explain a ball is pitched east at a speed of 40 km/h. the batter hits it west at a speed of 40 km/h.did the ball accelerate

1 answer:

5 0

Yep, uh-huh, indeed, indubitably, and you better believe it.
The ball did accelerate.

Acceleration means any one of these:
-- speeding up
-- slowing down
-- changing direction .

The direction of the ball changed from east to west.
That's acceleration right there.
_____________________________________

Acceleration means any change of velocity.
Velocity is speed and direction.

The velocity of the ball changed from (40 km/hr east) to (40 km/hr west).

The change in velocity was 80 km/hr to the west.

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a 10.0 kg ball falling at 10.0 m/s hits a mattress and comes to a complete stop in 1 s. what force does the mattress exert to st

eimsori [14]

Using Newton's second law;

F = ma, where m = mass, a = acceleration or deceleration

a = Δv/t = (v-u)/t, but v= 0, u = 10 m/s, t = 1.
Then,
a = (0-10)1 = -10 m/s^2

Substituting;
F = ma = 10*-10 = -100 N

The mattress exerts 100 N to stop the ball.

3 0

4 years ago

standing side by side, you and a friend step off a bridge and fall for 1.6s to the water below. your friend goes first, and you

Andrew [12]

(a) The distance will be more than 2.0 meters.

In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.

b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(1.6s)^2=12.56 m$

We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:

$S=\frac{1}{2}gt^2$

Using S=2.0 m,

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(2.0 m)}{9.81 m/s^2}}=0.64 s$

So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using

$t=1.6 s-0.64 s=0.96 s$

we find

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(0.96 s)^2 =4.52 m$

So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.

8 0

4 years ago

A wave has a frequency of 15,500 Hz and a wavelength of 0.20 m. What is the

Hoochie [10]

Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$