Explanation:
The orbital radius of the Earth is ![r_1=1.496\times 10^{11}\ m](https://tex.z-dn.net/?f=r_1%3D1.496%5Ctimes%2010%5E%7B11%7D%5C%20m)
The orbital radius of the Mercury is ![r_2=5.79 \times 10^{10}\ m](https://tex.z-dn.net/?f=r_2%3D5.79%20%5Ctimes%2010%5E%7B10%7D%5C%20m)
The orbital radius of the Pluto is ![r_3=5.91 \times 10^{12}\ m](https://tex.z-dn.net/?f=r_3%3D5.91%20%5Ctimes%2010%5E%7B12%7D%5C%20m)
We need to find the time required for light to travel from the Sun to each of the three planets.
(a) For Sun -Earth,
Kepler's third law :
![T_1^2=\dfrac{4\pi ^2}{GM}r_1^3](https://tex.z-dn.net/?f=T_1%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7BGM%7Dr_1%5E3)
M is mass of sun, ![M=1.989\times 10^{30}\ kg](https://tex.z-dn.net/?f=M%3D1.989%5Ctimes%2010%5E%7B30%7D%5C%20kg)
So,
![T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s](https://tex.z-dn.net/?f=T_1%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7B6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%201.989%5Ctimes%2010%5E%7B30%7D%7D%5Ctimes%201.496%5Ctimes%2010%5E%7B11%7D%5C%5C%5C%5CT_1%3D%5Csqrt%7B%5Cdfrac%7B4%5Cpi%5E%7B2%7D%7D%7B6.67%5Ctimes10%5E%7B-11%7D%5Ctimes1.989%5Ctimes10%5E%7B30%7D%7D%5Ctimes1.496%5Ctimes10%5E%7B11%7D%7D%5C%5C%5C%5CT_1%3D2%5Ctimes%2010%5E%7B-4%7D%5C%20s)
(b) For Sun -Mercury,
![T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s](https://tex.z-dn.net/?f=T_2%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7B6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%201.989%5Ctimes%2010%5E%7B30%7D%7D%5Ctimes%205.79%20%5Ctimes%2010%5E%7B10%7D%5C%20m%5C%5C%5C%5CT_2%3D%5Csqrt%7B%5Cdfrac%7B4%5Cpi%5E%7B2%7D%7D%7B6.67%5Ctimes10%5E%7B-11%7D%5Ctimes1.989%5Ctimes10%5E%7B30%7D%7D%5Ctimes%205.79%20%5Ctimes%2010%5E%7B10%7D%7D%5C%20m%5C%5C%5C%5CT_2%3D1.31%5Ctimes%2010%5E%7B-4%7D%5C%20s)
(c) For Sun-Pluto,
![T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s](https://tex.z-dn.net/?f=T_3%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7B6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%201.989%5Ctimes%2010%5E%7B30%7D%7D%5Ctimes%205.91%20%5Ctimes%2010%5E%7B12%7D%5C%5C%5C%5CT_3%3D%5Csqrt%7B%5Cdfrac%7B4%5Cpi%5E%7B2%7D%7D%7B6.67%5Ctimes10%5E%7B-11%7D%5Ctimes1.989%5Ctimes10%5E%7B30%7D%7D%5Ctimes%205.91%20%5Ctimes%2010%5E%7B12%7D%7D%5C%5C%5C%5CT_3%3D1.32%5Ctimes%2010%5E%7B-3%7D%5C%20s)
Lead-207 can change into Lead-208.
When Lead-207 is bombarded with neutrons, the atom acquires the neutron. Adding a neutron only changes the mass number of the atom. This does not involve change in the identity of the atom. So, Lead-207 can change into Lead-208 without change in its chemical properties.
Power = Work done / Time taken.
Work done = mgh
Mass, m = 33kg ( Am presuming it is 33 kg).
h = 85 m.
Work done = 33 * 9.81* 85 = 27517.05 J.
Time taken.
Since object was dropped from height, it fell under gravity.
Using H = ut + (1/2) * gt^2. u = 0.
H = 1/2 gt^2.
t = (2H/g) ^ (1/2)
t = (2*85/9.81) ^ 0.5 = 4.1628 s.
Power = 27517.05 / 4.1628 = 6610.23 Watts.
= 6610 W to 3 S. f.
Answer:
The one with the less mass
Explanation:
<span> Impulse and momentum are very important factors for engineers when thh Materials•Egg ... the force exerted on the ground) will increase the amount of time that the force ... us to spread the force of the egg evenly for the egg to survive the collision. The less velocity your egg has, the safer the egg will be.</span><span>
</span>