How fast will an object with a mass of #24 kg# accelerate if a force # 12 N# is constantly applied to it?

Answer 1

#"0.5 m/s"^2#

based on Newton's Second Law

#"a" = "F"/"m" = "12 N"/"24 kg" = "0.5 m/s"^2#
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Answer 2

To find the acceleration of the object, use Newton's second law of motion, which states that force equals mass times acceleration (F = ma).

Given: Mass (m) = 24 kg Force (F) = 12 N

Using the formula F = ma, solve for acceleration (a):

a = F / m a = 12 N / 24 kg a = 0.5 m/s²

So, the object will accelerate at a rate of 0.5 meters per second squared when a force of 12 Newtons is applied to it constantly.

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Answer 3

The acceleration of an object can be calculated using Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for calculating acceleration is:

[ a = \frac{F}{m} ]

Where:

  • ( a ) is the acceleration,
  • ( F ) is the force acting on the object, and
  • ( m ) is the mass of the object.

Substituting the given values into the formula:

[ a = \frac{12 \text{ N}}{24 \text{ kg}} ]

[ a = \frac{1}{2} \text{ m/s}^2 ]

So, the object with a mass of 24 kg will accelerate at a rate of ( \frac{1}{2} ) m/s² when a force of 12 N is constantly applied to it.

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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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