Hello can some please guide me through these? I don;t think i\'m understanding t

Question

Hello can some please guide me through these? I don;t think i'm understanding the concept. I appreciate it much! I actually really want to learn. I'd like to see the works to be showed! Thank you!

A rock of mass m1 = 0.58 kg is tied to another rock with a mass m2 = 0.63 kg with a string of length L1 = 0.13 m. The rock m2 is tied to another string of length L2 = 0.22 m, and the pair of rocks is slung around in a circle, making two complete revolutions in one second. In this problem, you should neglect gravity and assume the motion is in the horizontal plane.

a) What is the tension in the string connecting the two rocks?

b) What is the tension in the other string?

1)  A planet has twice the mass of the earth but the same radius. What is the acceleration of gravity on the surface of this planet?

the same as on earth

half as much as on earth

one-forth as much as on earth

twice as much as on earth

four times as much as on earth

** Since the A= V^2/r would the mass not matter? so the accerlation could be the same?

2)  A planet has twice the radius of the earth but the same mass. What is the acceleration of gravity on the surface of this planet?

the same as on earth

half as much as on earth

one-forth as much as on earth

twice as much as on earth

four times as much as on earth

** WOuld this be half as much??

A satellite orbits the earth at a distance equal to twice the earth's radius above the surface of the earth (the total distance from the center of the earth is three times the radius of the earth).

3)  The satellite has a mass of 60 kg. What is the force of attraction between the earth and the satellite?

588 N

318 N

65.3 N

21.0 N

7.26 N

4)  What is the centripetal acceleration of the satellite?

88.2 m/s2

29.4 m/s2

9.80 m/s2

3.27 m/s2

1.09 m/s2

5)  With what speed must the satellite move to follow a circular orbit?

4565 m/s

1987 m/s

567 m/s

222 m/s

18 m/s

6)  If the mass of the satellite was increased, what would need to happen to the speed to keep the same radius of orbit?

increase

decrease

remain the same

A block is swung around in a vertical circle.

7)  If the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in the string at the top of the circle?

2.9 N

3.6 N

4.8 N

6.1 N

7.7 N

8)  Is the tension greater or less or equal at the bottom of the circle?

greater at the bottom

less at the bottom

the same at the bottom

9)  If the velocity was increased, the tension at the top would:

increase

decrease

remain the same

10)  If the radius was increased, the tension would:

increase

decrease

remain the same

A car drives on a road which begins flat, then goes up and over the top of a hill and then goes down into a dip and back up to level ground.

11)  Compare the magnitude of the normal force acting on the car at each point:

Nflat > Ntop of hill > Nbottom of dip

Nflat > Nbottom of dip > Ntop of hill

Nbottom of dip > Ntop of hill > Nflat

Nbottom of dip > Nflat > Ntop of hill

Ntop of hill > Nbottom of dip > Nflat

Ntop of hill > Nflat > Nbottom of dip

12)  If the car is moving at 30 m/s at the bottom of the dip, its mass is 800 kg, and the radius of curvature for the dip is 20 m, what is the normal force acting on the car in the dip.

10459 N

19344 N

28625 N

37190 N

43840 N

13)  What is the acceleration of the car at the bottom of the dip?

0 m/s2

9.8 m/s2

35.2 m/s2

45.0 m/s2

54.8 m/s2

14)  What is the minimum speed at the top of the hill (assuming R = 20 m here, too) for the car to leave the road?

10 m/s

12 m/s

14 m/s

16 m/s

18 m/s

15)  What is the acceleration of the car if it just begins to leave the road?

0 m/s2

4.9 m/s2

9.8 m/s2

14.7 m/s2

19.6 m/s2

Explanation / Answer

1) twice as much as on earth
Beacuse,

acceleration due to gravity on earth, g = G*Me/Re^2

acceleration due to gravity on a planet, g_planet = G*M/R^2

= G*2*Me/Re^2

= 2*(G*Me/Re^2)

= 2*g

2) half as much as on earth

3) 65.3 N

at the given height, g = G*Me/(3*Re)^2

= g/9

= 9.8/9

= 1.09 m/s^2

so, F = m*g

= 60*1.089
= 65.3 N

4) 1.09 m/s^2

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