Answer:
4370.83044 m²
Explanation:
= Net work done = 570 kW
= Low temperature resrvoir = 20 °C
A = Area of solar collector
Amount of solar heat received =
= 0.315A
![\eta_{th}=\eta_{Carnot}=1-\frac{T_l}{T_h}\\\Rightarrow \eta_{th}=1-\frac{20+273}{500}\\\Rightarrow \eta_{th}=1-0.586\\\Rightarrow \eta_{th}=0.414](https://tex.z-dn.net/?f=%5Ceta_%7Bth%7D%3D%5Ceta_%7BCarnot%7D%3D1-%5Cfrac%7BT_l%7D%7BT_h%7D%5C%5C%5CRightarrow%20%5Ceta_%7Bth%7D%3D1-%5Cfrac%7B20%2B273%7D%7B500%7D%5C%5C%5CRightarrow%20%5Ceta_%7Bth%7D%3D1-0.586%5C%5C%5CRightarrow%20%5Ceta_%7Bth%7D%3D0.414)
![\eta_{th}=\frac{W_{net}}{Q_h}\\\Rightarrow Q_h=\frac{W_{net}}{\eta_{th}}\\\Rightarrow Q_h=\frac{570}{0.414}\\\Rightarrow Q_h=1376.81159\ kW](https://tex.z-dn.net/?f=%5Ceta_%7Bth%7D%3D%5Cfrac%7BW_%7Bnet%7D%7D%7BQ_h%7D%5C%5C%5CRightarrow%20Q_h%3D%5Cfrac%7BW_%7Bnet%7D%7D%7B%5Ceta_%7Bth%7D%7D%5C%5C%5CRightarrow%20Q_h%3D%5Cfrac%7B570%7D%7B0.414%7D%5C%5C%5CRightarrow%20Q_h%3D1376.81159%5C%20kW)
![Q_h=0.315A\\\Rightarrow A=\frac{Q_h}{0.315}\\\Rightarrow A=\frac{1376.81159}{0.315}\\\Rightarrow A=4370.83044\ m^2](https://tex.z-dn.net/?f=Q_h%3D0.315A%5C%5C%5CRightarrow%20A%3D%5Cfrac%7BQ_h%7D%7B0.315%7D%5C%5C%5CRightarrow%20A%3D%5Cfrac%7B1376.81159%7D%7B0.315%7D%5C%5C%5CRightarrow%20A%3D4370.83044%5C%20m%5E2)
The theoretical minimum solar collector area required is 4370.83044 m²
Answer:
B : It will avoid the distortion from the atmosphere and give scientists more accurate data
Explanation:
The answer is B because 13 m/s is a greater acceleration than 10 m/s in the same amount of time.
Answer:
Speed is greater in trial 2
Explanation:
"<em>Indicate whether the speed of the block relative to the table when the block reaches the bottom of the plane is greater in trial 1 or trial 2.</em>"
The initial potential energy of the block is converted into kinetic energy, with no work done by friction.
In Trial 1, the ramp is free to move, so both the block and the ramp have kinetic energy.
PE = KE + KE
mgh = ½ mv₁² + ½ Mv²
In Trial 2, the ramp is fixed, so only the block has kinetic energy.
PE = KE
mgh = ½ mv₂²
Setting the expressions equal:
½ mv₁² + ½ Mv² = ½ mv₂²
Therefore, v₂ must be greater than v₁.
Answer:
F= 4.3N
the question (for seo and search engine optimization to get this question to the top)
Calculate the force of gravity on the 1-kg mass if it were 3.2×106m above Earth's surface (that is, if it were one and a half Earth radii from Earth's center).
F= 4.3N