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laila [671]
3 years ago
11

The combination of an applied force and a friction force produces a constant torque of 36.0 N⋅m on a wheel rotating on a fixed a

xis. While the force acts for 6.00 s, the angular velocity of the wheel increases from 0 to 10.0 rad/s. The force is removed and the wheel comes to rest in 60.0 s
a. Find the moment of inertia of the wheel.
b. Find the magnitude of the torque due to friction.
c. Find the total number of rotations during the 66.0 s.
d. Find the Kinetic energy of the wheel at 6.00 s when the force is removed.
Physics
1 answer:
butalik [34]3 years ago
7 0

Answer:

D. Find the kinetic energy of the wheel at 6.00 z when the force is removed.

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a person hangs three pictures on the wall. the pictures all weigh the same. picture 1 and picture 2 are at the same height above
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Picture 1 and 2 or am I wrong and it's not that obvious.
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What happens to the temperature of a substance when it is changing state of matter?
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Solid substances have molecules held tightly and close together
Liquid substances have molecules moving loosely
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As moleclues get further apart, i.e. As a substance changes state from solid to liquid to gas, molecules gain kinetic energy and vibrate/move more. This means they gain heat energy (the averge energy a substance has) so the temperature increases

Substances exist in different states at different temperatures and different substances will exist in different states at the same temperature. This is to do with the forces between molecules and how much heat (energy) is required to break them
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3 years ago
Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de
-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

s(t)^2={h^2+x^2}...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

2s(t)s'(t)=2x(t)x'(t)

x'(t)=\dfrac{s(t)s'(t)}{x(t)}

When x = 40, s(t)=\sqrt{40^2+4^2}=40.19\ m

x'(t)=\dfrac{-240s(t)}{x(t)}

x'(t)=\dfrac{-240\times 40.19}{40}

x'(t)=-241.14\ m/s

So, the speed of the airplane is 241.14  m/s. Hence, this is the required solution.

8 0
4 years ago
5. A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has (2 points)
suter [353]
What’s the rest of the question or is that it?
4 0
3 years ago
An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

6 0
3 years ago
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