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2 years ago

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A horizontal jet of water is made to hit a vertical wall with a negligible rebound. If the speed of water from the jet is 'v', t

he diameter of the jet is 'd' and the density of water is 'p' , then what is the force exerted on the wall by the jet of water?

1 answer:

klasskru [66]2 years ago

3 0

Answer:

F = π/4 ρ d² v²

Explanation:

Force is mass times acceleration:

F = ma

Acceleration is change in velocity over change in time:

F = m Δv / Δt

Since there's no rebound, Δv = v.

F = m v / Δt

Mass is density times volume:

F = ρ V v / Δt

Volume per time is flow rate:

F = ρ Q v

Flow rate is velocity times cross sectional area:

F = ρ (v A) v

F = ρ A v²

Area of a circle is pi times radius squared, or pi/4 times diameter squared:

F = ρ (π/4 d²) v²

F = π/4 ρ d² v²

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Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

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2 years ago

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit

Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle $\theta_{a}$ , since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are $n_{red} = 2.410$ for red light and $n_{blue} = 2.450$ for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has $n_{air} = 1.000$ .

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

$\theta_{crit} = 24.09^{0}$

Explanation:

Only the blue refracted ray is related to the critical angle in this question

$n_{air} = 1.000$

$n_{blue} = 2.450$

The relationship between the critical angle( $\theta_{crit}$ ), $n_{air}$ and $n_{blue}$ can be given as $sin \theta_{crit} = \frac{n_{air} }{n_{blue} }$

$sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\ \theta_{crit} = 24.09^{0}$

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2 years ago

Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose

dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by $h=7\ m$

Suppose the dolphin jump with an initial velocity of $u$

so u is given by $u^2=2\cdot g\cdot h$

If dolphin launches at an angle $\theta$ then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

$R=\frac{u^2\sin 2\theta }{g}$

substitute the value of $u^2$

$R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}$

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Range will be maximum for $\theta =45^{\circ}$

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2 years ago

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OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

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The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

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$a_t = ar$

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Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

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Svetradugi [14.3K]

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