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Jobisdone [24]
2 years ago
12

The first hill of a roller coaster is 42.0 meters high. The bottom of the first hill is 4.2 meters off the ground.The roller coa

ster cart has a mass of 4500 kg.
Calculate its potential energy at the top of the first hill.





Calculate its potential energy at the bottom of the first hill.





Challenge- What is the velocity of the cart at the bottom of the first hill?
Physics
1 answer:
Karolina [17]2 years ago
7 0

Answer:

1.89*10⁶J

Explanation:

because 4500kg * 42m*10m/s² is 1890000J

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What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing a=76m^2

We know that density of air d=1.29kg/m^3

Speed at top surface v_2=290m/sec and speed at bottom surface v_1=150m/sec

According to Bernoulli's principle force is given by

F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N

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3 years ago
Which of the following ways of writing 1000w is incorrect?
xenn [34]

Answer:

the third one is incorrect

Explanation:

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2 years ago
A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

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The initial temperature of a bomb calorimeter is 28.50°
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The answer is 5,170 J
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