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Tanzania [10]
3 years ago
6

11. A roller coaster car is moving slowly up a long ramp to the very top of the roller coaster. It goes over the top and starts

down a steep slope and it picks up speed rapidly as it rolls down an almost vertical slope. Four seconds later, at the bottom of the steep slope, its velocity is 28 m/s. The average acceleration during those 4 seconds was 6 m/s2 . What was the initial velocity of the roller coaster car as it started down the slope?
Physics
1 answer:
DiKsa [7]3 years ago
3 0

Answer:

a = (v2 - v1) / t

a t = v2 - v1

v1 = v2 - a t

v1 = 28 m/s - 6 m/s^2 * 4 s = 4 m/s

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A stretched rubber band is an example of which type potential energy
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Answer: elastic potential energy

Explanation:

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2 years ago
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A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
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We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
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3 years ago
Gina made a poster for plastic recycling week and included this information on her poster:
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Gina should put “rubber tires” under “Synthetic.”

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Science help please!
OLga [1]

Answer:

104 N

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m = 1300 kg

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3 0
3 years ago
A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling
KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

                    = 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0
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