A 40g bullet traveling at 450 m/s passed through a board and comes out traveling 300m/s. The board is 7.6cm thick. What is the force of friction applied by the wood to the bullet?
Answer:
f = 29605.2 [N]
Explanation:
To solve this problem we must use the following equation of kinematics, then use Newton's second law.
[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]
where:
Vf = final velocity = 300 [m/s]
Vo = initial velocity = 450 [m/s]
a = acceleration [m/s²]
x = distance = 7.6 [cm] = 0.076 [m]
Now replacing and clearing a
2*a*0.076 = (450² - 300²)
a = 740131.57 [m/s²]
Now using Newton's second law which tells us that the force on a body is equal to the product of mass by acceleration.
f = m*a
where:
f = friction force [N]
m = mass = 40 [g] = 0.04 [kg]
f = 0.04*740131.57
f = 29605.2 [N]
A car with a mass of 1.5 × 103 kilograms is traveling west at a velocity of 22 meters/second. It hits a stationary car with a mass of 9.0 × 102 kilograms. If the collision is inelastic, what is the final direction and approximate velocity of the two cars?
Answer:
A. 14 meters/second to the west
Explanation:
The collision is inelastic, which means that:
- only the total momentum is conserved (not the kinetic energy)
Answer:
14 meters/second to the west
Explanation:
That's the answer!
The engine of a model airplane must both spin a propeller and push air backward to propel the airplane forward. Model the propeller as three 0.30-m-long thin rods of mass 0.040 kg each, with the rotation axis at one end.
What is the moment of inertia of the propeller?
How much energy is required to rotate the propeller at 5800 rpm? Ignore the energy required to push the air.
Solution :
Given :
Length of the propeller rods, L =0.30 m
Mass of each, M = 0.040 kg
Moment of inertia of one propeller rod is given by
[tex]$I=\frac{1}{3}\times M \times L^2$[/tex]
Therefore, total moment of inertia is
[tex]$I=3 \times \frac{1}{3}\times M \times L^2$[/tex]
[tex]$I=M\times L^2$[/tex]
[tex]$I=0.04\times (0.3)^2$[/tex]
[tex]$0.0036 \ kg \ m^2$[/tex]
Now energy required is given by
[tex]$E=\frac{1}{2}\times I \times \omega^2 $[/tex]
where, angular speed, ω = 5800 rpm
[tex]$\omega = 5800 \times \frac{2 \pi}{60} $[/tex]
= 607.4 rad/s
Therefore energy,
[tex]$E=\frac{1}{2}\times 0.0036 \times (607.4)^2 $[/tex]
= 664.1 J
The moment of inertia of the propeller is 0.0036 kgm² and the energy required is 663.21 J
Energy required for propeller:Given that the mass of the propellers is m = 0.040kg,
and their length is L = 0.30m
The moment of inertia of a rod with the rotation axis at one end is given by :
[tex]I = \frac{1}{3}m L^2[/tex]
so for 3 propellers:
[tex]I=3\times\frac{1}{3}\times(0.04)\times(0.3)^2[/tex]
I = 0.04 × 0.09
I = 0.0036 kgm²
Now, the frequency is given f = 5800 rpm
so anguar speed, ω = 5800×(2π/60)
ω = 607 rad/s
Energy required:
E = ¹/₂Iω²
E = 0.5 × 0.0036 × (607)² J
E = 663.21 J
Learn more about moment of inertia:
https://brainly.com/question/15248039?referrer=searchResults