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2 years ago

1. You push a deak across the room. A force was applied. The object moves as a result of a force.

Physics

1 answer:

FinnZ [79.3K]2 years ago

6 0

Answer:

what is the question?

Explanation:

I'm not sure if this was a question because you gave the answer lol

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The energy an object acquires when it is exposed to a force is called _____ energy

r-ruslan [8.4K]

I'm pretty sure the energy an object acquires when exposed to a force is known was potential energy.

4 0

2 years ago

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

3 years ago

Find the voltage across the 15 Ω resistor. .<br> 35 Ω<br> 20 Ω<br> 15 Ω<br> 10 V

almond37 [142]

Answer:

35 Ω

Explanation:

3 0

2 years ago

A rocket is launched from the launch pad at a speed of 700 km/hr.

Juliette [100K]

Answer:

Explanation:

Vm = Δs/Δt

700km/h = Δs/1.5h

700 = Δs/1.5

$\frac{700}{1} = \frac{S}{1.5}$

S = 700 x 1.5

S = 1050 Km

*(S = Δs)

Answer: The rocket will have traveled 1050 Km

Hope this help ☺

5 0

3 years ago

1. You push a deak across the room. A force was applied. The object moves as a result of a force.​

1. You push a deak across the room. A force was applied. The object moves as a result of a force.