Question
A force of 53 N is applied to an 11 kg object. What is the object’s acceleration?
*
1 point
583 N
4.8 m/s2
21 m/s2
*
1 point
583 N
4.8 m/s2
21 m/s2
Answers
GPT-4o mini
To find the acceleration of the object, we can use Newton's second law of motion, which states that:
\[ F = ma \]
where:
- \( F \) is the force applied (in Newtons),
- \( m \) is the mass of the object (in kilograms),
- \( a \) is the acceleration (in meters per second squared).
We can rearrange the formula to solve for acceleration:
\[ a = \frac{F}{m} \]
Given:
- \( F = 53 \, \text{N} \)
- \( m = 11 \, \text{kg} \)
Now, substituting the values:
\[ a = \frac{53 \, \text{N}}{11 \, \text{kg}} \]
\[ a = 4.8181 \, \text{m/s}^2 \]
Rounding it off to one decimal place, the acceleration is approximately:
\[ a \approx 4.8 \, \text{m/s}^2 \]
So the answer is **4.8 m/s²**.
\[ F = ma \]
where:
- \( F \) is the force applied (in Newtons),
- \( m \) is the mass of the object (in kilograms),
- \( a \) is the acceleration (in meters per second squared).
We can rearrange the formula to solve for acceleration:
\[ a = \frac{F}{m} \]
Given:
- \( F = 53 \, \text{N} \)
- \( m = 11 \, \text{kg} \)
Now, substituting the values:
\[ a = \frac{53 \, \text{N}}{11 \, \text{kg}} \]
\[ a = 4.8181 \, \text{m/s}^2 \]
Rounding it off to one decimal place, the acceleration is approximately:
\[ a \approx 4.8 \, \text{m/s}^2 \]
So the answer is **4.8 m/s²**.