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3 years ago

4. Charge may be measured in the

Physics

1 answer:

Maslowich3 years ago

4 0

Answer:

A 0.25 kg baseball is in contact with a bat for 0.0075 s and exerts a force of 18,500 N. What impulse is delivered to the ball?

33 kg •

140 kg •

4600 kg •

74,000 kg •

Explanation:

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The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t

⇒ α = (0 rev/min - 113 rev/min) / (60 min)

⇒ α = - 1.883 rev/min²

6 0

3 years ago

What happens to the energy of a wave as it moves away from its source?

Eddi Din [679]

Answer:

As the particles move further away from their normal position (up towards the wave crest or down towards the trough), they slow down.

Explanation:

This means that some of their kinetic energy has been converted into potential energy – the energy of particles in a wave oscillates between kinetic and potential energy. Hope that this helps you and have a great day :)

3 0

3 years ago

A simple pendulum is made by tying a 2.44 kg stone to a string 4.57 m long. The stone is projected perpendicularly to the string

alexdok [17]

Answer:

$v_{max}=8.2226m/s$

Explanation:

The problem is solved using the law of conservation of energy,

$mgL(1-cos\theta)+\frac{1}{2}mv^2_0=\frac{1}{2}mv^2_{max}$

$v_{max}=\sqrt{2gL(1-cos\theta)+v^2_0}$

$v_{max}=\sqrt{2(9.8)(4.57)(1-cos(69.4))+8^2}$

$v_{max}=8.2226m/s$

8 0

3 years ago

A 72-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.61 m,

chubhunter [2.5K]

Answer:

2849.98 J

Explanation:

From the question,

Work done by the boy = change Potential energy of the boy + change in kinetic energy of the boy

W = ΔP + ΔK..................... Equation 1

Where W = work done by the boy, ΔP = change in potential energy of the boy, ΔK = Change in kinetic energy of the boy.

But,

ΔP = mgΔh.................... Equation 2

ΔK = 1/2mΔv²................. Equation 3

Where m = mass of the boy, Δh = change in height of the boy, Δv = change in velocity of the boy.

Substitute equation 2 and 3 into equation 1

W = mgΔh+1/2mΔv²................. Equation 4

Given: m = 72 kg, Δh = 1.61 m, Δv = 8.5-1.6 = 6.9 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 72(9.8)(1.61)+1/2(72)(6.9²)

W = 1136.016+1713.96

W = 2849.98 J

8 0

3 years ago

A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What

iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

$\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s$

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

$a=\omega^2 r$

At the instant t = 0,

$\omega=3.93 rad/s$

So the centripetal acceleration of the tip of the blades is

$a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2$

5 0

3 years ago