Answer:
E = 12262 V/m
Explanation:
given,
Charge of the particle, q = +4.0 μC
mass of the ball, m = 5 g
Electric field, E = ?
Force due to weight
F = m g
Force due to electric field
F =q E
To balance the weight of particle both the forces must be equal
Electric field exerted on the ball will be equal to
m g = q E


E = 12262 V/m
Hence, the electric field acting in the particle is equal to E = 12262 V/m
The normal force is always (underline, bold) is always perpendicular to the surface an object is sitting on. If the object is on an inclined plane, then the normal will not be vertical but it will be perpendicular to the angle of the incline.
The diagram below (left) shows a normal force (GH) that is not vertical, but it is perpendicular to the surface. The object on the right is the more usual normal a mass on a table top.
The vertical line on the right is the normal and it points up.
Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km