A force of 120.0 N is applied to a 3.00 kg block. What is the average acceleration of the block?

What is the angular displacement of the wheel between t = 5 s and t = 15 s?

Mkey [24]

The question seems incomplete. The complete text is:

a)What is the angular displacement of the wheel between t = 5 s and t = 15 s

b)What is the angular velocity of the wheel at 15 s

And it refers to the attached figure.

a) 25 rad

The graph shown represent the angular position of the wheel at different times.

Therefore, we can simply calculate the angular displacement between two times by calculating the difference between the angular position at t2 and the angular position at t1.

At $t_1 = 5 s$ , the angular position from the graph is $\theta_1 = 100 rad$

At $t_2 = 15 s$ , the angular position from the graph is $\theta_2 = 125 rad$

Therefore, the angular displacement is

$\Delta \theta= \theta_2 - \theta_1 = 125-100 = 25 rad$

2) -5.0 rad/s

For a angular displacement vs time graph, the angular velocity at any time is simply equal to the slope of the curve at that time.

Here we want to calculate the angular velocity at t = 15 s, so we have to calculate the slope at that time.

By noting that the slope is constant in the last part of the motion, we find that the slope between 10 s and 20 s is:

$\frac{\Delta \theta}{\Delta t}=\frac{100 rad - 150 rad}{20 s - 10 s}=-5.0 rad/s$

This slope is constant between 10 s and 20 s, so the angular velocity of the wheel at t = 15 s

$\omega = -5.0 rad/s$

5 0

3 years ago

What rock type would most likely be found in the location marked by the black rectangle in the diagram below?

pychu [463]

There is no diagram below so I can't answer the question

5 0

3 years ago

Read 2 more answers

A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42

elena-s [515]

You said "<span>A rocket's acceleration is 6.0 m/s2.".

That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.

If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.

How long does it take to reach 42 m/s ?

Well, how many times does it have to go 6 m/s FASTER
in order to build up to 42 m/s ?

That's just (42/6) = 7 times.

Writing it correctly, with the units and everything, it looks like this:

(42 m/s) / (6 m/s</span>²)

= (42/6) (m/s) / (m/s²)

= (42/6) (m/s · s²/m)

= 7 seconds

7 0

4 years ago

How did you manage to overcome the difficulties that arose?

Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

4 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is