I believe it’s A, i could be wrong tho 3
Acceleration is the rate of change of velocity
To define acceleration, We need to know more about motion.
Motion: This can be defined as the change in position of a body from one point to another. When an object accelerates, it undergoes motion.
<u>Definition</u>
Acceleration can be defined as the rate of change of velocity. The S.I unit of acceleration is meter-per-squared seconds. (m/s²)
The formula of acceleration is
- a = (v-u)/t................. Equation 1
⇒ Where:
- a = acceleration
- u = initial velocity
- v = final velocity
- t = time
Hence, Acceleration is the rate of change of velocity
Learn more about acceleration here: brainly.com/question/605631
Answer:
68.8 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of box = 18 Kg
Coefficient of friction (μ) = 0.39
Force of friction (F) =?
Next, we shall determine the normal force of the box. This is illustrated below:
Mass (m) of object = 18 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal force (N) =?
N = mg
N = 18 × 9.8
N = 176.4 N
Finally, we shall determine the force of friction experienced by the object. This is illustrated below:
Coefficient of friction (μ) = 0.39
Normal force (N) = 176.4 N
Force of friction (F) =?
F = μN
F = 0.39 × 176.4
F = 68.796 ≈ 68.8 N
Thus, the box experience a frictional force of 68.8 N.
The force result in stretching the spring 10.0 centimeters is 2.5N.
<h3>
What is Hooke's law?</h3>
If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
Learn more about hooke's law.
brainly.com/question/13348278
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