MCAT Question of the Day

Correct Answer: A. R * C * ln(6/5) * s

You can answer this question quickly by dimensional analysis. The natural log is dimensionless yielding us 2 set of 2 choices. We are looking for time, units of seconds. One electrical value that contains seconds is current.

This leaves answers A and B. You should know with a discharging capacitor at time RC = Tc the value of the discharged capacitor is 37% (e^-1) its initial value. 0.37 != (25/30) leaving answer choice A.
For those who would like to remember the discharge formula its:

All you would need to do is plug in the values and rearrange until you arrive at answer choice A.
300 volts or proportionally 0.4 A/kg is a good estimate of how much charge is needed to restart the heart. You can calculate a lower 10 ohms resistance for the heart, from the required current and voltage.

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Correct Answer: B. 10 m/s

First apply the flow equation: A1v1 = A2v2, where A2 and v2 are the cross section and velocity of the top surface of the water:

A1v1 = A2v2
v1*pi* (1 * 10^-2 m)² = v2*pi*(2m²)

v1 = 4m²/ (1 * 10^-4 m²) * v2
Solving for v1 gives us v1 = (4 * 10⁴ m²/s) v2

Though this does not give us an exact answer yet, it DOES give us important info! That is the velocity of the top surface of the water (v2) is insignificant compared to the velocity of the water coming out of the hole (v1). This allows us to set v2 = 0 in our second equation, Bernoulli’s equation:
p1 + pgy1 + (1/2)pv1² = p2 +pgy2 + (1/2)pv2²
Not only can we set v2 = 0, but we can also recognize that in a 5 meter-high column of water, the pressure at the bottom (p2) and bottom (p1) will be essentially the same, due to the fact that the additional pressure added by 5 meters of water is insignificant compared with the atmospheric pressure at the top (p1). Thus, the pressure terms can also be neglected, and the Bernoulli equation simplifies to give us:

pgy1 + (1/2)pv1² = pgy2
v1² = 2g(y2 – y1)
Solving for v1 gives:

v1 = Sqrt(2(10m/s²)*5m)) = 10 m/s, or answer B.

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Correct Answer: B. 9 m/s

The water will exit at a velocity which is proportional to the height it has fallen, according to the equation v = sqrt(2gh). This can be derived from the basic equation for velocity and displacement, v2 = 2a * s, where a is acceleration, and s is displacement (or in this case height). Thus, the height the water has fallen in this problem is 4 meters, and v = 8.85m/s, which we round up to 9.

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Correct Answer: A. I only

We are looking for something with more energy than UVB radiation. We 1st need to know where UVB radiation lies on the electromagnetic spectrum. Recall that UVB is ultra violet, meaning that it exists just beyond the visible color violet.

What helps me remember the EM spectrum is the last word of each part. We have UV rays and IR waves. The rays are grouped together on one side of visible light, and the waves are grouped together on the other side. Now that we have an order we need a direction. Rays are more energetic than visible light. You can think about how long it takes for an ocean wave to pass and how quickly a sunray can hit you. This leaves only X-rays and gamma rays to having more energy than UVB.

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Correct Answer: D. The collision is inelastic, and the total kinetic energy in the system will be lower after the collision.

An elastic collision is one in which all of the kinetic energy is conserved in the collision. However, elastic collisions do not exist at large scale in the real world, and in this case the collision is not elastic. Instead, the collision is inelastic, meaning some kinetic energy is lost from the impact, usually as heat, sound, or vibration. The kinetic energy is lost as a result of the contact between the two balls, not because of the friction interaction between the grass and the ball, so answer B is not correct.

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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.

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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.

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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.

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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.

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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!

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