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Ksivusya [100]
3 years ago
6

A car’s 30.0-kg front tire is suspended by a spring with spring constant k=1.00x10^5 N/m. At what speed is the car moving if was

hboard bumps on the road every 0.750 m drive the tire into a resonant oscillation?
Physics
1 answer:
Taya2010 [7]3 years ago
6 0

we know the equation for the period of oscillation in SHM is as follows:

T = 2 * pi * sqrt(mass/k)

we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).

since d = v*T, we can say v = d/t = d * f

the final equation, after combining everything, is as follows:

v = d/(2 * pi) * sqrt(k/m)

by plugging everything in

v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))

We find our velocity to be:

v = 6.89 m/s

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A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt
gizmo_the_mogwai [7]

Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

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Inertia is the resistance of an object to a change in its....
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9. If everything in the universe – including atoms and particles stop moving, does time stop? Or does time continue even if ever
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Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference
strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

n_{air}=1

\theta_{liquid} = 19.38\°

\theta_{air}35.09\°

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

n_1sin\theta_1 = n_2sin\theta_2

n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}

n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}

Replacing the values we have:

n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}

n_liquid = 1.7323

Therefore the refractive index for the liquid is 1.7323

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