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KiRa [710]
4 years ago
14

A train composed of a small engine car and a massive cargo car are connected as they move along a track. The speed of the small

engine car is ____(blank 1)___ the speed of the massive cargo car. The magnitude of the acceleration of the small engine car is ____(blank 2)___ the magnitude of the acceleration of the massive cargo car because ____(blank 3)___ .
Physics
1 answer:
stepan [7]4 years ago
7 0

Answer:

The speed of the small engine car is <u>_equal to_</u> the speed of the massive cargo car. The magnitude of the acceleration of the small engine car is <u>_equal to_</u> the magnitude of the acceleration of the massive cargo car because <u>_the cars speeds cannot change by different amounts_</u> .

Explanation:

The two cars are connected, hence, they move with the same speed at all points in time.

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In heat conduction analysis, the conversion of ____ energy into ____ energy in solids is called heat generation. check all that
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4 0
2 years ago
A student walks 5.00 m at an angle of 53.0° then walks another 8.00 m at angle of 130°. what is the magnitude of the student's f
dmitriy555 [2]
This question can easily be answered using a scientific calculator via complex mode using the form r∠α wherein r is the magnitude and α is the angle. So, incorporating vector addition through scientific calculator:

5∠53° + 8∠130 = 10.34∠101.9°

Therefore, the student's displacement is 10.34 m, 101.9°. In other words, the student is heading 78.1° (180 - 101.9 = 78.1)  south of east.
6 0
4 years ago
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Two carts are free to slide along the frictionless track shown in figure below. Cart A of mass m1 = 8 kg is released from 12m. A
Harrizon [31]

The height reached by the two carts after collision is determined as 5.34 m.

<h3>Initial velocity of Cart A</h3>

Apply the principle of conservation of mechanical energy.

K.E = P.E

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

<h3>Final velocity of the two carts after the collision</h3>

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

8(15.34) + 4(0) = v(8 + 4)

122.72 = 12v

v = 10.23 m/s

<h3>Height reached by both carts</h3>

Apply the principle of conservation of mechanical energy.

P.E = K.E

mgh = ¹/₂mv²

h = v²/(2g)

h = (10.23²) / (2 x 9.8)

h = 5.34 m

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