<span>Earth (and hence the observer) moves.</span>
The answer would be erin out of all of them thank me later :)
No, because the distance-time would show a constant velocity but the velocity-time graph shows an increasing velocity.
Answer:
Explanation:
Vm = Δs/Δt
700km/h = Δs/1.5h
700 = Δs/1.5
S = 700 x 1.5
S = 1050 Km
*(S = Δs)
Answer: The rocket will have traveled 1050 Km
Hope this help ☺
At surface,
v = kq/r
And potential energy of an electron is given by,
PE = -ev = -ekq/r
At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C
Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s
