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olga55 [171]
3 years ago
7

Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th

at the +x-axis is directed eastward and the +y-axis is directed northward. Find the magnitude of the velocity vector of the truck relative to the SUV.
Physics
1 answer:
mario62 [17]3 years ago
5 0

Answer:25.61 m/s

Explanation:

Given

truck is moving eastbound with a velocity of 16 m/s

Velocity of truck v_t=16\hat{i}

SUV is moving south with a velocity of 20 m/s

Velocity of SUV in vector form v_s=-20\hat{j}

Velocity of truck relative to the SUV

v_{ts}=v_{t}-v_s

v_{ts}=16\hat{i}-(-20\hat{j})

Magnitude of relative velocity is

|v_{ts}|=\sqrt{16^2+20^2}

|v_{ts}|=25.61\ m/s                                                  

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The mechanical energy isn't conserved. Some energy is lost to friction.

Option A.

<h3><u>Explanation:</u></h3>

The mechanical energy is defined as the energy of a body which it achieves by virtue of its position and velocity. The mechanical energy are of two types - potential energy and kinetic energy. The potential energy is the energy of the body which it achieves by means of its relative position and is directly proportional to the height of the body from its relative plane. Whereas the kinetic energy of the body is achieved by virtue of its velocity and is directly proportional to the square of velocity of the body.

As the mountaineer is skiing down the slope of a mountain, the potential energy of the person is gradually changing into his kinetic energy. Had it been in an ideal situation, the potential energy lost would have been just equal to the kinetic energy gained by the person. But there's friction which opposes the speed of the body and reduces the velocity. Thus the kinetic energy will be lost to some extent and the energy won't be conserved.

4 0
3 years ago
A man of mass m 1 5 70.0 kg is skating at v1 5 8.00 m/s behind his wife of mass m 2 5 50.0 kg, who is skating at v2 5 4.00 m/s.
ehidna [41]

Answer:

A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)

B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist

C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf

Say mass of husband is m1

Mass of the wife is m2

Velocity of the husband is v1

Velocity of the wife is v2

According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf

The momentum equation is

m1v1+m2v2= (m1+m2)vf

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vf= {(m1v1) +(m2v2)}/(m1+m2)

E. Substituting our given data

vf=

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vf=91060+137700/4120

vf=228760/4120

vf=55.52m/s

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3 years ago
A 60 g ball of clay is thrown horizontally at 40 m/s toward a 1.5 kg block sitting at rest on a frictionless surface. the clay h
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The solution for this problem is:
Let u denote speed. 

Equating momentum before and after collision: 
= 0.060 * 40 = (1.5 + 0.060) u 
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
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I really need help for this question
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