Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.
Answer:
Solving for time :
(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known
2. The unknown constant that we want to solve)
s = (1/2)(u+v)t <--- one of the formulas
from linear motion
s (distance) = 0.05m
u (initial velocity) = 100m/s
v (final velocity) = 0 m/s (it stops)
t (time taken for change in velocity) = to be found
0.05 = (1/2)(100+0)t
t = 0.001 seconds
Solving for the resistant force :
Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.
When the bullet stops :
F net = 0
F r = F imp
F r = (mu -mv)/t
F r = (0.01x100-0.01x0)/0.001
F r = 1/0.001
F r = 1000N
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
A. The switch within the closed circuit is opened.
If the switch is opened current will stop flowing, as the current will meet a gap, that is not connected by a conductor.
