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arlik [135]
3 years ago
5

Arturo dreams about travelling to faraway stars someday. He read that the closest star to Earth is Proxima Centauri, which is 4.

25 light years away. He also read that the Law of Special Relativity says that the mass of an object increases as its speed increases. It approaches infinity as it approaches the speed of light. How does this most likely affect the goals of the United States space program?
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

Answer:

It will keep the goals from being comeplet when they wanted it to.

Explanation:

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What’s the equation for the alpha decay of plutonium-244.
kvv77 [185]

Answer:

Pu-244 and 94  alpha decay = U -240 and 92

8 0
3 years ago
A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o
Furkat [3]
            <span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

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4 0
3 years ago
Read 2 more answers
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s
loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

m*V_i = m_1V_1+m_2V_2

Here,

m = Total mass (8Kg at this case)

m_1=m_2 = Mass each part

V_i = Initial velocity

V_2 = Final velocity particle 2

V_1 = Final velocity particle 1

The initial kinetic energy would be given by,

KE_i=\frac{1}{2}mv^2

KE_i = \frac{1}{2}8*5^2

KE_i = 100J

In the end the energy increased 100J, that is,

KE_f = KE_i KE_{increased}

KE_f = 100+100 = 200J

By conservation of the moment then,

m*V_i = m_1V_1+m_2V_2

Replacing we have,

(8)*5 = 4*V_1+4*V_2

40 = 4(V_1+V_2)

V_1+V_2 = 10

V_2 = 10-V_1(1)

In the final point the cinematic energy of EACH particle would be given by

KE_f = \frac{1}{2}mv^2

KE_f = \frac{1}{2}4*(V_1^2+V_2^2)

200J=\frac{1}{2}4*(V_1^2+V_2^2)(2)

So we have a system of 2x2 equations

V_2 = 10-V_1

200J=\frac{1}{2}4*(V_1^2+V_2^2)

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: V_1 = 10m/s

Then replacing in (1) we have that

PART B: V_2 = 0m/s

8 0
3 years ago
If the real gi tract is 9 meters long, the model you created is what fraction of the real one
Sedbober [7]

Answer:

a

Explanation:

a

7 0
4 years ago
A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a fric
MAVERICK [17]

∑ <em>F</em> (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) <em>a</em>

==>   <em>a</em> = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²

7 0
3 years ago
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