How fast will an object with a mass of #8 kg# accelerate if a force of #63 N# is constantly applied to it?

Answer 1

#7.875ms^(-2)#

Using Newton's second Law

#F=ma#
#F=63N#the force applied
#m=8kg#the mass
#a=#acceleration

in this case

63=8a#

#:.a=63/8=7.875ms^(-2)#
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Answer 2

With my chosen interpretation of the question, the acceleration is #7.9 m/s^2#.

If the net force on this object is 63 N, it is a simple application of #F_"net"=m*a#. But, is that 63 N one of 2 or more forces on this object?
It could be that the direction of this force is directly up and that the 8 kg object is in on Earth where #g=-9.8 m/s^2#. In that situation, we would find that the object is heavier than 63 N and it would therefore accelerate downwards more slowly than in free-fall.

However, I will answer your question in the most straightforward way possible: there is no friction and a 63 N force is applied horizontally.

#F_"net"=m*a#.
#63 N = 8 kg*a#
#a = (63 N)/(8 kg) = 7.9 N/(kg) = 7.9 m/s^2#

Hope this is helpful, Steve.

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Answer 3

The acceleration of the object can be calculated using Newton's second law of motion: ( F = ma ), where ( F = 63 ) N (force) and ( m = 8 ) kg (mass). Rearranging the formula to solve for acceleration, ( a = \frac{F}{m} ), substituting the given values, yields the acceleration of the object.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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