A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg,
1 answer:
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Answer:
The complete question contains 2 parts such that
part a: Where the probe is submerged in water. In this case the wattage of heater is 90.47 W
part b: Where the probe is on a vessel. In this case the wattage of heater is So 121.34 W
Explanation:
Part a
For the first part when it is assumed that the probe is submerged in the water. In this case the heat loss is only due to the convective heat transfer which is given as
Here
- Q is the heat loss which is to be calculated
- A is the area of the body which is given as
. It is calculated as
- Tbody is 10 C
- Tsurr is 0C
- h is the convection coefficient of water which is 18 W/m^2K
So the wattage of heater is 90.47 W
Part b
Now when the probe is above sea water now, the losses are both because of convection and radiation now the loss is given as
Here
- Q_con is the heat loss due to convection which is to be calculated
- A is the area of the body which is given as 0.5026 m^2
- Tbody is 10 C
- Tair is -10C
- h is the convection coefficient of water which is 10 W/m^2K
Radiation loss is given as
Here
- Q_rad is the heat loss due to radiation which is to be calculated
- A is the area of the body which is given as 0.5026 m^2
- Tbody is 10+273=283K
- Tsky is 0+273=273K
- ε is the emissivity of the shell which is 0.85
- τ is the coefficient which is
So the total value of heat loss is
So the wattage of heater is 121.34 W
Answer:
The velocity of the cart with respect to the ground is
if we consider North the positive y-direction and East the positive x-direction.
Explanation:
We have for relative motion the following expression:
Where is the velocity of the cart with respect to the ground,
is the velocity of the cart with respect to the plane and
is the velocity of the plane with respect to the ground.
We find that:
Thus: