Answer:
Force, 
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :
Finally, it stops, v = 0
F = -22713.92 N
So, the magnitude of the force that stops the bullet is 
Answer:
Don't you worry, 'cause everything's gonna be alright, ai-a'ight
Be alright, ai-a'ight
Explanation:
<span>net work = change in kinetic energy
for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2
Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction
10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²
now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2
since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)
10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8
they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × 
slug/ft³
so here P1 + 
= P2 + 
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + = P2 +
0 + 
solve it we get
P2 = 87.9 lb/ft³
