A body of mass m has weight
F = GMm/r²
on the surface of the Earth, where G is the universal gravitational constant, M is the mass of the Earth, and r is it's radius.
If the weight is to be halved, then we have
1/2 F = 1/2 GMm/r² = (1/√2)² GMm/r² = GMm/(√2 r²)
so the distance between the body and the planet's center needs to be
√2 × 6.4 × 10⁶ m ≈ 9.1 × 10⁶ m
Answer:
a. Ssystem > 40 J/K
Explanation:
Given that
The entropy of first block = 10 J/K
The entropy of second block = 30 J/K
When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40 J/K.
Therefore the answer is a.
Ssystem > 40 J/K
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
Answer:
increase, GPE, KE, KE, KE, PE, PE, KE, KE,
Explanation:
