Answer:
The magnitude and direction of electric field midway between these two charges is
along AB.
Explanation:
Given that,
First charge 
second charge 
Distance = 20 cm
We need to calculate the electric field
For first charge,
Using formula of electric field

Put the valueinto the formula


Direction of electric field along AB
We need to calculate the electric field
For second charge,
Using formula of electric field

Put the valueinto the formula


Direction of electric field along AO
We need to calculate the net electric field at midpoint



Direction of net electric field along AB
Hence, The magnitude and direction of electric field midway between these two charges is
along AB.
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V
Answer:
Total Work done =0.65 joule
Explanation:
Work done is given Mathematically as
W=F *d
Where w=work done in joules
F=applied force
d= distance moved
The work done to move the toy accros the first meter is
W1=0.5*1
W1=0.5joule
The work done to move the toy across the next 2m at an angle of 30° is
.W2=0.5*2cos30
W2=0.5*2*0.154
W2=0.154joule
Hence total work done is
W1+W2=0.5+0.154
Total Work done =0.65 joule
Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s