All categories

3 years ago

A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?

Physics

1 answer:

goldfiish [28.3K]3 years ago

5 0

<u>Answer:</u>

Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 $m/s^2$ , we need to calculate displacement when time = 7.5 seconds.

Substituting

$s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter$

So ball will move 92.8125 meter along the cliff in 7.5 seconds.

You might be interested in

Not a assignment question, just wanna know if scientists know what’s beyond the edge universe

Bond [772]

No they don’t. All the pictures we see if the universe are computer generated. Well, most of them are. We don’t exactly know how the milky way looks. It’s just very good guess.

8 0

2 years ago

Read 2 more answers

A cheetah can go from a state of resting to running at 20 m/s in just two seconds what is the cheetah's average acceleration?

Sunny_sXe [5.5K]

It's C i believe. To solve it we just take 20 and divide it by 2. Which gives us the average of 10 m/s

-Steel jelly

8 0

3 years ago

Read 2 more answers

A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

Anna35 [415]

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

$h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2$

The direction of the acceleration of the ball is downwards

Learn more here: brainly.com/question/15407740

4 0

2 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

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

Andrei [34K]

Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

7 0

3 years ago