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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:

yarga [219]3 years ago

8 0

We use the relation:

Power = force x velocity
Power = 100 x 45
Power = 4,500 Watts

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3 years ago

What is the molecular geometry or of the molecular molecule

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Molecular Geometry is the three-dimensional structure or arrangement of atoms in a molecule.

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3 years ago

What exerts a centripetal force on a person running around a curve?

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Answer:

The inertial force of the body

Explanation:

Everybody that is moving in a curved path has an inertial force called centrifugal force.

The counterforce of the centrifugal force is called the centripetal force. It also acts on every rotating body.

This force is always directed towards the center of the origin of the curve.

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3 years ago

I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?

Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

$\frac{1}{2} m {v}^{2} = mgh$

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

$\sin(15) = \frac{h}{d} \\ or \: h = d \sin(15)$

Plugging this into the energy conservation equation and cancelling m, we get

${v}^{2} = 2gd \sin(15)$

Solving for d,

$d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m$

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3 years ago

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

4 years ago