of the motor is 4210 W.
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
Voltages in series can be added together if the voltages are aiding each other.
a. True
b. False
light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in one year.
Answer:
The distance traveled in 1 year is: [tex]3.143*10^{16}ft[/tex]
Explanation:
Given
[tex]s = 982,080,000 ft/s[/tex] --- speed
[tex]t = 32,000,000 s[/tex] --- time
Required
The distance traveled
This is calculated as:
[tex]Speed = \frac{Distance}{Time}[/tex]
So, we have:
[tex]Distance = Speed * Time[/tex]
This gives:
[tex]Distance = 982,080,000 ft/s * 32,000,000 s[/tex]
[tex]Distance = 982,080,000 * 32,000,000ft[/tex]
[tex]Distance = 3.143*10^{16}ft[/tex] -- approximated
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Where does the extra energy go?
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Then the extra energy goes in overcoming the gravitational acceleration and heating up of body etc
A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright fringes spaced 6.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 6.30 mm apart.
Required:
What is the wavelength of light produced by the pointer?
Answer:
Wavelength = [tex]\lambda_p = 3.986 * 10^{-6}[/tex] m
Explanation:
As we know
Fringe width (w) = [tex]\frac{D*\lambda}{d}[/tex]
where
[tex]\lambda[/tex] is the wavelength
D is distance between source and screen
d is the distance between two slits
[tex]\frac{D}{d} = \frac{y}{\lambda}[/tex]
[tex]\frac{D}{d} = \frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\lambda_p = \frac{y_p* \lambda_r}{y_r} \\\lambda_p =\frac{6.30 * 10^{-3} * 632.8 * 10^{-9}}{6 *10^{-3}} \\\lambda_p = 3.986 * 10^{-6}[/tex]m
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
As a new electrical engineer for the local power company, you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. Besides plenty of wire, you have two strong magnets that can produce a constant uniform magnetic field of 1.5 T over a square area with a length of 10.2 cm on a side when the magnets are separated by a distance of 12.8 cm . The basic design should consist of a square coil turning in the uniform magnetic field. To have an acceptable coil resistance, the coil can have at most 400 loops.
What is the minimum rotation rate of the coil so it will produce the required voltage? Express your answer using two significant figures.
Answer:
The rotation rate is 15.3 rad/s.
Explanation:
maximum voltage, V = 120 V
Magnetic field, B = 1.5 T
length, L = 10.2 cm
width, W = 12.8 cm
Number of loops, N = 400
Let the rate of rotation is w.
The maximum voltage is given by
V = N B A w
120 = 400 x 1.5 x 0.102 x 0.128 x w
w = 15.3 rad/s
Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More
Answer:
Less
Explanation:
Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.
How much work can a motor with a power output of 0.70 hp do in 2 s?
Answer:
the work done by the motor is 1,044 J.
Explanation:
Given;
the output power of the motor, P = 0.7 hp
duration of the work, t = 2 s
The relationship between horse-power and watt is given as;
1 hp = 745.7 W
0.7 hp = ?
0.7 hp = 522 W = 522 J/s
The work done by the motor is calculated as;
W = Power x time
W = 522 J/s x 2 s
W = 1,044 J
Therefore, the work done by the motor is 1,044 J.
PLEASE HELPPPPPP!!!!!!
Explanation:
Since gravitational forces are inversely proportional to the square of their distances, tripling the distance means the reduction of the force by a factor of 9. That means the force experienced by the two objects will be 1/9 smaller than before.
Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce the sound intensity by a factor of 305. If, when the jet is fired up, the sound intensity level experienced by the crew members wearing protective earplugs is 79 dB, determine the sound intensity level they would experience without the earplugs.
Answer:
the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
Explanation:
Given the data in the question;
sound intensity reduced by the factor, m = 305
the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB
Now, using the expression of sound intensity level;
L = 10log( [tex]I_0[/tex] )
where [tex]I_0[/tex] is the intensity at L level
so we substitute
79 = 10log( [tex]I_0[/tex] )
[tex]I_0[/tex] = [tex]10^{7.9[/tex]
Now, expression for actual intensity;
[tex]I[/tex] = m[tex]I_0[/tex]
where [tex]I[/tex] is the actual intensity
so we substitute
[tex]I[/tex] = 305 × [tex]10^{7.9[/tex]
Next, we write the expression of sound intensity level for reduced intensity;
L' = 10log( [tex]I[/tex] )
So we substitute
L' = 10log( 305 × [tex]10^{7.9[/tex] )
L' = 10log( 24227011159.09058 )
L' = 103.8 dB
Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 32.9 mm. What is the closest object that can be photographed
Answer:
p = 6.64 cm
Explanation:
For this exercise we use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image
[tex]\frac{1}{p} = \frac{1}{f} - \frac{1}{q}[/tex]
1 / p = 1 / 2.2 - 1/3.29
1 / p = 0.15059
p = 6.64 cm
therefore the farthest distance from the object is 6.64 c