.
<h3>Explanation</h3>
The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

where,
the total power emitted,
the surface area of the body,
the Stefan-Boltzmann Constant, and
the temperature of the body in degrees Kelvins.
.
.
.
Keep as many significant figures in
as possible. The error will be large when
is raised to the power of four. Also, the real value will be much smaller than
since the emittance of a human body is much smaller than assumed.
Answer:
c)At a distance greater than r
Explanation:
For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance between the satellite and the Earth's centre
v is the speed of the satellite
Re-arranging the equation, we write

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.
Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is
c)At a distance greater than r
Answer:
500kg
Explanation:
mass = newtons/force divided by the acceleration rate
m = 30,000/60
m = 500
Answer:
The tank is losing

Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume
≅ 0 ;
then
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = 
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : 
