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CaHeK987 [17]
4 years ago
15

The Top Thrill Dragster stratacoaster at Cedar Point Amusement Park in Ohio uses a hydraulic launching system to accelerate ride

rs from 0 to 54 m/s (120 mi/hr) in 3.8 seconds before climbing a completely vertical 420-foot hill. Determine the net force required to accelerate an 86-kg man.
Physics
1 answer:
Mariulka [41]4 years ago
3 0

Answer:

A. 1222,10 N

Explanation:

First, we determine the value of the acceleration that the hydraulic launching system must guarantee. Let’s remember that the definition of acceleration is the rate of change of velocity with respect to time. Therefore:

a_hyd = ∆v/∆t = (54 m/s-0)/(3,8 s-0) = 14,21 m/s^2  

According to Newton’s second law:

∑F_net = m*a = 86 kg*14,21 m/s^2 = 1222,10 N

So, the net force required to accelerate an 86-kg man is 1222,10 N

Have a nice day! :D

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B

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A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
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Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

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  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

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  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

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The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

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