Answer:
I'm not %100 sure but it might be d
Explanation:
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 $[/tex]
= 664.1 J
The moment of inertia of the propeller is 0.0036 kgm² and the energy required is 663.21 J
Energy required for propeller:Given that the mass of the propellers is m = 0.040kg,
and their length is L = 0.30m
The moment of inertia of a rod with the rotation axis at one end is given by :
[tex]I = \frac{1}{3}m L^2[/tex]
so for 3 propellers:
[tex]I=3\times\frac{1}{3}\times(0.04)\times(0.3)^2[/tex]
I = 0.04 × 0.09
I = 0.0036 kgm²
Now, the frequency is given f = 5800 rpm
so anguar speed, ω = 5800×(2π/60)
ω = 607 rad/s
Energy required:
E = ¹/₂Iω²
E = 0.5 × 0.0036 × (607)² J
E = 663.21 J
Learn more about moment of inertia:
https://brainly.com/question/15248039?referrer=searchResults
A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is
shown in the figure above. The value for each force is shown in the figure. What is the acceleration of the student-bicycle system?
m
А
0
2
B
0.2"
C) 5
D
25"
Solution:
The gravitational force that acts on the bicycle system is
[tex]$F_{g} = 500 \ N$[/tex]
Now the force, that is the gravitational force is related to mass of the system and the acceleration due to gravity of the system, 'm' and 'g' respectively.
Therefore, we can write
[tex]$F_g=mg$[/tex]
500 = m x 10 (since , g = 10 m/s-s)
∴ m = 50 kg
Now the net vertical force acting on the student bicycle system is 0. And the vertical acceleration of system is also 0. The total horizontal force acts to the right of the system. So by Newton's 2nd law of motion, we can write
[tex]$F_f = ma$[/tex]
[tex]$a=\frac{F_f}{m}$[/tex]
[tex]$=\frac{250}{50}$[/tex]
Therefore [tex]$a= 5 \ m/s^2$[/tex]
Hence (C) is correct option.
You are looking straight down on a magnetic compass that is lying flat on a table. A wire is stretched horizontally under the table, parallel to and a short distance below the compass needle. The wire is then connected to a battery so that a current I flows through the wire. This current causes the north pole of the compass needle to deflect to the left. The questions that follow ask you to compare the effects of different actions on this initial deflection.
If the wire is lowered farther from the compass, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection?
Answer:
It is smaller
Explanation:
in a distance vs time graph what does the slope represent
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!
A mason stretches a string between two points 70 ft apart on the same level with a tension of 20 lb at each end. If the string weighs 0.18 lb, determine the sag h at the middle of the string.
Answer:
Explanation:
Let tension be T in each string . Let angle with horizontal be θ in the middle
The sum of vertical components of tension of two string will balance the weight
2Tsinθ = mg
2 x 20 sinθ = .18
sinθ = .18 / 40 = .0045
θ = .25783 degree
If sag be y
y / 70 = tan .25783
y = 70 x tan.25783
= .3150 m
= 31.50 cm .
3. A car travels at 30mph for 2hours and then 40mph
for 3hours in the same direction. What is the average
speed of the vehicle's entire trip?
Answer:
1. 60/90
2.67/63
Explanation:
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]
What is the force acting an object that is accelerating at 3.6 m/s/s and has a mass of 15 kg?
Answer:
54 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 15 × 3.6
We have the final answer as
54 NHope this helps you