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Dmitrij [34]
3 years ago
14

An electric motor moves a chain that pulls the roller coaster car 180 meters to the top of the first hill. The chain exerts a fo

rce of 2,524 newtons on the car. How much kinetic energy does the car gain from the work done on it? Use the formula Change in KE = F × d.
Physics
2 answers:
nirvana33 [79]3 years ago
4 0

Answer:

Explanation:

The work done on an object is equal to its change in kinetic energy: Change in KE = F × d.

Plug the values for F and d into the formula and solve:

Change in KE = 2,524 × 180

= 454,320 joules

The roller coaster gains 454,320 joules of energy from the work done on it by the chain.

MrRa [10]3 years ago
3 0
The energy added here is potential energy since it is moving upward 180 meters in a gravitational field.  This is then turned into KE when it rolls down.  2524N x 180m = 454,320J
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What is the efficiency of a machine?​
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Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work).

Explanation:

It is a measure of how well a machine reduces friction.

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2 years ago
Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?
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Two 8.0 Ω lightbulbs are connected in a 12 V parallel circuit. What is the power of both glowing bulbs?
kati45 [8]

Answer:

96w

Explanation:

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8 0
2 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0
3 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
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