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Tresset [83]
3 years ago
15

Can Jet Fuel burn through Steel Pipes?

Physics
1 answer:
Free_Kalibri [48]3 years ago
8 0
Burning jet fuel is not hot enough to fully melt steel. Burning jet fuel, however, is more than hot enough to weaken steel.
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Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0
3 years ago
Technician A says that the starter motor used to crank diesel engines can draw up to 400 amps of current. Technician B says that
Aleks04 [339]

Answer: Option A : Technician A

Explanation:

The statement/observation, "that the starter motor used to crank diesel engines can draw up to 400 amps of current" made by Technician A is correct.

A diesel engine uses up to 400+ Amperes of electricity to start up a diesel engine in the ignition chamber of motor engine.

8 0
3 years ago
Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the conta
hodyreva [135]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file has a detailed solution of the given problem.

4 0
3 years ago
A student throws a baseball horizontally at 25 meters per second from a cliff 45 meters above the level ground. Approximately ho
Delicious77 [7]
First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed. 
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m 
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s 
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use: 
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s 
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
8 0
3 years ago
A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int
nevsk [136]

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

5 0
3 years ago
Read 2 more answers
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