A) it occurs when earth is between the sun and the moon
The instrument would be a barometer.
<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
