# Homework Help: Three Physics Problems

1. Jun 27, 2005

### GJBenn85

Problem 1:

I need you to check the first problem. The answer makes me think I may have done something wrong.

1) A quantity of water equal to 2.5 x 10^10 cc falls from a rain cloud that is 1000 m high. If this rain falls in just 1 hour, how much power is generated?

The formula I used was p = F x d/t and my answer ended up being 66,666,666.67 N*m/sec, or 66,666,666.67 watts.

Problem 2:

2) The mass of an electron is 9 x 10^-31 kg. The mass of a proton is 1.7 x 10^-27 kg. The proton and electron are about 1.0 x 10^-10 m apart in a hydrogen atom. What force of gravitation exists between the proton and the electron of the hydrogen atom? What is the CENTRIPETAL acceleration of the electron?

Using Gm1m2/r2, I figured the force of gravitation to be 1.02 x 10^-47 N. However, I have no idea how to calculate the CENTRIPETAL acceleration. Also, am I right on the force of gravitation?

Problem 3:

3) The end of the lever of a tire jack travels 7 m for every centimeter that it lifts the car. If the car has a mass of 1000 kg and a force of 50 N is needed to lift the car, what is the efficiency of the jack?

Here is what I have so far and am stuck:

work done to lift car = F x d (work output)
(1000 kg)(9.8 N/kg) = 9800 N
winput = F2(pie)r
(50N)(2)(3.14)(3.5m) = 1099 N*m
woutput = 9800 N x ... ???

2. Jun 27, 2005

### whozum

Problem 1:

Power is the work done divided by the time interval it took to get the work done (1 hr), how much work is done on the raindrops to get them to fall? You can check this one yourself.

Problem 2:

The expression for centripetal acceleration is

$$a = \frac{mv^2}{r}$$ as it is for any centripetally accelerating body.

Problem 3:

Efficiency is the ratio of output to input work. To lift the car 1cm you must do $mg\Delta x [/tex] J of work which is [itex] 1000*9.8*0.01$ J. How much work do you do on the jack to lift the car 1 centimeter?

3. Jun 29, 2005

### GJBenn85

You are just confusing me even more.

Problem 2:

Without looking it up, I assume the formula a = mv^2/r is such that m = mass, v = velocity, and r = radius. Where in the problem do I find the radius and velocity?!

Problem 3:

I "think" I have already figured half of the information needed to determine the efficiency of the jack. Guidance would be more helpful than another question.

4. Jun 29, 2005

### Astronuc

Staff Emeritus
Centripetal acceleration is $$a = \frac{v^2}{r}$$

Centripetal force is $$f = \frac{mv^2}{r}$$

F = m a

5. Jun 29, 2005

### GJBenn85

Back to problem 1:

Is the answer to problem one 68,055,555 watts? Here is my process:

2.5 x 10^10 cc = 2.5 x 10^10 g = 25,000,000 kg * 9.8 N/kg = 245,000,000 N

(245,000,000N)(1,000m)/(3600 seconds) = 68,055,555 N*m/sec = 68,055,555 watts

Can someone please just tell me if I am right or wrong? I know the formula but it is my conversion of the 2.5 x 10^10 cc into newtons that I am unsure of. My conversion is based on a syllabus example but it was 3,000 cc instead of 2.5 x 10^10 cc.

Last edited: Jun 29, 2005
6. Jun 29, 2005

### Staff: Mentor

Looks OK to me. I suggest rounding off your answer to a reasonable number of significant figures.

7. Jun 29, 2005

### GJBenn85

I still do not have a clue how to calculate the centripetal acceleration for problem 2 so moving onto problem 3:

The end of the lever of a tire jack travels 7 m for every centimeter that it lifts the car. If the car has a mass of 1000 kg and a force of 50 N is needed to lift the car, what is the efficiency of the jack?

While the problem does not say, I assume 7 m is the circumference and I need to find the radius. Using the radius of 1.11xxx, I come up with a work output of 14Nm and a work input of 350 Nm, producing an efficiency of 4%. The efficiency seems to low so I am wondering if when calculating the work output, do I use .01 m for the distance or do I use .01m/7? Using .01m/7 for the distance gives an efficiency of 28% which seems more reasonable.

Problems 2 and 3 are the last to be done on this assignment and I can submit it. Maybe I am trying to read into the questions too much but they are driving me nuts!

8. Jun 29, 2005

### Staff: Mentor

Can you find the acceleration of the electron? (What direction does that acceleration point?)

Imagine the car is raised by 0.01m. What's the output work required to lift the car? Correspondingly, the end of the jack moves 7m. How much input work is required to move the jack that far? Compare.

9. Jun 29, 2005

### GJBenn85

Since the formula for centripetal acceleration requires a radius, do I use the distance between the proton and electron as the radius? What do I use for "v"?

Using only 0.01m as the distance, the output is 98 N*m. Input is 350 N*m if I am correct in using a radius of 1.11xxxx to find the work input. So the resulting efficiency would be 28%?

10. Jun 29, 2005

### Staff: Mentor

One more time:
(a) Can you find the acceleration of the electron? (F = ma)
(b) What is the direction of that acceleration?
(c) What does the word "centripetal" mean?

Sounds good to me. (No need to compute the radius. You are told that the end of the jack moves 7m and are given the force.)

11. Jun 29, 2005

### GJBenn85

F = ma, so do I find the acceleration of the electron by dividing the force of gravitation by the mass of the electron?

12. Jun 29, 2005

### Staff: Mentor

I just realized that the problem asks for gravitational force. Are you sure it doesn't ask for electrical force? (The gravitational force is negligible compared to the electrical force.)

In any case, the net force on the electron divided by its mass will give its acceleration.

13. Jun 29, 2005

### whozum

So sorry about mixing up centripetal acceleration and force. Should be more careful wth what I type.

14. Jun 29, 2005

### GJBenn85

The problem asks me to find the gravitational force and the centripetal acceleration.

Does an acceleration of 1.13 x 10^-17 sound right?

Last edited: Jun 29, 2005