A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was
1 answer:
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The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.
m mass
v velocity
h height
g = 9.81m/s²
The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)
m mass
v velocity
h height
g = 9.81m/s²
The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)
Answer:
(a)0.0675 J
(b)0.0675 J
(c)0.0675 J
(d)0.0675 J
(e)-0.0675 J
(f)0.459 m
Explanation:
15g = 0.015 kg
(a) Kinetic energy as it leaves the hand
(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J
(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.
(d) The gravitational energy at peak point would also be the same as 0.0675J
(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J
(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:
mgh = 0.0675
0.015*9.81h = 0.0675
h = 0.459 m
The ball would reach a maximum height of 0.459 m
Answer:
linear charge density = -9.495 × C/m
Explanation:
given data
revolutions per second = 1.80 ×
radius = 1.20 cm
solution
we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force
so
- q × E = m × w² × r ..................1
- 9 × ×
q = m × w² × r ............2
and w =
w =
w = 1.80 × ×
w = 11304000 rad/s
so here from equation 2
- 9 × ×
1.80 ×
= 1.672 ×
× 11304000² × 0.0120
linear charge density = -9.495 × C/m