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Zina [86]
2 years ago
13

A particular roller coaster has a mass of 3500 kg, a height of 4.0 m, and a velocity of 12 m/s. What is the kinetic energy? If n

eeded, use g= 10. m/s^2
Physics
1 answer:
Alenkinab [10]2 years ago
8 0

Explanation:

given

m=3500kg,h=4.0m,v=12m/s

ke=½mv²

½×3500×12²=3500×144÷2

=252,000j or 252j

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If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
  • force to move the object: F

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax
alexdok [17]

Answer:

Explanation:

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Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

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= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

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v = 36.25 m / s .

6 0
3 years ago
What is the momentum of a 0.1-kg mass moving with a speed of 5 m/s
Andrej [43]
This question is wrong because in momontum we will write acceleration instead of speed.     suppose acceleration is 5m/s2 then 
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