True !! Hope I helped you out a bit!
A = 4\pi r^2
A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2
A = 1.33*!0^{-5}MM^2
Mass of the displaced material. In water it would be the mass of the water that the volume of the ball displaces.
Answer:
a) 4 289.8 J
b) 4 289.8 J
c) 6 620.1 N
d) 411 186.3 m/s^2
e) 6 620.1 N
Explanation:
Hi:
a)
The kinetic energy of the bullet is given by the following formula:
K = (1/2) m * v^2
With
m = 16.1 g = 1.61 x 10^-2 kg
v = 730 m/s
K = 4 289.8 J
b)
the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:
W = ΔK = Kf - Ki = 4 289.8 J
c)
The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).
If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:
W = F * L
Where F is the net force and L is the length of the barrel, that is:
F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N
d)
The acceleration can be found dividing the force by the mass:
a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2
e)
The force will have a magnitude equal to c) and direction along the barrel towards the exit
Explanation:
003 (part 1 of 2)
Pressure is force divided by area.
P = F / A
P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)
P = 229,320 Pa
003 (part 2 of 2)
There are approximately 6895 Pa in 1 psi.
P = 229,320 Pa × (1 psi / 6895 Pa)
P = 33.3 psi
004 (part 1 of 2)
Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).
Impulse = change in momentum
F Δt = m Δv
F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))
F = 0.218 N
004 (part 2 of 2)
Pressure is force over area.
P = F / A
P = 0.218 N / 0.712 m²
P = 0.306 N/m²
