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blondinia [14]
2 years ago
8

A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte

rval between ticks of the spacecraft clock to be 3.2 s. What is the time interval (in seconds) that an earth observer measures
Physics
1 answer:
Afina-wow [57]2 years ago
6 0

Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

t = 3.2 / √( 1 - 0.36 )

t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

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Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
2 years ago
An action potential arriving at the presynaptic terminal causes what to occur?
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Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell

5 0
3 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

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Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

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∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

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4 0
2 years ago
24. A anvil with a mass of 60 kg falls from a height of 9.5 m. How fast is it going right
shepuryov [24]

So they give us this

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So we insert the thing that we know.

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Lastly we solve.

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6 0
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Explanation:

4 0
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