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mestny [16]
3 years ago
15

2. A 4.0 kg magnetic toy car traveling at 3.0 m/s east collides and sticks to a 5.0 kg toy magnetic car also traveling at 2.0 m/

s east. Calculate the final speed and direction of the magnetic car (coupled) system?
Physics
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:

2.44 m/s due East

Explanation:

From the question given above, the following data were obtained:

Mass of 1st car (m₁) = 4 Kg

Velocity of 1st car (u₁) = 3 m/s

Mass of 2nd car (m₂) = 5 Kg

Velocity of 2nd car (u₂) = 2 m/s

Final velocity (v) =?

The final velocity can be obtained as follow:

v(m₁ + m₂) = m₁u₁ + m₂u₂

v(4 + 5) = (4×3) + (5×2)

9v = 12 + 10

9v = 22

Divide both side by 9

v = 22/9

v = 2.44 m/s

Thus, the final velocity is 2.44 m/s.

Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.

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Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2
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Answer:

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Explanation:

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F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C

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A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90
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Answer:

a= 3.49 m/s^2

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