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3 years ago

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Problem 28.3 Suppose that you are planning a trip in which a spacecraZ is to travel at a constant velocity for exactly six month

s, as measured by a clock on board the spacecraZ, and then return home at the same speed. Upon your return, the people on earth will have advanced exactly one hundred years into the future. According to special relaEvity, how fast must you travel? Express your answer to five significant figures as a mulEple of c – for example, 0.95585c.

1 answer:

iragen [17]3 years ago

3 0

Answer:

v = 0.999981c m/s

Explanation:

Using the time dilation equation

$T = \frac{T_{0} }{\sqrt{1 - \frac{v^{2} }{c^{2} } } }$

T = stationary time = 100 years

T₀ = 11/12 years = 0.917 years

v = speed of travel in the space = ?

c = speed of light = 3 * 10⁸ m/s

$100 = \frac{0.917 }{\sqrt{1 - \frac{v^{2} }{(3*10^{8} )^{2} } } }\\$

$(0.917/100) ^{2} = 1 - \frac{v^{2} }{ 9 * 10^{16} }$

v = 299987395.57 m/s

v = 2.99 * 10⁸ m/s

v = 0.999981c m/s

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A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

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3 years ago

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

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3 years ago

Find the distance in nm between two slits that produces the first minimum for 410-nm violet light at an angle of 14.5°.

Galina-37 [17]

Answer:

820 nm

Explanation:

We are given that

Wavelength= $\lambda=410 nm$

$\lambda=410\times 10^{-9} m$

$1nm=10^{-9} m$

$\theta=14.5^{\circ}$

For first minimum therefore

m=0

We know that for destructive interference

$(m+\frac{1}{2})\lambda=dsin\theta$

Substitute the values

$(0+\frac{1}{2})\times 410\times 10^{-9}=dsin 14.5$

$d=\frac{410\times 10^{-9}}{2\times sin 14.5}$

$d=820\times 10^{-9} m=820 nm$

Hence, the distance between two slits that produces the first minimum=820 nm

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3 years ago

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Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$

Replacing,

$a =\frac{42N}{83000kg}$

$a =5.06*10^{-4}m/s^2$

The total speed change

$\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity }$ we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

$a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}$

We have that,

$t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }$

$t = 1403.16s$

Therefore the Rocket should be fired around to 1403.16s

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3 years ago

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.320 m2. at

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Remember, half of the energy in an EM wave is in the E field, the rest is in the B field. Thus, multiply E field energy by 2.
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3 years ago