The answer to this is formation of earth and our solar system
The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
<h3>What is Electric field?</h3>
Electric field is the physical field that surrounds a charge.
<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
- From Newton's second law, the acceleration the electron will be
a=F/m=qE/m
- where q= charge of electron
E= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
- The kinematics equation v²=v0²+2a(Δx)
- where v=final velocity of the electron
v0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
- Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2
=25×10¹²+10.52×10¹²
=35.52×10¹²
- Now velocity of electron=5.95×10⁶m/s.
Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
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Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required:
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
<u>τ = 4900 N.m</u>
<u></u>
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
<u>τ = 4900 N.m</u>
<u></u>
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
<u>τ = 4900 N.m</u>
<u></u>
Hence, the correct answer will be:
<u>e. The torque is the same for all cases.</u>
<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
