MCAT Question of the Day

Correct Answer: C. I and II

Solution: Recall power is energy over time. We will need to know what type of energy is used and then find it. Calculating the amount of energy comes down to remembering the formulas.

The Olympic cyclist exhibits kinetic energy.
[latex]K.E.=\frac{1}{2}mv^2,\ \Rightarrow K.E.=\left(\frac{1}{2}\right)\left(100\right){\left(\frac{200}{10}\right)}^2=20000\ {\rm J,\ }[/latex]
[latex]\Rightarrow {\rm P=}\frac{{\rm E}}{{\rm t}}?\left(\frac{{\rm 20000}}{{\rm 10}}\right){\rm =2000\ Watts}[/latex]

The Olympic weightlifter exhibits potential energy. Potential energy can be thought as the work of the force of weight. Work being distance times force.
[latex]P.E.=mgh,\ \Rightarrow P.E.=\left(200\right)\left(10\right)\left(4\right)=8000\ {\rm J}[/latex]
[latex]\Rightarrow {\rm P=}\frac{{\rm E}}{{\rm t}}?\left(\frac{{\rm 8000}}{{\rm 4}}\right){\rm =2000\ W}[/latex]

The laptop exhibit electrical power. Electrical power is current times voltage.
[latex]P=IV,\ V=IR\Rightarrow P=\frac{V^2}{R}\Rightarrow \frac{{20}^2}{4}=100\ {\rm W}[/latex]

Both Olympic athlete outperform the laptop at 2000 W.

2000 watts is thought to be the maximum power humans can achieve. Olympic level athletes push this limit every Olympic season.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. 400 N

We can set up a free body diagram representing the situation.

To solve this problem you could use the conservation of work and find the required work to lift himself up against gravity. The distance will be the hypotenuse times the sine, soh-cah-toa.

[latex]W=\overrightarrow{F}\cdot \overrightarrow{d},\ \ W_g=W_F\Rightarrow mg\frac{L}{2}sin\theta =F\frac{7L}{8}sin\theta \Rightarrow[/latex]
[latex]F=mg\frac{4}{7}[/latex]

You could also solve this problem with the understanding of torque. The person is virtually a ridged body, with the point of rotation placed at his feet. While he does this pushup he will be in rotational equilibrium. That means the if his butt move up, his head will also move up at the same rate. With that said the torque on his weight must equal the torque he exerts.

[latex]\overrightarrow{\tau }{\rm =}\overrightarrow{{\rm r}}{\rm \times }\overrightarrow{{\rm F}},\ \ {|\overrightarrow{\tau }}_F|=|{\overrightarrow{\tau }}_w|\Rightarrow F\frac{7L}{8}=mg\frac{L}{2}\ \Rightarrow[/latex][latex]F=mg\frac{4}{7}\Rightarrow \left(70\right)\left(10\right)\left(\frac{4}{7}\right)=400\ {\rm N}[/latex]

It is interesting to note that you only lift a little bit more than half your weight when you execute a pushup.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: A. 16 N-s

This question combines two principles, translational motion and impulse. The magnitude of impulse is given by the equation

J = Ft = m(vf-vi)

In this case, we are not given time (t) so to find the magnitude of impulse, we must focus on the change in momentum. The change in momentum is given with the equation

m(vf – vi) = 1 kg * (10 m/s – vi)

vf is the velocity directly after impact with the floor. vi is velocity directly before impact. It is important to note that vi is also the final velocity of the ball in the reference frame of throwing the ball to the ground. Thus,

vf² = vi² +2a?x = 4² + 2(10)(1) = 36

Our velocity right before impact is thus 6 m/s. Because velocity is a vector function, we must remember that this velocity is in the opposite direction of vf. Finally, we return to our previous impulse equation substituting
-6 m/s for vi and we see that the magnitude of impulse (J) is 16 N-s.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: D. ²⁶³Sg -> ²⁵⁹Rf

Alpha decay results in the ejection of an alpha particle from the nucleus. The alpha particle contains two protons and two neutrons. The two protons would decrease the atomic number by two and the total mass number of the atom would decrease by four. The only choice that satisfies both of these requirements is D, ²⁶³Sg -> ²⁵⁹Rf.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. has a fixed pitch but increases in intensity.

This is a question about the Doppler effect. If the source of a sound moves towards the observer, then the perceived frequency will increase. Just like in life, the ambulance seems to get louder as it gets closer (increase intensity), but then makes much less noise after it passes. However, if the ambulance’s speed is constant, its higher pitch will remain constant as well.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: D. 180 J/s

From P = IV, we get P = (1.5A)(120V) = 180 W = 180 J/s. One watt is equal to 1 joule per second. Be sure you understand units in physics!

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. 4.9 m/s²

The acceleration imparted by the force of gravity to an object on africtionless inclined plane is a = g sin (theta). Therefore, if theta = 30, we find: a = g*sin(theta) = 9.8*sin(30) = 4.9 m/s².

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: D. 3400 Hz

Using the equation (lambda)*(f) = v, we find that: f = v/(lambda) = (340 m/s) / (0.1 m) = 3400 Hz.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer:C.Parabolic

If air resistance is neglected, the trajectories of all projectiles fired near the surface of the earth follow a parabolic path because the horizontal velocity is constant while the vertical acceleration is also constant.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. E points to the right.

By convention, electric field vectors always point away from positive source charges and toward negative ones. Thus, in the region between the given point charges, the electric field points to the right: away from the positive and toward the negative.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.