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Viefleur [7K]
3 years ago
9

A railway car having a total mass of 5.8 × 105 kg, moving with a speed of 9.1 km/h, strikes another car that has a mass of 8.7 ×

105 kg and is initially at rest. The speed of the coupled cars after the collision is
A) 9.1 km/h.
B) 7.2 km/h.
C) 3.6 km/h.
D) 1.8 km/h.
E) 4.2 km/h
Physics
1 answer:
Jobisdone [24]3 years ago
3 0

Answer:

Final speed of both the cars, V = 3.28 km/h

Explanation:

It is given that,

Mass of the railway car, m_1=5.8\times 10^5\ kg

Initial speed of the railway car, u_1=9.1\ km/h=2.52\ m/s

Mass of another car, m_2=8.7\times 10^5\ kg

Initial speed of another car, u_2=0

To find,

The speed of the coupled cars after the collision.

Solution,

It is a case of inelastic collision in which the linear momentum before and after the collision remains same. Let V is the coupled velocity of both of the cars. So,

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{m_1u_1}{(m_1+m_2)}

V=\dfrac{5.8\times 10^5\times 2.52778}{(5.8\times 10^5+8.7\times 10^5)}

V = 1.011 m/s

or

V = 3.28 km/h

So, the speed of the coupled cars after the collision is 3.28 km/h. Hence, this is the required solution.

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A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C
MAVERICK [17]

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

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3 years ago
The loaded cab of an elevator has a mass of 3.0 x 10 3 kg and moves 200 m up the shaft in 20 s at constant speed. At what averag
Alexeev081 [22]

The average rate at which the cable does work is 294,000 J/s.

The given parameters:

  • <em>mass, m = 3000 kg</em>
  • <em>height, h = 200 m</em>
  • <em>time of motion, t = 20 s</em>

The average rate at which the cable does work is calculated as follows;

P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{3000 \times 9.8 \times 200}{20} \\\\P = 294,000 \ J/s

Thus, the average rate at which the cable does work is 294,000 J/s.

Learn more about energy and power here: brainly.com/question/13387946

4 0
3 years ago
A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
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The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision  = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2

v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0
3 years ago
A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
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