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vlada-n [284]
4 years ago
9

The Starship Enterprise speeds past an asteroid at 0.920c. If an observer on the asteroid sees 10.0 seconds pass on her watch, h

ow long would that time interval be if measured by an observer on the Enterprise?
Physics
1 answer:
lukranit [14]4 years ago
5 0

Answer:

25.52 seconds

Explanation:

Speed of space ship = 0.92 c = v

c = Speed of light

Time observed from asteroid = Δt = 10 seconds

Time dilation

\Delta t'=\frac{\Delta t}{\sqrt{1-\frac {v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-\frac {0.92^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-0.92^2}}\\\Rightarrow \Delta t'=25.52\ s

∴ Time interval be if measured by an observer on the Enterprise would be 25.52 seconds

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Suppose the voltage source for a series RL-circuit were given as V0sin(ωt) instead of V0cos(ωt). Write an expression for the cur
Gala2k [10]

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = \mathbf{\sqrt{R^2+L^2}}

\mathbf{z = \sqrt{R^2+(Lw)^2}}

Current amplitude

\mathbf{I_o = \dfrac{V_o}{z}} \\ \\  \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}

a)

Given that:

V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\  L = 0.52 H

∴

I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}

\mathbf{I_o= 0.0562}  \\ \\ \mathbf{I_o = 56.2 \ mA}

b)

Phase constant :

tan  \ \phi = \dfrac{L \omega}{R } \\ \\  tan  \ \phi = \dfrac{0.52 \times 51}{21} \\ \\  tan \phi = 1.263

\text{Phase constant : }\phi = tan^{-1} (1.263)   \\ \\  \phi = 51.6^0\\ \\\text{To radians} \phi  = 51.6 \times \dfrac{\pi}{180} \\ \\  \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}

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3 years ago
Can someone please help me?
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7 0
3 years ago
A rocket is fired upwards with an acceleration of 32 m/s2 . If the 200kg rocket experiences an air resistance force of 12000N, w
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Answer: 20360 N

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6 0
3 years ago
A block slides to a stop as it goes 47 m across a level floor in a time of 6.35 s. a) What was the initial velocity? b) What is
UNO [17]

Answer :

(a) The initial velocity is, 14.8 m/s

(b) The acceleration is, -2.33m/s^2

Explanation :

By the 1st equation of motion,

v=u+at     ...........(1)

where,

v = final velocity = 0 s

u = initial velocity

t = time = 6.35 s

a = acceleration

The equation 1 will be:

0=u+a(6.35}

u=-6.35a       ..........(2)

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2     ...........(3)

where,

s = distance = 47 m

Now substitute equation 2 in 3, we get:

47=(-6.35a)\times (6.35)+\frac{1}{2}\times a\times (6.35)^2

By solving the term, we get:

a=-2.33m/s^2

The acceleration is, -2.33m/s^2

Now we have to calculate the initial velocity.

Using equation 2, we gte:

u=-6.35a

u=-6.35s\times (-2.33m/s^2)

u=14.8m/s

The initial velocity is, 14.8 m/s

5 0
3 years ago
What function does the shape of the star-nosed mole’s nose serve?
horsena [70]
Hello!

The shape of the star-nosed mole's nose helps it to take away the excess heat.

Letter b)

Hugs!
6 0
3 years ago
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