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DENIUS [597]
3 years ago
7

What pressure, in millimeters of mercury (mm Hg), is equivalent to 2.13 atmospheres

Physics
1 answer:
Sati [7]3 years ago
6 0
The "Standard Atmosphere" is 760 mm Hg . 2.13 times that pressure is (2.13 x 760) = 1,618.8 mm Hg.
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A magnetic field would be produced by a beam of
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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

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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,

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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

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Explanation:

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Answer:

Explanation:

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