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monitta
3 years ago
8

A ______ is a massive object with gravitational effect so strong that even light cannot escape it. A)black hole

Physics
2 answers:
ruslelena [56]3 years ago
4 0
A. Black hole is the answer


Dovator [93]3 years ago
4 0
The answer is A) Black Hole

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What is the force on a 52 coulomb charge in a field of 5 N/C
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3 years ago
A rifle bullet leaves the barrel at 1500 meters per second. If it does not slow down or hit anything, how far can it travel in h
galben [10]

Since we are dividing a second in 2, we can simply divide the amount of meters by 2.

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7 0
3 years ago
A railroad cart with a mass of m1 = 11.6 t is at rest at the top of an h = 10.9 m high hump yard hill.
leonid [27]

The final common speed of the two carts will be 69.3 m/sec.The momentum conservation principle is applied.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Unit conversion:

1 metric tons = 1000 kg

Given data;

m₁= 11.6 metric ton =11600 kg

m₂ = 23.2 metric ton =  23200 kg

Let v represent the combined velocity of the two carts once they are connected, and let u represent the starting velocity of cart 1 when it reaches the bottom.

Considering energy conservation;

\rm  m_1 g h = \frac{1}{2} m_1 \times  u^2 \\\\ u^2 = 2gh\\\\ u^2 = 2  \times 9.8 \times 10.6 \\\\  u = 207.972 \ m/s

From the conservation of momentum principle;

\rm m_1 \times  u = (m_1 + m_2) v\\\\ 11600 \times  207.972 = (11600 + 23200) \times v \\\\ v = 69.3  \ m/s

Hence, the final common speed of the two carts will be 69.3 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

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7 0
2 years ago
Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The
Kay [80]

Answer:

25.33 rpm

Explanation:

M = 100 kg

m1 = 22 kg

m2 = 28 kg

m3 = 33 kg

r = 1.60 m

f = 20 rpm

Let the new angular speed in rpm is f'.

According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.

Initial angular momentum = final angular momentum

(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =

                                  (1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'

(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

2660 = 105 x f'

f' = 25.33 rpm

8 0
3 years ago
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