At what height does a 1000-kg mass have potential energy of 1J relative to the ground?
1 answer:
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Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
Answer:
a)η = 69.18 %
b)W= 1210 J
c)P=3967.21 W
Explanation:
Given that
Q₁ = 1749 J
Q₂ = 539 J
From first law of thermodynamics
Q₁ = Q₂ +W
W=Work out put
Q₂=Heat rejected to the cold reservoir
Q₁ =heat absorb by hot reservoir
W= Q₁- Q₂
W= 1210 J
The efficiency given as
η = 69.18 %
We know that rate of work done is known as power
P=3967.21 W
Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
=
So the lines terminating at - 2 micro coulomb
=
So, the number of filed lines terminates at - 2 micro Coulomb are 6.