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Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop. She drops a wooden block with a mass of 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. What's the acceleration of the wooden block when it hits the sensor? Use [tex]F = ma[/tex].

A. [tex]2.45 \, \text{m/s}^2[/tex]
B. [tex]4.4 \, \text{m/s}^2[/tex]
C. [tex]5.4 \, \text{m/s}^2[/tex]
D. [tex]9.8 \, \text{m/s}^2[/tex]


Sagot :

To determine the acceleration of the wooden block when it hits the sensor, we start by utilizing the formula related to Newton's Second Law of Motion, which is represented by:

[tex]\[ F = m \times a \][/tex]

Here:
- [tex]\( F \)[/tex] is the force,
- [tex]\( m \)[/tex] is the mass, and
- [tex]\( a \)[/tex] is the acceleration.

In this problem:
- [tex]\( F = 4.9 \)[/tex] newtons,
- [tex]\( m = 0.5 \)[/tex] kilograms.

We need to solve for the acceleration [tex]\( a \)[/tex]. Rearranging the formula to solve for [tex]\( a \)[/tex] gives us:

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

Substituting the given values into the equation:

[tex]\[ a = \frac{4.9 \, \text{N}}{0.5 \, \text{kg}} \][/tex]

Now, we perform the division:

[tex]\[ a = 9.8 \, \text{m/s}^2 \][/tex]

Therefore, the acceleration of the wooden block when it hits the sensor is:

[tex]\[ \boxed{9.8 \, \text{m/s}^2} \][/tex]

The correct answer is [tex]\( \text{D} \, 9.8 \, \text{m/s}^2 \)[/tex].
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