Is the variable you change, independent, I, something I change.
- The data for the first part of the experiment support the first hypothesis.
- As the force applied to the cart increased, the acceleration of the cart increased.
- Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.
When we state something about the results on the basis whether the observed data supports the original hypothesis, we say that we are concluding the results.
What is the relationship between force and acceleration based on Newton's 2nd law?
Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
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Answer:
A
Explanation:
A and B are isotopes of one another but the same element
C and D are isotopes of one another but the same element
However, A and B have a different proton count than C and D, indicating different elements because the proton count is equivalent to the atomic number.
By using the second law of Newton, the frictional force is 200N.
We need to know about the second law of Newton (force) to solve this problem. The total force applied an object is proportional to the mass of object and acceleration. It can be defined as
∑F = m . a
where F is force, m is mass and a is acceleration.
From the question above, we know that
F1 = 200N
v = constant therefore (a = 0 m/s²)
By using second law of Newton, we get
∑F = m . a
F1 - Ffriction = m . 0
200 - Ffriction = 0
Ffriction = 200 N
Hence, the frictional force is 200N.
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Thank you for posting your question here at brainly. I would say yes to the above question. <span>Work done is the force applied multiplied by the distance travelled. </span><span>Wd = F x d. </span><span>So if d increases, Wd increases also. I hope the answer will help you. </span>
