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Liula [17]
3 years ago
5

A remote-controlled car’s wheel accelerates at 22.2 rad/s2 . If the wheel begins with an angular speed of 11.0 rad/s, what is th

e wheel’s angular speed after exactly twelve full turns? Answer in units of rad/s.
Physics
1 answer:
wariber [46]3 years ago
4 0

The angular speed remained constant after twelve turns at 11 rad/s.

Explanation:

As per the Newtons equation of motion, the final velocity can be determined by using the below equation, provided ,the displacement and acceleration with initial velocity is given.

v^{2}= u^{2}+2as

Since, the present case follows circular motion, the displacement will be equal to angular displacement. And the velocity is equal to angular velocity.

As the car revolves complete twelve turns, then the displacement will be zero, since the final position is equal to the initial position.

So,

v^{2}= (11)^{2}+2*22.2*0\\\\v = 11 rad/s

Thus, the angular speed remained constant after twelve turns at 11 rad/s.

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An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

v=\omega r

where

\omega is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s

r = 0.600 m

So the speed of the ball is

v=(51.0 rad/s)(0.600 m)=30.6 m/s

In situation 2), we have

\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s

r = 0.900 m

So the speed of the ball is

v=(37.9 rad/s)(0.900 m)=34.1 m/s

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) 1561 m/s^2

The centripetal acceleration of the ball is given by

a=\frac{v^2}{r}

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2

(c) 1292 m/s^2

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2

5 0
3 years ago
Help! Please???
Andre45 [30]
Well for the The weather on the coast of Virginia is probably warm and dry. A cold front would likely bring <span>rain or rainstorms. would be B Rainstorms</span>
3 0
3 years ago
Why is it impossible to create a perpetual motion machine?
Tatiana [17]

Answer:

It is impossible to create a perpetual motion machine because some of the energy will always be lost in the conversion  and therefore it will eventually stop.

Correct Answer : Option B

Explanation:

The perpetual motion machine is impossible machine as it is hypothetical working machine which would be in motion for an indefinite time in continuity of the motion. The continuity of motion for an indefinite time would mean that the working principle of the machine would never allow the dissipation of energy from the machine and all the machine would ultimately reserve all the energy and be converting it into forms without any loss.

This working principle is violation of first and second laws of thermodynamics and hence a machine will eventually lose some of its energy in every conversion of the working cycle and hence there will be a time where the machine would be stopped, and hence a perpetual motion machine cannot be made.

8 0
3 years ago
A car is moving South at a constant speed of 60 miles per hour. The force of the
Anastaziya [24]

Answer: 750 N

Explanation:

The net force is 1200 - 450 = 750 N

As we are told the speed is constant, then this force must be increasing the car's potential energy by climbing a hill.

F = mgsinθ

If we knew the car mass, we could find the hill slope angle.

If we knew the hill slope angle, we could find the car mass.

6 0
3 years ago
A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum
qwelly [4]

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

π

2α

≤x≤+

π

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(c) Find the particle’s average position.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

6 0
3 years ago
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