A waterskier is being pulled at a constant velocity of 15.8m/s. The drag force on the boat is 280N.  What is the foce applied to the boat by the propellor?

I don't even know where to start with this one. help please.

On mars, the acceleration due to gravity is 3.71m/s2. What would a rock's velocity be 0.5 sec. after you dropped it on mars?

What is the formula for this problem

Dear Professor Cram:

An executive flew his corporate jet to a meeting in a city 1500 kilometers away. After traveling the same amount of time on return flight, the pilot mentioned that they still had 300 km to go. The air speed of the plane was 600 km/hr for both legs. Assuming that the wind direction was parallel to the flight path and constant all day, how fast was the wind blowing?

Ekta, Middlesex Community College

Ekta, thank you for this physics question involving vectors. Let's see how we can tackle it.

I'm assuming that the constant 600 km/hr was the instrument output of the plane, not the actual as-observed-from-the-ground speed. Thus, the trip out was clocked at 600 km/hr plus the tailwind, while the return trip was at 600 km/hr minus the tail wind.

In this problem, the actual speed of the plane is a vector combination of plane speed and wind speed. On the first trip, both speeds are working together so we have a vector sum; on the return trip, the vectors are working against each other. (Reality check -- because the plane made it there and partway back, that means the plane speed vector is larger than the wind speed vector.)

According to the question, during the first leg the travel time can be expressed as:

• time = distance/speed, or
• time = 1500/(600 + x)

where x is the speed of the wind. On the return leg, the same amount of time elapsed but the plane still had 300 km to go.

• time = distance/speed, or
• time = (1500-300)/(600 - x), or
• time = 1200/(600 - x)

Since these two times are equal, we can set both equations equal to each other and solve for x:

• 1500/(600 + x) = 1200/(600 - x), or
• (1500)(600 - x) = (1200)(600 + x), or
• 900,000 - 1500x = 720,000 + 1200x, or
• 900,000 - 720,000 = 1200x + 1500x, or
• 180,000 = 2700x, or
• x = 66 2/3

The wind was blowing at 66 2/3 km/hr.

Dear Professor Cram:

I have a question about series circuits. There are 3 resistors, R1 is 10kohms/.5W and R2 is 3.3kohms/.250W and the third is unknown. If the circuit draws a current of 1.5mA from a 27V source, how do i calculate the ohmic value of the third resistor?

Nicholas, Mohawk College

Thank you for your question, Nicholas. Physics is not normally my forte, but I can help you with this one.

Our first step is to determine the power and resistance for the circuit as a whole. We are given the power in volts and current in milliamps, so to find the power we should convert milliamps to amps:

• 1.5mA x 1A/1000mA = 0.0015A

With the current now expressed in Amps, we can calculate the power for this circuit:

• power = current x volts, or
• power = 0.0015A x 27V, or
• power = 0.0405W

Now let's find the total resistance in the series circuit by using Ohm's law:

• resistance = volts/current, or
• resistance = 27V/0.0015A, or
• resistance = 18,000 ohms or 18kohms

Now comes the fun part. The resistance values given in the question are not the resistance in this circuit but resistance ratings based on wattage, as evidenced by the ohms/W units. To find their resistance in this circuit, we'll need to convert these values to ohms/0.0405W as our next step:

• r1/0.0405W = 10000ohms/.5W
• r1 = (10,000ohms)(.0405W)/(.5W), or 810 ohms

• r2/0.0405W = 3300ohms/.25W
• r2 = (3,300ohms)(.0405W)/(.25W), or 534.6 ohms

Now for the big finish. We know the total resistance of the series circuit (18,000 ohms). We know the resistance of two of the three resistors. We know that in a series circuit, the total resistance is the sum of all the resistors in series. Thus, the third resistor is:

• r3 = 18,000 ohms - (810 ohms + 534.6 ohms), or
• r3 = 16,655.4 ohms

I hope this helps. Let us know if you need anything else.

Dear Professor Cram:

I'm not really sure how to tackle a problem like this... An inclined plane rests at a 37 degree angle with a box weighing 50N on it as a vector down. I'm supposed to find components of the force parallel and perpendicular to the plane. What I've come up with is to find the sides of an imaginary triangle using the force vector vs the angle (x = 50(tan(37)) from which I can find the hypotenuse, but I'm not really sure where to go from here. I could really use some pointers on this one :)

Jeff S.

Thank you for your excellent physics question. The inclined plane has been the bane of many a physics student, but I'll try to help you see your way through this problem.

You had a good start but fumbled a bit. (Sorry, watched alot of football this past weekend...) The inclined plane is a triangle with an acute angle of 37° and a vertical side of 50N.

The question is asking for the components of force parallel and perpendicular to the plane; because the object is being pushed up the inclined plane, the 50N force down must be counterbalanced by an equal force upwards. Thus, the perpendicular force is 50N (up).

For the parallel force, you are correct in wanting to use the tangent to calculate a side of the triangle. Remember how the tangent of an angle is the opposite side over the adjacent side? Well, the opposite side to the 37° angle is 50N, while the adjacent side is the parallel force we're looking for. Solving the tangent equation for the adjacent side:

• tan 37° = 50N/Adjacent, or
• Adjacent = 50N/(tan 37°), or
• Adjacent = 66.35N

I hope this helps. Let us know if you need anything else.