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lorasvet [3.4K]
3 years ago
5

Beth moves a 15 N book 20 meters in 10 seconds. How much power was produced?

Physics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

30 Watts

Explanation:

 Power = Work/Time

Work = Force*Distance

Power = Force * Distance / Time

Power = 15 N * 20 meters / 10 sec

Power = 30 Watts

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Newton's first law of motion is connected with inertia :)
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Doppler Effect: A stationary source produces a sound wave at a frequency of 50 Hz. The wave travels at 100 feet per second. A ca
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Explanation:

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Why is there more potential for wind energy in the upper himalayan in nepal?​
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A car is moving South at a constant speed of 60 miles per hour. The force of the
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As we are told the speed is constant, then this force must be increasing the car's potential energy by climbing a hill.

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If we knew the car mass, we could find the hill slope angle.

If we knew the hill slope angle, we could find the car mass.

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The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di
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Answer and Explanation:

Let:

x(t)=Acos(\omega t+ \phi)

The equation representing a simple harmonic motion, where:

x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

\phi = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

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3 years ago
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