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3 years ago

A ball at the end of a string of 2.2m length rotates at

Physics

2 answers:

Sergio039 [100]3 years ago

4 0

Answer:

The frequency of motion is 5.8 Hz.

Explanation:

frequency of motion of any object is defined as the number of times the object repeats it's motion in 1 second.

mathematically frequency equals $f=\frac{1}{T}$

where,

'T' is the time it takes for the object to complete one revolution. Since it is given that the ball completes 5.8 revolutions in 1 seconds thus the time it takes for 1 revolution equals $\frac{1}{5.8}s$

Hence $T=\frac{1}{5.8}s$

thus the frequency equals

$\frac{1}{\frac{1}{5.8}}s^{-1}\\\\f=5.8s^{-1}=5.8Hz$

Sunny_sXe [5.5K]3 years ago

3 0

Answer:

5.8 Hz.

Explanation:

<u>Given:</u>

The length of the string to which the ball is attached, L = 2.2 m.

The number of revolutions, the ball is making in unit time = 5.8 rev/s.

The frequency of the motion of a rotating object is defined the total number of revolutions the object makes in unit time. It is measured in units of Hertz (Hz) or per second.

It is given that the ball is making 5.8 revolutions in one second, therefore, its frequency of motion would be the same, i.e., 5.8 Hz.

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geniusboy [140]

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

$s = ut + \frac{1}{2}a {t}^{2} \\ 0.8 = 1.7 \times t \\ \frac{0.8}{1.7} = t \\ t = 0.47s$

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

$s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} \times 10 \times (0.47) ^{2} \\ s = 1.1045m$

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2 years ago

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

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2 years ago

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lidiya [134]

Answer:

The answer would be 0.04ohms.

Explanation:

Hopefully this helps

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pishuonlain [190]

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