A spring scale shows a net force of 0.8 N acting on a 1.5-kg mass. What happens to the acceleration of the object if the net force is decreased to 0.2 N?(1 point)
The acceleration increases to four times its original value, or about 0.13 m/s2.
The acceleration decreases to a quarter of its original value, or about 0.13 m/s2.
The acceleration increases to four times its original value, or about 2.13 m/s2.
The acceleration decreases to a quarter of its original value, or about 2.13 m/s2.
B) The acceleration decreased to *a quarter* of its original value, or about 0.13 m/s^2.
1. 5 feet
2. The cat began its journey in bed and ended its journey 10 meters to the right of its bed.
3. Distance is a scalar quantity, and displacement is a vector quantity.
4. A downward-sloping line segment.
5. 50 N; the slab will move upward.
6. 98 N
7. 11.34 kg * m/s
8. The acceleration decreases to a quarter of its original value, or about 0.13 m/s2.
9. Segments 2 and 3 have equal periods of time but the force during segment 2 is different than the force during segment 3.
10. Velocity and position
11. Decreasing the mass of the airplane will result in the airplane accelerating more.
12. The is both a positive correlation and causation.
13. The drag on the helicopter is directed downward and to the south.
14. Dynamic equilibrium
15. Between 200,000 and 300,000 km/s
16-18 essay questions
Thanks!!!!
Mj is correct!
Well, if the net force decreases to 0.2 N, the acceleration won't be able to get a promotion. It will actually decrease to a quarter of its original value, or about 0.13 m/s². So, no need to start planning any celebration for the acceleration!
To determine the relationship between net force and acceleration, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass.
The formula for Newton's second law is as follows:
F_net = ma
Where:
F_net is the net force applied to the object (in Newtons)
m is the mass of the object (in kilograms)
a is the acceleration of the object (in meters per second squared)
In the given scenario, with a net force of 0.8 N acting on a 1.5 kg mass, we can rearrange the formula to solve for acceleration:
a = F_net / m
Substituting the given values, we get:
a = 0.8 N / 1.5 kg
a ≈ 0.53 m/s^2
Now, we can determine the new acceleration if the net force is decreased to 0.2 N. Again, using the same formula:
a' = F_net' / m
Substituting the new net force, we get:
a' = 0.2 N / 1.5 kg
a' ≈ 0.13 m/s^2
Comparing the original acceleration (0.53 m/s^2) with the new acceleration (0.13 m/s^2), we see that the acceleration decreases to a quarter of its original value.
Therefore, the correct answer is:
The acceleration decreases to a quarter of its original value, or about 0.13 m/s^2.
F = m*a
0.8 = 1.5a
a = 0.53 m/s^2.
F = m*a
0.2 = 1.5a
a = 0.13 m/s^2
a = 0.13 m/s^2.