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svp [43]
3 years ago
14

Desperado, a roller coaster built in Nevada, has a mass of 800 kg. It also has a vertical drop of 225 feet down the first hill.

The roller coaster is designed so that the speed of the cars at the end of this drop is 80 mi/h. Assume the cars are at rest at the start of the drop. How much work is done by friction on the car as it drops down the hill
Physics
1 answer:
weeeeeb [17]3 years ago
6 0

Answer:

the work done by friction on the car is 524,582 J.

Explanation:

Given;

mass of the roller coaster, m = 800 kg

distance moved by the coaster, d = 225 ft = 68.58 m

final velocity of the coaster, v = 80 mi/h = 35.76 m/s

The time taken for the coaster to drop down the hill is calculated as;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\ \times \ 68.58}{9.8} } \\\\t = 3.74 \ s

The work done by friction on the car is calculated as;

W =F\ \times \ d = \frac{mv}{t} \times \ d\\\\W = \frac{800 \ \times \ 35.76  }{3.74} \times \ 68.58\\\\W = 524,582 \ J

Therefore, the work done by friction on the car is 524,582 J.

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how much does a bookshelf weigh if the movers are pushing it at a speed of 10 m/s^2 by applying 100 N force
Delicious77 [7]

Answer:

10 kg

Explanation:

Assuming a frictionless surface, then force F=ma where F is the applied force, m is the mass and a is acceleration. Making m the subject of the formula then m=\frac {F}{a}

Substituting 100 N for the applied force F and 10 m/s^2 for acceleration a then the value of m will be m=\frac {100}{10}=10\ kgs

Therefore, in terms of kilograms, the bookshelf weighs 10 Kg

7 0
2 years ago
A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?
lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos\theta, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, \theta = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0
2 years ago
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
8 0
3 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i
alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

d=vt+\frac{1}{2}at^2

Now by substituting values in, we get

d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2

d = 19.796m

7 0
2 years ago
190 kg of water is to be raised by a water pump to a height of 25 meters from the bottom of a well in 60 seconds. What should be
ladessa [460]

Answer:

776.6 w

1.04 hp

Explanation:

given:

Mass, m = 190kg

height change, h = 25m

time elapsed, t = 60 s

acceleration due to gravity, g = 9.81 m/s²

Potential energy required raising 190 kg of water to a height of 25m

= mgh

= 190 x 9.81 x 25

= 46,597.5 J

Power required in 60 s

= Energy required ÷ time elapsed

= 46,597.5 ÷ 60

= 776.6 Watts  (Use conversion 1 W = 0.00134102 hp)

= 776.6 w x 0.00134102 hp/w

= 1.04 hp

6 0
2 years ago
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