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kirill115 [55]
3 years ago
10

If an object with more mass is pushed with the same force as an object with less mass the object with more mass will accelerate?

Quickly? At the same rate? Or slowly?
Physics
2 answers:
Bezzdna [24]3 years ago
6 0

Now really ! You can do this one in your head.

Imagine ...

-- First, you push on a golf ball with one pound of force, and you watch to see how it picks up speed.

-- Then, you push on a school bus with one pound of force, and you watch to see how it picks up speed.

Now please, in your mind, which one speeds up easier ?

And don't tell me it's because one is white and the other one is yellow. It's the MASS !

expeople1 [14]3 years ago
5 0

Answer:

smaller acceleration, so lower change in velocity

Explanation:

To answer this question we examine the equation that relates mass with force and with acceleration: F=m*a.

Since we want to know what happens to the acceleration, we solve for it in the equation: a=\frac{F}{m}

Notice that we are asked what happens when the force applied is the same, but now it is applied in an object with more mass (M).

We therefore would have to compare our initial form:

a=\frac{F}{m} with the new one: a=\frac{F}{M} wher the denominator is a larger quantity, therefore making our division/quotient smaller. Then, we conclude that the acceleration will be smaller, and therefore the change in velocity of the object will be lower.

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A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

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a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

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b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

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    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
This is a measure of the quantity of matter.
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The measure of the quantity of matter would be mass. Mass is measured in kilograms. I hope this helped!:)

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