Answer:

Explanation:
We are given that
Initial kinetic energy of an electron=K
Distance=d
Final velocity=v=0
Charge,q=-1e
We have to find the magnitude of electric field.
Work done=
Using the formula
Work done=
Using work energy theorem
Work done=Final K.E-Initial K.E=0-K
Work done=-K
Substitute the values
-K=-eEd
K=eEd

Hence, the magnitude of the electric field=
Answer:
0Nm, no work is done.
Explanation:
Work done is defined as the Force per distance meaning force times the distance moved in the direction of the force.
Now the body of mass 50g has a centripetal force acting on it directed towards the centre. Now in actuality the body stays along the circle it doesn't really move to the centre of the circle.
Hence the force doesn't move a distance, and so from the definition of work done;
F×d ; d =0
Hence work done = mv2/r × 0= 0Nm
It has three significant figure
Answer:
The drag coefficient is
Explanation:
From the question we are told that
The density of air is 
The diameter of bottom part is
The power trend-line equation is mathematically represented as

let assume that the velocity is 20 m/s
Then


The drag coefficient is mathematically represented as

Where
is the drag force
is the density of the fluid
is the flow velocity
A is the area which mathematically evaluated as

substituting values


Then

Answer:
he would have to run at a 7 and a half mile to get 8 miles in an hour
Explanation: