Answer:
The force exerted by the ball on the bat has a magnitude of 100 N and its direction is exactly opposite to that of the force exerted by the bat on the ball.
Explanation:
Recall that Newton's third law tells us that : "For every action, there is an equal and opposite reaction."
Therefore if the bat acts on the ball with a force of 100 N, the ball acts on the bat with a similar magnitude of force (100 N) but direction opposite to the original force.

Answer:
L= 2 mH
Explanation:
Given that
Frequency , f= 10 kHz
Maximum current ,I = 0.1 A
Maximum energy stored ,E= 1 x 10⁻⁵ J
The maximum energy stored in the inductor is given as follows

Where ,L= Inductance
I=Current
E=Energy
Now by putting the values in the above equation


L=0.002 H
L= 2 mH
We know that frequency f is given as

C=Capacitance , f=frequency ,L=Inductance
Now by putting the values






Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.
Explanation:
We have,
Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.
1 mile = 1.6 km
45 miles = 72.42 km
74 miles = 119.0 km
1 hour 15 min means 1.25 hours
Average speed of Ajoba is :

Average speed of Prav,

Difference in average speed of Ajoba and Prav is :

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.
Answer:
Volume = 1,015 acre-feet (Approx)
Explanation:
Given:
Rain = 1.7 in
Time = 30 min
Area = 29 km²
Find:
Volume in acre-feet
Computation:
1 km = 1,000 m
1 m = 3.28 feet
1 km² = 247.105 acre
d = 1.7 in = 1.7 / 12 = 0.14167 ft
Area = 29 × 247.105 = 7,166.045 acre
Volume = 7,166.045 acre × 0.14167 ft
Volume = 1,015 acre-feet (Approx)
Answer:
When they are connected in series
The 50 W bulb glow more than the 100 W bulb
Explanation:
From the question we are told that
The power rating of the first bulb is 
The power rating of the second bulb is 
Generally the power rating of the first bulb is mathematically represented as

Where
is the normal household voltage which is constant for both bulbs
So

substituting values

Thus the resistance of the second bulb would be evaluated as

From the above calculation we see that

This power rating of the first bulb can also be represented mathematically as

This power rating of the first bulb can also be represented mathematically as

Now given that they are connected in series which implies that the same current flow through them so

This means that

So when they are connected in series

This means that the 50 W bulb glows more than the 100 \ W bulb