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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 :

GinnyAnsweridn
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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