PHYSICS COLLEGE

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.

Answers

Answer 1

Answer:

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 :

$I = (1)/(3)m L^2$

so for 3 propellers:

$I=3*(1)/(3)*(0.04)*(0.3)^2$

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:

brainly.com/question/15248039?referrer=searchResults

Answer 2

Answer:

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

$I=(1)/(3)* M * L^2$

Therefore, total moment of inertia is

$I=3 * (1)/(3)* M * L^2$

$I=M* L^2$

$I=0.04* (0.3)^2$

$0.0036 \ kg \ m^2$

Now energy required is given by

$E=(1)/(2)* I * \omega^2$

where, angular speed, ω = 5800 rpm

$\omega = 5800 * (2 \pi)/(60)$

= 607.4 rad/s

Therefore energy,

$E=(1)/(2)* 0.0036 * (607.4)^2$

= 664.1 J

Related Questions

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.078 V exists across the membrane. The thickness of the membrane is 7.1 x 10-9 m. What is the magnitude of the electric field in the membrane?
Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by magnetic forces. Michelle is a(n)
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound produced by the pipe, in SI units, is closest to:
Light has wavelength 600 nm in a vacuum. it passes into glass, which has an index of refraction of 1.5. what is the frequency of the light inside the glass
An ordinary drinking glass is filled to the brim with water (268.4 mL) at 2.0 ° C and placed on the sunny pool deck for a swimmer to enjoy. If the temperature of the water rises to 32.0 ° C before the swimmer reaches for the glass, how much water will have spilled over the top of the glass? Assume the glass does not expand.

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95* 10^(-3)\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600* 10^(-9)\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=(d_v)/(\lambda)$

$N=(1.24705)/(6* 10^(-7))$

$N\approx2.0784* 10^(6)$ .......................(1)

Now we find the wavelength of the light wave in the glass:

$n=(\lambda)/(\lambda')$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=(600* 10^(-9))/(\lambda')$

$\lambda'=375* 10^(-9)\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=(x)/(\lambda')$

$N'=(2.95* 10^(-3))/(375* 10^(-9))$

$N'=7.8667* 10^(3)$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.

(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?

Answers

Answer:

a) $\lambda=2.95x10^(-6)m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=(c)/(\lambda)$

If we replace the last equation into the E formula we got:

$E=h(c)/(\lambda)$

And if we solve for $\lambda$ we got:

$\lambda =(hc)/(E)$

Using the value of the constant $h=4.136x10^(-15) eVs$ we have this:

$\lambda=(4.136x10^(15)eVs (3x10^8 (m)/(s)))/(0.42eV)=2.95x10^(-6)m$

$\lambda=2.95x10^(-6)m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^(-6)m$

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.What is the battery's internal resistance?

Answers

Answer:

0.46Ω

Explanation:

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $(V^(2) )/(R)$

=> R = $(V^(2) )/(P)$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $(V^(2) )/(P)$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $(12.0^(2) )/(4)$

R₁ = $(144)/(4)$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $(V^(2) )/(P)$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $(12.0^(2) )/(4)$

R₂ = $(144)/(4)$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$(1)/(R_(X) )$ = $(1)/(R_1)$ + $(1)/(R_2)$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$(1)/(R_X)$ = $(1)/(36)$ + $(1)/(36)$

$(1)/(R_X)$ = $(2)/(36)$

Rₓ = $(36)/(2)$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $(11.7)/(18)$

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $(0.3)/(0.65)$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer:

$R_i_n_t=0.45 \Omega$

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

$R_i_n_t=((V_N_L)/(V_F_L) -1)R_L$

Where:

$V_F_L=Load\hspace{3}voltage=11.7V\nV_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\nR_L=Load\hspace{3}resistance$

As you can see, we don't know the exactly value of the $R_L$ . However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

$(12)/(11.7) =(4)/(P)$

Solving for $P$ :

$P=(11.7*4)/(12) =3.9W$

Now, the electric power is given by:

$P=(V^2)/(R_b)$

Where:

$R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb$

So:

$R_b=(V^2)/(P) =(11.7^2)/(3.9) =35.1\Omega$

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

$(1)/(R_L) =(1)/(R_b) +(1)/(R_b) =(2)/(R_b) \n\n R_L= (R_b)/(2) =(35.1)/(2)=17.55\Omega$

Finally, now we have all the data, let's replace it into the internal resistance equation:

$R_i_n_t=((12)/(11.7) -1)17.55=0.45\Omega$

What is an inexpensive, portable, and common way to assess body fat in the fitness industry?DEXA
Bioelectrical impedance
Skinfold testing
Hydrostatic weighing

Answers

Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.