charge must be equal to 5.74 ×10⁻⁵
In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.
→ Fnet =0
→ mg = qE
substituting the values we get :
0.00345 × 9.81 = q × 590
→ q = 5.74 ×10⁻⁵
Hence the charge must be equal to 5.74 ×10⁻⁵.
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Answer:
there are 25 kg objective travelling at 2m/s to the right.
The answer is 789.25 which you’d subtract 2011.25-122.2 I think sry if I’m wrong
To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.
Kepler's third law tells us that the period is defined as
The given data are given with respect to known constants, for example the mass of the sun is
The radius between the earth and the sun is given by
From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun
Therefore:
Substituting in Kepler's third law:
Therefore the period of this star is 3.8years
Velocity is displacement/time
(Displacement is the overall change in distance)
So you’ll want to divide 200 by 25, which should give you:
8 m/s
