Answer:
5.7 x 10^12 C
Explanation:
Let the charge on earth and moon is q.
mass of earth, Me = 5.972 x 10^24 kg
mass of moon, Mm = 7.35 x 10^22 kg
Let d be the distance between earth and moon.
the gravitational force between them is
The electrostatic force between them is
According to the question
1 % of Fg = Fe
q = 5.7 x 10^12 C
Thus, the charge on earth and the moon is 5.7 x 10^12 C.
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
Answer:
B) collision is inelastic because they stick together after collision and share a common final velocity Vf
C) M1V1 + M2V2 = (M1 + M2)Vf
D) Vf = 6.33m/s
E) force = 3040N
Explanation:
Detailed explanation and calculation is shown in the image below
The work done is the same as the amount of energy increase. The formula for kinetic energy is 122
1
2
m
v
2
.
The initial KE of the car is 12(1000)×202=200,000
1
2
(
1000
)
×
20
2
=
200
,
000
joules.
The final KE of the car is 12(1000)×302=450,000
1
2
(
1000
)
×
30
2
=
450
,
000
joules.
The difference between these is the amount of work done: 450,000−200,000=250,000
450
,
000
−
200
,
000
=
250
,
000
joules.
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