Answer: The motion of the object will remain the same
Explanation:
Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as


Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
False... I hope that helps ;)
Answer:
<h3>14.97m/s</h3>
Explanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J