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3 years ago

5 Complete the following sentences.

Physics

1 answer:

finlep [7]3 years ago

8 0

Answer: During a conversation, you are talkative or don’t make a sound.

Explanation:

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Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . Assu

Step2247 [10]

To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

$W=\tau \Delta\theta$

In the case of rotational kinetic energy we know that

$KE = \frac{1}{2}I\omega^2$

PART A) $\theta$ is given in revolutions and needs to be in radians therefore

$\theta = 12rev(\frac{2\pi rad}{1rev})$

$\theta = 24\pi rad$

Replacing in the work equation we have to

$W=\tau \Delta\theta$

$W= (25)(24\pi)$

$W = 1884.95J$

PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

$\tau = I \alpha$

Rearrange for the angular acceleration,

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{25}{50}$

$\alpha = 0.5rad/s$

From the kinematic equations of angular motion we have,

$\omega_f^2=\omega_i^2+2\alpha\theta$

$\omega_f^2=0+2*0.5*24\pi$

$\omega_f=\sqrt{0+2*0.5*24\pi}$

$\omega_f = 8.68rad/s$

In this way the rotational kinetic energy would be given by

$KE = \frac{1}{2}I\omega_f^2$

$KE = \frac{1}{2}(50)(8.68)^2$

$KE = 1883.56J$

3 0

3 years ago

Read 2 more answers

Help

arlik [135]

I think the answer is C

7 0

3 years ago

A variable that is not changed.

AleksAgata [21]

Answer:

A controlled variable does not change during a experiment

Explanation:

it's c

5 0

3 years ago

If you speed up from the rest to 12m/s in 3 seconds, what is your acceleration?

azamat

I believe that your acceleration would be 4 m/s

7 0

4 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 32.3

olga_2 [115]

Answer:

$0.00986m/s^2$

Explanation:

I think there are some repetitions in the question which may be due to typographical errors. The correct question should have been as stated below:

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 32.3 m/s . It then flies a further distance of 41300 m, and afterwards, its velocity is 43.1 m/s . Find the airplane's acceleration.

This problem could be solved using the third equation of a uniformly accelerated motion, since the airplane is said to accelerate at a constant (uniform) rate. The equation is given by;

$v^2=u^2+2as$

where v is the final velocity, u the initial velocity, a acceleration and s distance.

$u=32.3m/s\\ v=43.1m/s\\s=41300m\\a=?$

We substitute these values into the equation and then solve for the unknown;

$43.1^2=32.3^2+2(a)(41300)\\43.1^2-32.3^2= 82600a\\1857.61-1043.29=82600a\\814.32=82600a\\Hence\\a=814.32/82600=0.00986m/s^2$

8 0

4 years ago