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rodikova [14]
3 years ago
9

You are flying your personal rocketcraft at 0.9c from Star A toward Star B. The distance between the stars, in the stars' refere

nce frame, is 1.0 ly. Both stars happen to explode simultaneously in your reference frame at the instant you are exactly halfway between them.
The question asked of the aforementioned scenario is "Do you see the flashes simultaneously?" and what is the time difference between the two?
Physics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

See explanation.

Explanation:

If both stars explode in simultaneously in the <em>your </em>frame of reference then obviously you will see the two flashes simultaneously, and therefore, the time difference between the events would be zero.

If however, the stars exploded simultaneously in their frame of reference, then you would not observe the flashes simultaneously.  Then the time difference between the events will not be zero, rather, you will observe star B exploding first and star A after.

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What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the pro
Novay_Z [31]

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) ×  (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

6 0
3 years ago
We think of our muscular system as the system that helps us move around. It also functions on an internal level by helping food
trapecia [35]
Stomach;small intenstines

7 0
3 years ago
The name dinitrogen tetroxide tells you that this compound contains
Leto [7]
The answer to that is c hope it helps
5 0
4 years ago
A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h
raketka [301]

Answer:

\rho = 1848.03 kg m^{-3}

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height  of water  = 20 cm  =0.2 m

Pressure  p = 1.01300*10^5 Pa

pressure at bottom

P =  P_{fluid} + P_{h_2 o}

P   = P_{fluid}  + \rho g h

P_{fluid}  = P - \rho g h

                 = 1.01300*10^5 - 1000*0.2*9.8

                 = 99340 Pa

p_{fluid}  = P_{atm} + \rho g h_{fluid}                       h_[fluid} = 0.307m

99340 = 104900 + \rho *9.8*0.307

\rho = 1848.03 kg m^{-3}

5 0
4 years ago
The two forces of a 3rd law pair always act on different bodies. The two forces of a 3rd law pair always act on different bodies
svetoff [14.1K]

1. The two forces of a 3rd law pair always act on different bodies.

TRUE

SO above statement is TRUE because Newton's III law is valid for two different objects where one object will exert force and other body will exert reaction force on it.

2. Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.)

FALSE

This is false because the force of action and Reaction is same on two different objects but for finding acceleration we need to divide the force by mass of two objects and since the mass of two bodies may be different so we can say that acceleration may be different.

3. Part G According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is

(i) greater in magnitude and antiparallel to the force on the earth due to the moon.

(ii) greater in magnitude and parallel to the force on the earth due to the moon.

(iii) equal in magnitude but antiparallel to the force on the earth due to the moon.

(iv) equal in magnitude and parallel to the force on the earth due to the moon.

(v) smaller in magnitude and antiparallel to the force on the earth due to the moon.

(vi) smaller in magnitude and parallel to the force on the earth due to the moon.

Since Earth and moon is an isolated system so here Newton's III law is valid due to which we can say that two forces are equal in magnitude but opposite in sign

so correct answer will be

equal in magnitude but antiparallel to the force on the earth due to the moon.

4 0
3 years ago
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