Answer:
given,
mass of copper = 100 g
latent heat of liquid (He) = 2700 J/l
a) change in energy
Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (300 - 4)
Q = 11153.63 J
He required
Q = m L
11153.63 = m × 2700
m = 4.13 kg
b) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (78 - 4)
Q = 2788.41 J
He required
Q = m L
2788.41 = m × 2700
m = 1.033 kg
c) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (20 - 4)
Q = 602.90 J
He required
Q = m L
602.9 = m × 2700
m =0.23 kg
Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
Answer: 9130 joules
Explanation:
Workdone by wheelbarrow = ?
Time = 11 seconds
Power = 830 watts
Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.
i.e Power = (workdone/time)
830 watts = Workdone / 11 seconds
Workdone = 830 watts x 11 seconds
Workdone = 9130 joules
Thus, 9130 joules of work is required to get the wheelbarrow across the yard.
F=ma
For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left
For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left
Therefore, the lighter (5kg) cart experiences a greater acceleration.
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