The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from
1 answer:
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Answer:
(a)
M = 1.898 x 10^27 kg
(b)
v = 13.74 km/s
(c) E = 0.28 N/kg
Explanation:
Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s
Radius, r = 6.71 x 10^8 m
G = 6.67 x 10^-11 Nm^2/kg^2
(a)
M = 1.898 x 10^27 kg
(b) Let v be the orbital velocity
v = 13739.5 m/s
v = 13.74 km/s
(b) The gravitational field E is given by
E = 0.28 N/kg
Answer:
D) the second at the doorknob
Explanation:
The torque exerted by a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation
is the angle between the direction of the force and d
In this problem, we have:
- Two forces of equal magnitude F
- Both forces are perpendicular to the door, so
- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force
--> therefore, the 2nd force exerts a greater torque