MCAT Question of the Day

Correct Answer: D. 25 m/s

F = ma
a = F/m = (10 N/ 2.0kg) = 5.0 m/s²

V = V₀ + at
V = at
V = (5 m/s²)(5.0 sec) = 25 m/s

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Correct Answer: D. 125 m

The initial horizontal speed is irrelevant. Use this equation: x = (1/2)g(t)² = (1/2)(10)(5)² = 125 m.

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Correct Answer: B. 10 m.

All of the kinetic energy will be transferred into potential energy at the maximum height of the cannonball. The trick is to not replace KE with (1/2)mv², the kinetic energy is already given! Therefore:

KE -> PE -> KE = PE -> KE = mgh -> h = KE/mg = 500 J / (5 kg)(10 N/kg) = 10 m.

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Correct Answer: D. halved.

The centripetal force of the car is mv²/r, where m is the car’s mass, v is the velocity, and r is the radius of the circular track. If we double the radius while keeping everything else uniform, the centripetal force will be halved because they are inversely proportional.

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Correct Answer: A. 20 N

There are two forces acting on the string. The buoyant force of the ball upward, the tension in the string downward, and the mass of the ball. Since the ball is stationary, these forces must cancel. Thus, T + w = B, which implies:

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Correct Answer: A. 10^-5

Therefore:

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Correct Answer: C. 10 mi*h^-1*s^-1

The formula for acceleration is velocity divided by time:

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Correct Answer: C. 6000 N-m

Assuming that the water stream is tangent to the wheel, it will be perpendicular to the radius vector at the point of contact. Therefore, the torque of the force of the water stream is Torque = r*F = (20m)(300 N) = 6000 N-m.

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Correct Answer: A. N decreases as x increases, but not proportionally.

The normal force is the force that pushes up against on object. For an inclined plane, the formula for normal force is N = m*g*cos(x). The mass of the sled and gravity stay the same, so N must decrease as x increases due to the cosine function. Try 30 degrees for x and then 60 degrees for x in a calculator, you’ll see that the normal force decreases, but does not decrease by a factor of 2.

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Correct Answer: A. Electric potential will only flow from a high potential to a lower potential.

Electric current moves in the direction positive charges move in. Therefore, the electric flow of current flows from the point of higher potential to lower potential, similar to how water falls from the top of the waterfall to the bottom.

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