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2 years ago

14

A volleyball is served at a speed of 8 / at an angle 35° above the horizontal. What is the speed of the ball when received by th

e opponent at the same height?

1 answer:

bezimeni [28]2 years ago

7 0

The speed of the ball is how hard u hit it

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An engineer is investigating potential energy with two

miv72 [106K]

Answer:

Since they move in the same direction, share the same magnetic qualities, and move one unit identically
We can safely say its D

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2 years ago

Read 2 more answers

26. A baseball traveling 38 m/s is caught by the catcher. The catcher takes 0.1 seconds to stop the ball. What is the accelerati

Furkat [3]

Answer:

The ball has an acceleration of -380 m/s², this means the ball slows down

An acceleration of -380 m/s² is the equivalent of 38.736 g's

Explanation:

Step 1: Data given

Velocity of the baseball at time t=0 = 38 m/s

At time t, the ball stops. This means v = 0

time before stops = 0.1s

Step 2: Calculate the acceleration

v= v0+at

with v= the velocity of the ball at time t = 0. v= 0

with v0 = the velocity of the ball at time t=0. v0 = 38 m/s

with a= the acceleration in m/s²

with t = time in seconds

0 = 38 + a*0.1

a = -380 m/s²

The ball has an acceleration of -380 m/s², this means the ball slows down

An acceleration of -380 m/s² is the equivalent of 38.736 g's

5 0

2 years ago

What is the part of the earth that you walk on called?

olasank [31]

The part of the Earth that you walk on is called the CRUST.

6 0

2 years ago

A cheetah, which is the fastest land mammal, can accelerate at 9 m/s from rest to in 31 m/s. How long does

Anna007 [38]

31 m/s ÷ 9 m/s² = 3.44 s

Time = Change in velocity divided (÷) by acceleration.

8 0

2 years ago

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance <em>s</em><em> </em><em> </em>between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0

2 years ago