Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer:
Average recoil force experienced by machine will be 200 N
Explanation:
We have give mass of each bullet m = 50 gram = 0.05 kg
There are 4 bullets
So mass of 4 bullets = 4×0.05 = 0.2 kg
Initial speed of the bullet u = 0 m/sec
And final speed of the bullet v = 1000 m/sec
So change in momentum 
Time is given per second so t = 1 sec
We know that force is equal to rate of change of momentum
So force will be equal to 
So average recoil force experienced by machine will be 200 N
Answer: Remember speed is distance divided by time, so if he travels 1000 m in 7.045 s, his speed is
(1000 m)/(7.045 s) = 141.9 m/s.
Note there are 1609 metres in a mile, or 1 mi = 1609 m, so m = 1/1609 mi, or
141.9/1609 mi/s = 0.08822 mi/s. Now, note that 1 h = 3600 s, so the speed is
0.08822*3600 mi/h = 317.6 mi/h.
Answer:
6.03 mV
Explanation:
length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,
Magnetic field due to solenoid
B = μ0 n i = μ0 N i / L
dB/dt = μ0 N / L x di / dt
dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s
Induced emf, e = rate of change of magnetic flux
e = dΦ / dt = A x dB / dt
e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV
