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antoniya [11.8K]
3 years ago
10

What is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?

Physics
2 answers:
maksim [4K]3 years ago
8 0

Answer:

M=75

Explanation:

33750=1/2x M x30²

30x30=900

1/2x900=450

33750/450=75

nikklg [1K]3 years ago
5 0
The Energy is Kinetic Energy.

Kinetic Energy = 1/2*mv²,  Where m is mass in kg, v is velocity in m/s

Energy is 33750 Juoles,  v = 30m/s

1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²)     Using a calculator

m = 75 kg

Mass of object is 75 kg.
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A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin
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Answer: The total momentum before the docking maneuver is mV_{1}+MV_{2} and after the docking maneuver is (m+M) U

Explanation:

Linear momentum p (generally just called momentum) is defined as mass in motion and is given by the following equation:  

p=m.v  

Where m is the mass of the object and v its velocity.

According to the conservation of momentum law:

<em>"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision </em>p_{i} <em>will be the same as the total momentum of these same two objects after the collision </em>p_{f}<em>". </em>

<em />

p_{i}=p_{f}

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the<u> initial momentum</u> is:

p_{i}=mV_{1}+MV_{2}

Where:

m is the mass of the vehicle

V_{1} is the velocity of th vehicle

M is the mass of the space station

V_{2} is the velocity of the space station

And the <u>final momentum</u> is:

p_{f}=(m+M)U

Where:

U is the velocity of the vehicle and space station docked

6 0
3 years ago
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Special relativity can be used to study an object in which frame of reference? A. A frame of reference that has no change in vel
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A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r
cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

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Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0
2 years ago
The attraction will vary directly with the separation between the charges.
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If the pendulum took longer to complete one oscillation, how would the graph change?
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Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

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