All categories

2 years ago

A 1.5 kg baseball is pitched at 38 m/s and is hit by the batter. It heads directly back at the pitcher at 27 m/s.

Physics

1 answer:

lesantik [10]2 years ago

5 0

Answer:

The initial velocity of the ball is +38.0 m/s. The final velocity is -27.0 m/s.

momentum = mass x velocity

(mass x final velocity) – (mass x initial velocity)

(1.5)(-27) – (1.5)(38) = -97.5

The impulse applied to the ball is -97.5 kg m/s

impulse = force x time interval

force = impulse/time interval

-97.5/0.45 s = - 216.6

The force applied was 216.6 N.

You might be interested in

How much time does it take for a spinning baseball with an angular speed of 38 rad/s to rotate through 15 degrees

Nadusha1986 [10]

Answer:

the time of motion of the ball is 6.89 ms.

Explanation:

Given;

angular speed, ω = 38 rad/s

angular distance, θ = 15 degrees

Angular distance in radian;

$\theta = 15^0 \times\frac{2\pi \ rad}{360^0} = 0.2618 \ rad$

Time of motion is calculated as;

$time = \frac{angular \ distance}{angular \ speed} \\\\t= \frac{\theta}{\omega} = \frac{0.2618 \ rad}{38 \ rad/s} \\\\t = 6.89 \ \times 10^{-3} \ s\\\\t = 6.89 \ ms$

Therefore, the time of motion of the ball is 6.89 ms.

3 0

3 years ago

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its

Savatey [412]

Answer:

a). $E_{kA}=1.2635 J$

b). $V_{B}=4.535\frac{m}{s}$

c). Δ $E_{t}=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J$

b).

$E_{kB}=\frac{1}{2}*m*v_{B} ^{2}$

$V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}$

c).

net work= EkA+EkB

$E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J$

3 0

3 years ago

What is the difference between a chemical change and a physical change? *

slavikrds [6]

Answer:

I do belive that it is B hrs cn I an gn

7 0

3 years ago

Ron has not used heroin for several days. He has a

katen-ka-za [31]

I wanna say withdrawal? Ron needs to go to rehab

6 0

3 years ago

What keeps the particles of a liquid from spreading out to fill an entire container in the same way that gas particles do?

Sonbull [250]

b. the forces of attraction among them limit their motion.

7 0

4 years ago

Read 2 more answers