What Are the Correct Responses to These Basic Physics Questions?

[salman19753]

Homework Statement

1. The acceleration of an object undergoing uniform circular motion is constant in
a. magnitude only b. direction only
c. both magnitude and direction d. neither magnitude nor direction

2. The centripetal force on a car rounding a curve on a level road is provided by
a. gravity b. friction between its tires and the road
c. the torque applied to its steer d. its brakes



3. The radius of the path of an object in uniform circular motion is doubled. The centripetal force needed if its speed remains the same is
a. half as great as before b. the same as before
c. twice as great as before d. four times as great as before




4. The centripetal force that keeps a stone attached to a string in a vertical circle is given by:
a. the tension of the string
b. the centrifugal force pulling down
c. the tension plus the weight of the stone
d. the tension minus the weight of the stone




5. The centripetal force needed to keep the Earth in orbit is provided by
a. inertia b. its rotation on its axis
c. the gravitational pull of the sun d. the gravitational pull of the moon







6. The gravitational acceleration ‘g’ of an object
a. has the same value everywhere in space
b. has the same value everywhere on the earth’s surface
c. varies somewhat over the earth’s surface
d. is greater on the moon because of its smaller diameter





7. If the Earth were three times farther from the sun than it is now, the gravitational force exerted on it by the sun would be
a. three times as large as it is now b. nine times as large as it is now
c. one-third as large as it is now d. one-ninth as large as it is now




8. Newton realized that:
a. the force between two objects is inversely proportional to the distance between them
b. the force between two objects is inversely proportional to the square of the distance between them
c. the force between two objects is directly proportional to the distance between them
d. the force between two objects is directly proportional to the square of the distance between them.


9. Two masses, m1 and m2, are separated by a distance r. The force of attraction between the two masses is FG = 150 N. If m1 is tripled, what is the new FG?
a. 150 N b. 300 N
c. 75 N d. 450 N


10. Two masses, m1 and m2, are separated by a distance r. The force of attraction between the two masses is FG = 150 N. If r was halved and both masses doubled, what is the new FG?
a. 2400 N b. 1200 N
c. 150 N d. 450 N





My attempts at answers:
1. a
2. c
3. a
4. a
5. c
6. b
7. d
8. b
9. d
10. a

 

[Yenaled]

#2 - Think again. Think about: if this force were absent, the car wouldn't be able to round the curve.

#6 - The Earth is not a perfect sphere.

#10 - [Edit: Nevermind, the answer you supplied is correct.]

 

[salman19753]

#2 - Think again. Think about: if this force were absent, the car wouldn't be able to round the curve.

#6 - The Earth is not a perfect sphere.

#10 - First of all, you didn't factor in the fact that the r is halved (look at question 7). Secondly, how does FG depend on each mass? (look at question 9)

#2- I believe that it would be the friction between the tires and the road because without the friction, you would just continue to go straight.

#6- I believe that it would vary on some places in Earth, because of the different distances from the Earth's center

#10- I believe that it will be 1200 N because I forgot to take into account halving it.

By the way Thank You So Much for your help!

 

[Yenaled]

#2 - Yes!

#6 - Correct.

#10 - [sorry, this is correct]

 

[salman19753]

#2 - Yes!

#6 - Correct.

#10 - No. Remember the inverse square law: when the distance is halved, the force is reduced by a factor of 4. Also, remember that force is proportional to mass, not the square of mass. When a mass is doubled, the force is doubled. When two masses are each doubled, what happens to the force?

Does the force stay the same? At 150 N?

 

[Yenaled]

When both masses are doubled, the force is quadrupled.

 

[Yenaled]

My apologies -- the original answer was correct.

When the radius is halved, the force increases by a factor of 4. And when both masses are doubled, the force is quadrupled. So the force ultimately increases by a factor of 16.

 

[salman19753]

My apologies -- the original answer was correct.

When the radius is halved, the force increases by a factor of 4. And when both masses are doubled, the force is quadrupled. So the force ultimately increases by a factor of 16.

Okay so the original 2400 N is correct right?

 

[Yenaled]

Yes. Sorry for the confusion.