If your velocity is negative your position is going down and your displacement is negative.

A ball undergoes uniform acceleration of 2.50 m/s2 (beginning at rest). How far does the ball travel during that 4s time interva

-Dominant- [34]

Answer:

The ball travels a distance of 20 m in the time interval of 4 s

Explanation:

Using s = ut + 1/2at² where s = distance travelled by the ball, u = initial velocity of ball = 0 m/s (since it starts from rest), a = acceleration of the ball = 2.50 m/s² and t = time = 4 s.

Substituting the variables into the equation, we have

s = ut + 1/2at²

s = 0 × 4 s + 1/2 × 2.50 m/s² × (4 s)²

s = 0 + 1/2 × 2.50 m/s² × 16 s²

s = 1/2 × 40 m

s = 20 m

So, the ball travels a distance of 20 m in the time interval of 4 s.

4 0

4 years ago

An archer pulls back the string of a bow to release an arrow at a target. Which kind of potential energy is transformed to cause

Morgarella [4.7K]

It is elastic energy

8 0

4 years ago

Read 2 more answers

Which of the following Illustrates 2 resistors in a series circult? A, B, C, D.

Kaylis [27]

Answer:

It's either B or D, I'm not positive which it is

Explanation:

6 0

2 years ago

A team of students develop a model to describe why structural changes to genes affect proteins. The team provides two groups of

PSYCHO15rus [73]

Answer:

The team's model is incorrect because they provide the same materials to build the same product.

Explanation:

5 0

4 years ago

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$