Answer:
A) 4037.2[km]; B) 21472[m/s]
Explanation:
A)
This part can be solved using the principle of energy conservation, where kinetic energy will be equal to potential energy. We will define the kinetic energy and the potential energy.
Now we have to match both equations in this way the mass value is canceled and we can clear h.
B)
To solve this problem we can use the kinematic equations, but first we must identify the initial data:
yo = 2.35*10^7[m]
y = 0 [m] when hit the ground [m]
vo = 0 [m/s], It is essentially at rest...
Answer:
a. 2143 turns/m
b. 111.5 m
Explanation:
a. The minimum number of turns per unit length (N/L) can be found using the following equation:
Hence, the minimum number of turns per unit length is 2143 turns/m.
b. The total length of wire is the following:
Since each turn has length 2πr of wire, the total length is:
Therefore, the total length of wire required is 111.5 m.
I hope it helps you!
Answer:
r = 1.61 x 10^{11} m
Explanation:
energy radiated (H) = 2.7 x 10^31 W
surface temperature (T) = 11,000 k
assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}
we can find the radius of the star from the equation below
H = A x ε x σ x T^{4}
where area (A) = 4 x π x r^{2} (assuming it is a sphere)
therefore the equation becomes
H = 4 x π x r^{2} x ε x σ x T^{4}
2.7 x 10^31 = 4 x π x r^{2} x 1 x 5.67 x 10^{-8} x (11,000)^{4}
r =
r = 1.61 x 10^{11} m