We will have the following:
![\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20F%3DG%5Cfrac%7Bm_1m_2%7D%7Br%5E2%7D%5CRightarrow%20F%3D%5Cfrac%7B%286.67%5Cast10%5E%7B-11%7Dm%5E3%5Cast%20kg%5E%7B-1%7D%5Cast%20s%5E%7B-2%7D%29%2850kg%29%2850kg%29%7D%7B%280.3m%29%5E2%7D%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20F%3D1.852777778...%5Cast10%5E%7B-6%7DN%5CRightarrow%20F%5Capprox1.85%5Cast10%5E%7B-6%7DN%20%5Cend%7Bgathered%7D)
So, the force is approximately 1.85*10^-6 N.
Answer:
C
Im not totally sure but that's what my science teacher taught me sorry if is wrong
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.
D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .
-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
Answer:
![T_s = 6.8 degree C](https://tex.z-dn.net/?f=T_s%20%3D%206.8%20degree%20C)
Explanation:
As per thermal radiation we know that rate is heat radiation is given as
![\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)](https://tex.z-dn.net/?f=%5Cfrac%7BdQ%7D%7Bdt%7D%20%3D%20%5Csigma%20eA%20%28T%5E4%20-%20T_s%5E4%29)
here we know that
T = 34 degree C = 307 K
![A = 1.38 m^2](https://tex.z-dn.net/?f=A%20%3D%201.38%20m%5E2)
e = 0.557
![\sigma = 5.67 \times 10^{-8} W/m^2K^4](https://tex.z-dn.net/?f=%5Csigma%20%3D%205.67%20%5Ctimes%2010%5E%7B-8%7D%20W%2Fm%5E2K%5E4)
![\frac{dQ}{dt} = 120 J/s](https://tex.z-dn.net/?f=%5Cfrac%7BdQ%7D%7Bdt%7D%20%3D%20120%20J%2Fs)
now we have
![120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)](https://tex.z-dn.net/?f=120%20%3D%20%285.67%20%5Ctimes%2010%5E%7B-8%7D%29%280.557%29%281.38%29%28307%5E4%20-%20T_s%5E4%29)
![120 = (4.36 \times 10^{-8})(307^4 - T_s^4)](https://tex.z-dn.net/?f=120%20%3D%20%284.36%20%5Ctimes%2010%5E%7B-8%7D%29%28307%5E4%20-%20T_s%5E4%29)
![T_s = 279.8 K](https://tex.z-dn.net/?f=T_s%20%3D%20279.8%20K)
![T_s = 6.8 degree C](https://tex.z-dn.net/?f=T_s%20%3D%206.8%20degree%20C)