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3 years ago

A pitcher exerts 100.0 N of force on a ball with a velocity of 45 m/s. What is the pitcher's power?

Physics

1 answer:

Andreas93 [3]3 years ago

7 0

The pitcher's power is given by the formula P= fxv
P=100x 45m/s= 4500 Watts

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A satellite orbiting the moon very near the surface has a period of110 min. What is free-fall acceleration on the surface of the

Norma-Jean [14]

Answer:

1.54 m/s²

Explanation:

The free-fall acceleration is calculated as

g = w²r

Where w is the angular velocity of the satellite and r is the radius of the moon.

The angular velocity can be calculated as

$w=\frac{2\pi}{T}$

Where T is the period, so

T = 110 min = 110 x 60 s = 6600 s

Then,

$w=\frac{2\pi}{6600\text{ s}}=9.52\times10^{-4}\text{ rad/s}$

Finally, the radius of the moon is r = 1.7 x 10⁶ m, so the free-fall acceleration is

$\begin{gathered} g=w^2r \\ g=(9.52\times10^{-4})^2(1.7\times10^6) \\ g=1.54\text{ m/s}^2 \end{gathered}$

Therefore, the answer is 1.54 m/s²

5 0

2 years ago

A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is

Andrew [12]

Answer:

The correct one is that the force on B is half of the force on A

Explanation:

Because radius for the inside of the curve is half the radius for the outside and Car A travels on the inside while car B, travels at equal speed on the outside of the curve. Thus force on B will be half on A

3 0

3 years ago

A person of mass 80 kg stands at the center of a rotating merry-go-round platform of radius 3.5 m and moment of inertia 950 kg*m

Leya [2.2K]

Answer with Explanation:

We are given that

Mass of person,M=80 kg

Radius,r=3.5 m

Moment of inertia of merry,I= $950 kgm^2$

Angular velocity of platform= $\omega=0.85rad/s$

1.Law of conservation of angular momentum

$I\omega=I'\omega'$

Where $I'=950+80(3.5)^2$

Substitute the values

$950\times 0.85=(950+80(3.5)^2)\omega'$

$\omega'=\frac{950\times 0.85}{950+80(3.5)^2}=0.418rad/s$

Angular velocity when the person reaches the edge =0.418 rad/s

2.Rotational kinetic energy of the system before the person's walk

$K_i=\frac{1}{2}I\omega^2=\frac{1}{2}(950)(0.85)^2=343.2 J$

3.Rotational kinetic energy of the system after the person's walk

$K_f=\frac{1}{2}(950+80(3.5)^2)\times (0.418)^2$

$K_f=168.6 J$

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3 years ago

Which of the following is the most likely end for a star that is small to average in size? A. blue main sequence B. red supergia

likoan [24]

"White dwarf" is the one among the following choices given in the question that is <span>the most likely end for a star that is small to average in size. The correct option among all the options that are given in the question is the last option or option "D". I hope that this is the answer that has actually helped you.</span>

7 0

3 years ago

Read 2 more answers

A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.140 m2. The coil is now stretched to hav

DENIUS [597]

Answer:

Magnitude of induced emf will be 87.5 volt

Direction of induced emf will be clockwise

Explanation:

We have given number of turns in the coil N = 50

Initial area $A=0.140m^2$