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Flura [38]
3 years ago
14

A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of

____. A. positive acceleration B. velocity C. negative acceleration D. speed
Physics
2 answers:
Neko [114]3 years ago
5 0
The answer is letter C. <span>A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of negative acceleration.

In physics, acceleration can be described as an objects change of velocity. When an object gains velocity, it is positive acceleration, and negative acceleration for the opposite.

</span>

Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.

<span> </span>

hjlf3 years ago
5 0
<span>Negative Acceleration is when the acceleration of an object slows down, like throwing something in the air. It will get slower going up, and then it will stop, that is the negative acceleration, and then the object will come back down, and will speed up, which is positive acceleration.
</span>
A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of <span>negative acceleration.
Best Choice : </span><span> C. negative acceleration</span>

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Explain how you know that gasoline burning in a car engine is a chemical change.?
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fuel was consumed at a certain rate of 0.05Kg\s in a rocket engine and ejected as a gas with a speed of4000m\s . Determine the t
ivann1987 [24]

Answer:

Thrust = 200 N

Explanation:

The engine thrust can be found by using the following formula:

Thrust = mv

where,

m = mass flow rate of the fuel = 0.05 kg/s

v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

Thrust = (0.05\ kg/s)(4000\ m/s)

<u>Thrust = 200 N</u>

5 0
2 years ago
A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
Find the final velocity
Verizon [17]

Answer: v = 4.4 m/s

Explanation:

In the absence of friction, the total mechanical energy will be constant

KE₀ + PE₀ = KE₁ + PE₁

0 + mg(6) = ½mv₁² + mg(5)

½mv₁² = mg(6 - 5)

v = √(2g(1)) = 4.4 m/s

3 0
3 years ago
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