Answer:
Grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules = 54.3 g
Explanation:
Heat required to increase temperature
H = mcΔT
m = mass of material
C = specific heat of material
ΔT = Change in temperature.
Here we need to find how many grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules
That is
H = 12500 J
Specific heat of water, C = 4186 J/kg°C
ΔT = 75 - 20 = 55
Substituting
12500 = m x 4186 x 55
m = 0.0543 kg
m = 54.3 g
Grams of water can be heated from 20.0°C to 75°C using 12500.0 Joules = 54.3 g
Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
You are correct it is translucent because opaque let’s no light through but transparent lets almost all light through
Answer:
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Explanation: