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tensa zangetsu [6.8K]

2 years ago

10

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

ress your answer in terms of any or all of m, F, and t.

1 answer:

olganol [36]2 years ago

6 0

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, $a=\frac{F}{m}$

Final velocity , $v=a+ut$

Substitute the values

$v=0+\frac{F}{m}t=\frac{F}{m}t$

We know that

Momentum, p=mv

Using the formula

$p=m\times \frac{F}{m}t=Ft$

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avanturin [10]

Answer:

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4 0

2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

2 years ago

A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e

m_a_m_a [10]

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

$10 years = \frac{2d}{v}$

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

$0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }$

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

$0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }$

$0.64 = 1-\frac{v^{2} }{186282^{2} }$

solving for v, gives = 35 288 miles/s

4 0

2 years ago

NEED ANSWER PLEASE!!!!

olganol [36]

Answer:A, Concave

Explanation:

3 0

2 years ago

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a

fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

$sin \theta = \frac{n \lambda}{a}$

where

$\theta$ is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

$\lambda$ is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because $\theta$ increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) $10.8^{\circ}$

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

$sin \theta = \frac{(1) \lambda}{a}$

in this case we have

$\lambda=54 cm = 0.54 m$ is the wavelength

$a=5.7 m$ is the width of the opening

Solving the equation, we find

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}$

(c) $0.26^{\circ}$

Here we can apply the same formula used before, but this time the wavelength of the signal is

$\lambda=13 mm=0.013 m$

so the angle corresponding to the first minimum is

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}$

5 0

2 years ago