Answer:
6.4 × 10^7 W/m^2
Explanation:
The sun is assumed a sphere and therefore can be calculated by the surface area of a sphere.
surface area of sphere = 4πr^2 where r is the radius of the earth in m.
r = 6.96 ×10^8 m
substitute the value into the formula
surface area of the sun = 4 × 3.142 × ( 6.96 × 10^8)^2
the sun radiates at a rate of 3.90 × 10^26 W
to calculate the rate of energy emitted per square meter of the sun, we need to divide the rates at which the sun radiates energy per second by the surface area of the sun
that is P/surface area = 3.90 × 10^26 / ( 4 × 3.142 × ( 6.96 × 10^8)^2 = 6.4 × 10^7 W/m^2
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
Answer:
4 seconds
Explanation:
On the Y axis - you go to 12. Go directly horizontally till you hit the line. Then go directly down. If you do this correctly, you should hit 4 seconds on the X axis.
Answer:
F_total = 10000 N
Explanation:
For this exercise we use that the force is a vector and the best way to do the sum is that since the two tugs pull the boat with the same intensity and angle, the sum of the component of the force perpendicular to the movement becomes zero.
The components parallel to the movement of the tugs is
∑ F = 2 Fcos θ
let's calculate
F_total = 2 10000 cos 60
F_total = 10000 N
The particles which are in the field of forces means in the field line of forces they experience electric force.
Explanation: