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tamaranim1 [39]
2 years ago
11

A 535 kg roller coaster car began at rest at the top of a 93.0 m hill. Now it is at the top of the first loop-de-loop. This roll

er coaster’s track is nearly frictionless, so resistance can be ignored. Using g = 9.8 m/s2, what best describes the roller coaster car when it is at the top of the loop-de-loop?
Physics
1 answer:
Nikitich [7]2 years ago
3 0
The correct answer is <span>The car has both potential and kinetic energy, and it is moving at 24.6 m/s.</span>
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What is the velocity (in m/s) of a 550 kg roller coaster cart at the bottom of the track if it started with 990,000 J of gravita
sesenic [268]

Answer:

60m/s

Explanation:

initial energy = final energy

g.p.e = k.e

k.e = 0.5 × mass × velocity²

g.p.e = 990000J as per Question

990000Nm = 0.5 × 550 × V²

V² = 3600

V = 60m/s

4 0
2 years ago
Go around your house and hit (or tap) ten different objects. The objects should be different shapes and made of different materi
Blababa [14]

Answer:

 

Explanation:

go around your house and tap random objects. For example, a sink. What noise did it make? was it loud or quiet? was it soft or hard? I hope this helps

4 0
2 years ago
Read 2 more answers
A car is moving with the velocity of 90km/h. If the car come to rest after 10 seconds. Calculate the final velocity and distance
djyliett [7]

Answer:

you can learn from here

https://www.toppr.com/ask/en-bd/question/a-car-is-moving-with-a-velocity-of-10-ms-the-driver-sees-a-wall/

6 0
3 years ago
a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


3 0
3 years ago
Answers are - <br><br>A 235 N<br><br>B 376 N<br><br>C 271 N<br><br>D 188 N<br><br>E 470 N
amm1812

Answer:

C

Explanation:

8 0
3 years ago
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