Answer:
The horizontal distance traveled by the projectile is 15.23 m.
Explanation:
Given;
angle of projection, θ = 25⁰
initial velocity of the projectile, u = 15 m/s
time of flight, t = 1.12 s
The the travelling path of the object is calculated as the range of the projectile

Therefore, the horizontal distance traveled by the projectile is 15.23 m.
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,



Answer:
2.24 seconds
Explanation:
xf = xo + vo t + 1/2 at^2
45 = 0 + 15 t + 1/2 (4.5) t^2
2.25 t^2 + 15t - 45 = 0 Quadratic formula shows t = 2.24 seconds
Answer:0.00125 watts
Explanation:
resistance=50 ohms
Current=5 milliamps
Current=5/1000 milliamps
Current =0.005 amps
power=(current)^2 x (resistance)
Power=(0.005)^2 x 50
Power=0.005 x 0.005 x 50
Power=0.00125 watts