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3 years ago

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Choose all the answers that apply.

1 answer:

spin [16.1K]3 years ago

3 0

Answer:

B and C?

Explanation:

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Astronaut mark uri is space-traveling from planet x to planet y at a speed of relative to the planets, which are at rest relativ

lyudmila [28]

<span>From the point of view of the astronaut, he travels between planets with a speed of 0.6c. His distance between the planets is less than the other bodies around him and so by applying Lorentz factor, we have 2*</span>√1-0.6² = 1.6 light hours. On the other hand, from the point of view of the other bodies, time for them is slower. For the bodies, they have to wait for about 1/0.6 = 1.67 light hours while for him it is 1/(0.8) = 1.25 light hours. The remaining distance for the astronaut would be 1.67 - 1.25 = 0.42 light hours. And then, light travels in all frames and so the astronaut will see that the flash from the second planet after 0.42 light hours and from the 1.25 light hours is, 1.25 - 0.42 = 0.83 light hours or 49.8 minutes.

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3 years ago

The bodies in this universe attract one another name the scientist who propounded this statement

Colt1911 [192]

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7 0

2 years ago

The coefficient of linear expansion of ordinary glass is three times that of Pyrex glass. An ordinary glass rod and a Pyrex glas

DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length given as

ΔL = L α ΔT

Therefore

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}$

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}$

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

6 0

3 years ago

Assuming a roughly spherical shape and a density of 4000 kg/m3, estimate the diameter of an asteroid having the average mass of

vredina [299]

Explanation:

It is given that,

Density of asteroid, $\rho=4000\ kg/m^3$

Mass of asteroid, $m=10^{17}\ kg$

We need to find the diameter of the asteroid. The formula of density is given by:

$\rho=\dfrac{m}{V}$

V is the volume of spherical shaped asteroid, $V=\dfrac{4}{3}\pi r^3$

$r^3=\dfrac{3M}{4\pi \rho}$

$r^3=\dfrac{3\times 10^{17}\ kg}{4\pi \times 4000\ kg/m^3}$

$r^3=\sqrt{5.96\times 10^{12}}$

r = 2441311.12 m

Diameter = 2 × radius

d = 4882622.24 m

or

$d=4.88\times 10^6\ m$

Hence, this is the required solution.

8 0

3 years ago

At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit

matrenka [14]

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

Starting point. With the mouse in the center

L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

I = 1/3 M L²

Final point. When the mouse is at the end of the rod

$L_{f}$ = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

L₀ = L_{f}

I w₀ = (I + m L²) w

w = I / I + m L²) w₀

We substitute the moment of inertia

w = 1/3 M L² / (1/3 M + m) L² w₀

w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

w = 1/3 / (1/3 + 0.02) w₀

w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

w = 0.943 rad / s

3 0

3 years ago