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3 years ago

If a pitcher throws a baseball 10 m in 30 seconds - what would be the average speed?

Physics

2 answers:

allsm [11]3 years ago

7 0

Answer:

0.333 m/s

Explanation:

avg speed= (total distance)/(total time)

10m/30s

0.333 m/s

dedylja [7]3 years ago

4 0

0.333 m/s is not the answer

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A ball rolls from point A to point B. The total energy of the ball at point A isn’t the same as the sum of its potential energy

murzikaleks [220]

B. Some of the ball’s energy is transformed to thermal energy.

Hope this helps you!

3 0

3 years ago

Read 2 more answers

A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational p

Afina-wow [57]

Answer:

5760 J

Explanation:

From the question given above, the following data were obtained:

Mass of block = 48 kg

Height (h) = 12 m

Gravitational field strength (g) = 10 N/Kg

Gravitational potential energy (PE) =?

The gravitational potential energy stored by the block can simply be obtained as follow:

PE = mgh

PE = 48 × 10 × 12

PE = 5760 J

Therefore, the gravitational potential energy stored by the block is 5760 J

3 0

3 years ago

LARGER vibration produces a higher ............

balandron [24]

A or possibly C because the other options have nothing to do with the size of the vibration. If i was you I would answer with A

8 0

3 years ago

Explain how the thermal energy of a swimming pool compares to the thermal energy of a lake at the same temperature.

astra-53 [7]

There is more thermal energy in the lake because there is more water which is more thermal energy

3 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

3 years ago