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denis23 [38]
2 years ago
12

A hockey puck is pushed by a stick with a force of 750 newtons. The puck travels 2.0 meters in 0.30 seconds. How powerful is the

push? (Ignore frictional effects.)
Physics
1 answer:
nekit [7.7K]2 years ago
5 0
The work done by the force pushing the puck is equal to the product between the force and the distance covered:
W=Fd=(750 N)(2.0 m)=1500 J

And the power is the work done per unit time:
P= \frac{W}{t}
and using t=0.30 s, we find the power:
P= \frac{1500 J}{0.30 s}=5000 W
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A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the
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Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written asx(t)=Asin(\omega t+\phi)

'A' is the amplitude = 6 inches (given)

\omega =\sqrt{\frac{k}{m}} is the natural frequency of the system

At equilibrium we have

mg=kx\\\\k=\frac{mg}{x}

Applying values we get

k=40 lb/ft

thus natural frequency equals

\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}

Thus the equation of motion becomes

x(t)=6sin(8t+\phi)

At time t=0 since mass is at it's maximum position thus we have

A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})

Thus the position of mass at the given times is as follows

1) at \frac{\pi}{12} x(t)=5.99inches

2) at \frac{\pi}{8} x(t)=5.9909inches

3) at \frac{\pi}{6} x(t)=5.98397inches

4) at \frac{\pi}{4} x(t)=5.9639inches

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4 0
1 year ago
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g
mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

n_{r}=1.572

n_{b}=1.587

We need to calculate the angle for red wavelength

Using Snell's law,

n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}

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1.572\sin37=1\times\sin\theta_{r}

\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})

\theta_{r}=71.0^{\circ}

We need to calculate the angle for blue wavelength

Using Snell's law,

n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}

Put the value into the formula

1.587\sin37=1\times\sin\theta_{b}

\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})

\theta_{b}=72.7^{\circ}

We need to calculate the angle between the red and blue light

Using formula of angle

\Delta \theta=\theta_{b}-\theta_{r}

Put the value into the formula

\Delta \theta=72.7-71.0

\Delta \theta=1.7^{\circ}

Hence, The angle between the red and blue light is 1.7°.

8 0
2 years ago
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