Answer:
[tex]0.842\ \text{lb ft}[/tex]
[tex]0.1052\ \text{lb ft}[/tex]
Explanation:
d = Diameter of wheel = 6 in
r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]
t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]
w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]
[tex]t_2[/tex] = Time taken to slow down = 35 s
[tex]t_1[/tex] = Time taken to reach operating speed = 5 s
[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
Weight is given by
[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]
Mass is given by
[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]
Moment of inertia is given by
[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]
Angular acceleration while slowing down is given by
[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]
Frictional moment is given
[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]
Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]
Angular acceleration while speeding up is given by
[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]
Motor torque is given by
[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]
Motor torque is [tex]0.842\ \text{lb ft}[/tex].
A car is moving around a circular track of radius 25m with a speed of 30 m/s. What is the centripetal acceleration of the car? *
A 72 m/s2
B 36 m/s2
C 30 m/s2
D 6 m/s2
Answer:
I don't know how to do it the subject
The centripetal acceleration of the car when there is the radius and the speed is given so it should be considered as the option b. 36 [tex]m/s^2[/tex].
Calculation of the centripetal acceleration of the car:Since
The radius is 25 cm and the speed should be 30 m/s
Now the following formula should be used
Acceleration = speed^2/ radius
= 30^2 / 25
= 900 / 25
= 36 [tex]m/s^2[/tex]
Therefore, The centripetal acceleration of the car should be considered as the option b. 36 [tex]m/s^2[/tex].
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What is Newton's scientific view?
Answer:
Newton's first law of motion concerns any object that has no force applied to it.
Explanation:
three laws of motion and the law of universal gravitation.
Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.
Explanation:
As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.
Which particle needs to be added to this equation to show that the total numbers of neutrons and protons are not changed by the reaction? MARKLING BRAINLIEST 70 points must be correct!
Answer:
C.
Explanation:
Answer:A
Explanation:ap3x
would it be m/s or kg?
Answer:
m.s
Explanation:
Please answer this for 15 points please don’t put in a link.
Answer:
c. Double Replacement
Explanation:
As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.
Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2
and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.
A car weighing 1,500kg possesses 20 000 units of momentum. What would be the car's new momentum if it's velocity was tripled
Answer:
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Explanation:
light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and second (m=2) interference minima?
Answer:
3.35
Explanation:
Got it on Acellus
The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.
What is diffraction?Diffraction is the phenomenon of bending of waves through obstacles.
Given is the wavelength λ= 485 nm, silt width d = 8.32 *10⁻⁶ m, then the angle θ will be
d sinθ =mλ
for m=1, sin θ₁ = λ/d
for m=2, sin θ₂ = 2λ/d
Substitute the values into both expressions to find the angles,
sin θ₁ = 485 x 10⁻⁹ / 8.32 *10⁻⁶
θ₁ = 3.34°
and sin θ₂ = (2 x 485 x 10⁻⁹ )/ 8.32 *10⁻⁶
θ₂ = 6.7°
The angle between m =1 and m=2 will be
θ₂ -θ₁ = 6.7° - 3.34° =3.36°
Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.
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A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s
From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.
i.e.
[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]
Given that:
the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m∴
From the equation above, multiplying both sides with 2, we have:
[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]
Making (k) the subject of the formula;
[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]
[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]
k = 216.26 N/m
However, when compressed to one-half of the maximum distance; the speed is computed as follows:
x = 0.68/2 = 0.34 m
∴
[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]
[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]
[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]
25 - v² = 6.25
25 -6.25 = v²
v² = 18.75
[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]
v = 4.33 m/s
Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s
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The magnitude of the force can be determined as?
Answer:
the mass of the object multiplied by the acceleration of the object
Explanation:
N2L states that F = ma (force equals mass times acceleration).
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.70 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a) v = 1.1 m/s
b) A = 0.315 m
c) v = 1.1 m/s A= 0.15 m
Explanation:
a)
In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:[tex]v = \lambda * f (1)[/tex]
If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:⇒ T = 2.60 s * 2 = 5.20 s (2)Since we have now T, we can find the frequency f as follows:[tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]
Replacing f and λ in (1) we get:[tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]
b)
The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:[tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]
c)
Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:v = 1.10 m/s (6)
Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:[tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]
Which factor affects the color of the star?
Luminosity
Temperature
Apparent Magnitude
None of the aobve
Answer:
temperature
Explanation:
if you look at a hertzsprung-russel diagram. you can understand how I got that answer
5) Which statement about leaders is true?
A) Leaders always make the right decisions.
B) A leader keeps the team focused on achieving its goals.
C) A leader's opinion counts more than the opinions of the other team members.
D) all of the above.
Answer: D) All of the above
Explanation:
Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter. a) Determine the angular deflection of both wavelengths in the 1st and 2nd order.
Answer:
[tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength
[tex]\theta[/tex] = Angle
m = Order
Distance between grating is given by
[tex]d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}[/tex]
[tex]\lambda=656\ \text{nm}[/tex]
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}[/tex]
The first and second order angular deflection is [tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]\lambda=410\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}[/tex]
The first and second order angular deflection is [tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex].
