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The possible magnitude for the force of static friction on the stationary cart is 72.1 N.
The given parameters:
- <em>Applied force on the cart, F = 72.1 N</em>
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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.
F = ma
Static frictional force is the force resisting the motion of an object at rest.

where;
is the frictional force

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.
Learn more about Newton's second law of motion: brainly.com/question/25307325
Answer:
The answer is C. 120,000 J.
Explanation:
Answer:
Diameter of Newton’s 5th ring = 0.30 cm
Diameter of Newton’s 15th ring = 0.62 cm
Diameter of Newton’s 25th ring = ?
From Newton’s rings experiment we infer that
D2n+m − D2n = 4λmR
For the 5th and 15th rings we have
D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)
For 15th and 25th rings
D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)
We equate the two derivatives
Equation (2) = Equation (1)
D225 − D215 = D215 − D25
D225 = 2D215 – D25
Substituting the values into the equation
D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2
D25 = 0.8239 cm
I believe the answer is A