A solid conducting sphere with radius R = 0.390 m carries a net charge of +0.650 nC.
1 answer:
Answer:
38.5 N/C
Explanation:
The electric field generated by a charged sphere at a point outside the sphere is equivalent to the electric field generated by a single point charge, and it is given by
where
is the Coulomb's constant
Q is the net charge
r is the distance from the centre of the sphere
In this problem, we have
Substituting into the equation, we find
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Answer:
819.78 m
Explanation:
<u>Given:</u>
- OA = range of initial position of the airplane from the point of observation = 375 m
- OB = range of the final position of the airplane from the point of observation = 797 m
= angle of the initial position vector from the observation point =
= angle of the final position vector from the observation point =
= displacement vector from initial position to the final position
A diagram has been attached with the solution in order to clearly show the position of the plane.
Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:
The magnitude of the displacement vector =
Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.
Answer:
Explanation:
As we know that backpack is kicked on the rough floor with speed "v"
So here as per force equation in vertical direction we know that
so normal force on the block is given as
now the magnitude of kinetic friction on the block is given as
now when bag is sliding on the floor then net deceleration of the block due to friction is given as
now we know that bag hits the opposite wall at L distance away in time t
so we have