Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.
Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
Answer:
2.877 m/s
Explanation:
According to the laws of conservation of linear momentum,
the momentum of the moving objects before impact is equal to the momentum of the objects after impact (Assuming no external forces were applied)
Let both players are tackled and moving in V velocity
- M and m - masses of the players
- U and u - velocities of them respectively (both velocities are towards east direction )
momentum before impact = momentum after impact
→MU + →mu = →(M+ m )v
91.5 * 2.73 + 63.5 * 3.09 = (91.5 + 63.5) * V
→V = 2.877 m/s (To East)
Answer:
The correct answer would be B.
Explanation:
I believe the answer is B because for electrical energy you would need electric charges to be moved by kinetic energy.
How do you find instantaneous velocity
Select a point on a distance-time curve graph. Draw a tangent to the curve at that point. Tangent -> hypotenuse of right angled triangle. Opp/adjacent in graph units is vel at that point -> in distance and/or time
Answer:
T = mg/6
Explanation:
Draw a free body diagram (see attached). There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.
The center rotates about the point where the cords touch the edge. Sum the torques about that point:
∑τ = Iα
mgr = (1/2 mr² + mr²) α
mgr = 3/2 mr² α
g = 3/2 r α
α = 2g / (3r)
(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)
Now, sum of the forces in the y direction:
∑F = ma
2T − mg = m (-a)
2T − mg = -ma
Since a = αr:
2T − mg = -mαr
Substituting expression for α:
2T − mg = -m (2g / (3r)) r
2T − mg = -2/3 mg
2T = 1/3 mg
T = 1/6 mg
The tension in each cord is mg/6.
