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faust18 [17]
3 years ago
13

Assume that a team of engineers runs a friction test on the car and track. The team finds that the car loses 210 joules of energ

y over each meter of track. How much energy does the roller coaster lose from friction as it travels from the top of the first hill to the top of the second hill
Physics
2 answers:
Reika [66]3 years ago
0 0

The rate of energy loss from friction is 210 joules/meter of track. Multiply by the length of the track (320 meters) to find the total energy lost:

210 J/m × 320 m = 67,200 J

The roller coaster car loses 67,200 joules of energy as heat between the tops of the two hills.

diamong [38]3 years ago
0 0

Answer:

The rate of energy loss from friction is 210 joules/meter of track. Multiply by the length of the track (320 meters) to find the total energy lost:

210 J/m × 320 m = 67,200 J

The roller coaster car loses 67,200 joules of energy as heat between the tops of the two hills.

Explanation:

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stealth61 [152]

Answer :

(a) The acceleration  of the car is, -2m/s^2

(b) The distance covered by the car is, 150 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = ?

Now put all the given values in the above equation 1, we get:

5m/s=25m/s+a\times (10s)

a=-2m/s^2

The acceleration  of the car is, -2m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = -2m/s^2

Now put all the given values in the above equation 2, we get:

s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2

By solving the term, we get:

s=150m

The distance covered by the car is, 150 m

8 0
3 years ago
A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break
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Answer:

2.4 m/s^2

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Explanation:

We are given that

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Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

T=m(g+a)

Where g=9.8 m/s^2

Substitute the values

156=12.8(a+9.8)

9.8+a=\frac{156}{12.8}=12.2

a=12.2-9.8=2.4 m/s^2

Hence, the acceleration =a=2.4 m/s^2

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Answer:

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