Answer:
Explanation:
a. Landing height is
H=1.3m
Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft
u=1.3m/s
Velocity of lander at impact, i.e final velocity is needed
v=?
The acceleration due to gravity is 0.4 times that of the one on earth,
Then, g on earth is approximately 9.81m/s²
Then, g on Mars is
g=0.4×9.81=3.924m/s²
Then using equation of motion for a free fall body
v²=u²+2gH
v²=1.3²+2×3.924×1.3
v²=1.69+10.2024
v²=11.8924
v=√11.8924
v=3.45m/s
The impact velocity of the spacecraft is 3.45m/s
b. For a lunar module, the safe velocity landing is 3m/s
v=3m/s.
Given that the initial velocity is 1.2m/s²
We already know acceleration due to gravity on Mars is g=3.924m/s²
The we need to know the maximum height to have a safe velocity of 3m/s
Then using equation of motion
v²=u²+2gH
3²=1.2²+2×3.924H
9=1.44+7.848H
9-1.44=7.848H
7.56=7.848H
H=7.56/7.848
H=0.963m
The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m
Metal bar is aligned along East- West direction and it is dropped vertically down
So its velocity is along -Z direction,
Now the Earth's magnetic field is towards north so its towards +Y direction
now we have formula for force on a moving charge
v = - z direction
B = + y direction
now by the above formula
So force is towards East
so all positive charge is towards East end
Answer:
Explanation:
charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C
mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg
Radius, r = 4.5 cm = 0.045 m
Magnetic field, B = 1.20 T
(a) Let the speed is v.
v = 2.6 x 10^6 m/s
(b) Let T be the period of revolution
T = 1.09 x 10^-7 s
(c) The formula for the kinetic energy is
K = 2.25 x 10^-14 J
(d) Let the potential difference is V.
K = qV
V = 70312.5 V
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