A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration
Answer:
Explanation:
mass per unit length ρ = .100 / 1.65 = .0606 . kg /m
length of wire L = 1.65 m
For fundamental frequency , the expression is as follows
n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
L = 1.65 , T = 16 n and m = .0606
n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]
= 4.9 /s .
This is fundamental frequency .
other mode of vibration ( first three ) will be as follows
4.9 x 2 = 9.8 /s ,
4.9 x 3 = 14.7 /s .
the atom of an element x has 21protrons and 23neutrons. What is the
(a) Electron number
(b) Mass number
(c) Neutron number
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
A student using a stopwatch finds that the time for 10 complete orbits of a ball on the end of a string is 25 seconds. The period of the orbiting ball is
Answer:
T = 2.5 s
Explanation:
Given that,
Number of complete orbits = 10
Time, t = 25 seconds
We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.
So,
[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]
So, the period of the orbiting ball is equal to 2.5 seconds.
Which of these cubes absorb the most light?
Answera black cube or dark colors cause dark colors suck in heat
True or False
Microscopic organisms grow on the rocks near a volcano if the rocks are cooled to 120 degrees Celsius or less.
Explanation:
it's true
Kiss me if I'm wrong. But dinosaurs still exist right?
Answer:
True
Definetely true
Pls help ASAP
Imagine that Maritans launch a rocket toward the Earth at a great speed. While the
rocket is traveling toward us, it will appear
than it actually is.
O more blue
darker
larger
more red
Answer:
The rocket will appear larger than it actually is
What is wrong with the following momentum value: 25 kg*m/s
Answer:
25N/s or 25kg*m/s^2
Explanation:
It is written wrong because the unit of momentum is kgm/s^2 or N/s
If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron
Answer:
[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field
Explanation:
B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]
[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]
v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
Magnetic force is given by
[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]
The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.
Blocks A (mass 5.00 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 3.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line.
(a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
(b) Find the velocity of each block after they have moved apart.
Answer: i believe its B Find the velocity of each block after they have moved apart sorry
Explanation: have a nice day buddy
Make a poem about waves with 12 Lines and 3 Stanzas.
In a ocean full of storms
A new wave was born
Deep into that darkness flooding
Suddenly, I heard some pummeling
By the grave I saw the winds
And the sun just shined
Answer:
a friendly face that comes with waves,
the waves of all the memorial days,
and with these days we smile with pride,
as for the waves we used to ride,
given up the day has passed,
how it went away like an hour glass,
as if we knew the world was right,
just like the waves, oh so bright,
the time has come the days have passed,
the waves ashore the waves alast,
as if the friendly face was right,
the waves that rode, oh goodnight.
Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.
Answer:
The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒ [tex]W=e \Delta v[/tex]
and,
⇒ [tex]\frac{1}{2}mv^2=e \Delta v[/tex]
then,
⇒ [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]
⇒ [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]
⇒ [tex]=1.185\times 10^7 \ m/sec[/tex]
If an athlete runs the triathlon of 10 km in 2 hours, what is her average speed in kilometers per hour?
Answer: 5 km per hour
Explanation:
if in 10 km there is 2 hours, then 10 divided by 2 is 5.
which particle have a mass of 1 u
Answer:
Explanation:
proton
7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.
The tension in the strings are 31.47 and 19.25 N respectively.
Mass of the block, m = 3 kg
From the figure, consider the vertical components,
T₁ sin45° + T₂ sin30° = mg
(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4
Also, consider the horizontal components,
T₁ cos45° = T₂ cos30°
T₁/√2 = T₂ x√3/2
T₁ = T₂ x √3/2 x √2
So,
T₁ = 0.612T₂
Applying in the first equation,
(T₁/√2) + (T₂/2) = 29.4
(0.612T₂/1.414) + 0.5T₂ = 29.4
0.434 T₂ + 0.5 T₂ = 29.4
0.934 T₂ = 29.4
Therefore, the tension,
T₂ = 29.4/0.934
T₂ = 31.47 N
So, the tension,
T₁ = 0.612 T₂
T₁ = 0.612 x 31.47
T₁ = 19.25 N
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One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
Answer:
x = 0.455 L
Explanation:
For this exercise we must use the rotational equilibrium condition
Σ τ = 0
it has two forces, the first is perpendicular to the rod, so its stub is
τ₁ = F₁ L
the second force is applied with an angle, so we can use trigonometry to find its components
sin θ = F_parallel / F₂
cos θ = F_perpendicular / F₂
F_parallel = F₂ sin θ
F _perpendicular = F₂ cos θ
torque is
τ₂ = F_perpendicular x + F_parallel 0
the parallel force is on the rod therefore its distance is zero
we apply the equilibrium equation
τ₁ - τ₂ = 0
F₁ L = F₂ cos θ x
x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]
let's calculate
x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]
x = 0.455 L