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Inessa05 [86]
3 years ago
8

You are helping two friends from our class with a physics problem where a cart is pushed up a ramp. In examining the motion of t

he cart up the ramp, one friend says that the acceleration has to be negative because the cart is slowing down. The other friend says the acceleration can be positive or negative, but it depends on the motion detector. What do you think and what would you say to your friends to convince them of your point of view

Physics
2 answers:
stiks02 [169]3 years ago
8 0

Answer: Acceleration will have 2 components, vertical and horizontal.

Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.

Net-horizontal acceleration will  be equal to the horizontal component of the applied acceleration.

Explanation:

Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.

Scorpion4ik [409]3 years ago
3 0

Answer and Explanation:

The acceleration can either be positive or negative depending on the direction of the net force applied on the cart.

  • If the cart is moving up the ramp and the net force is also in the upward direction then the cart ascends with [positive acceleration.
  • If the cart is moving up the ramp and the net applied force is in the direction opposite to the motion of the cart then the cart decelerates, i.e., the acceleration of the cart is negative and the speed slows down.
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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
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Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

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3 years ago
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