An astronaut of mass m in a spacecraft experiences a gravitational force F=mg when stationary on the launchpad.
1 answer:
gravitational force is the attraction force of earth on an object which is near the surface of earth
It will not depend on the velocity or acceleration of earth
So it will not change while an object is stationary or it is moving with some acceleration
So here we will say that force will remain the same
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Answer:
r = 41.1 10⁹ m
Explanation:
For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal
∑ F = 0
F (Earth- probe) - F (Mars- probe) = 0
F (Earth- probe) = F (Mars- probe)
Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system
the distance from Earth to the probe is R (Earth-probe) = r
the distance from Mars to the probe is R (Mars -probe) = D - r
where D is the distance between Earth and Mars
M_earth (D-r)² = M_Mars r²
(D-r) = r
r ( ) = D
r =
We look for the values in tables
D = 54.6 10⁹ m (minimum)
M_earth = 5.98 10²⁴ kg
M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg
let's calculate
r = 54.6 10⁹ / (1 + √(0.642/5.98) )
r = 41.1 10⁹ m
Answer:
(a)
(b)
Explanation:
Part (a)
The total length of copper cord L=86.3 m
The cross sectional area A=1.71×10⁻⁶m²
The resistivity of copper p=1.72×10⁻⁸Ω
Thus the resistance of extension cord is
Part (b)
The resistance of trimmer Rt=17.9 ohms
When voltage of 120V is applied then the current I is passing through series circuit is
Thus the voltage across the trimmer is: