A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop

Answers

Answer 1

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length [tex]l[/tex] = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance [tex]R_{sol[/tex] = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

[tex]R_o[/tex] = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) =  [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]

so

ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )

ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]

ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

Answer 2

Answer:

1.28 *10^-5 A

Explanation:

Same work as above answer. Needs to be more precise


Related Questions

The primary purpose of a switch in a circuit is to ___________.

A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use

Answers

Answer:

store a charge for later use

What is measurement

Answers

Answer:

Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.

Explanation:

A scientific method can be defined as a research method that typically involves the use of experimental and mathematical techniques which comprises of a series of steps such as systematic observation, measurement, and analysis to formulate, test and modify a hypothesis.

Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.

Basically, the dimensions include important parameters such as width, height, length, area, volume, circumference, breadth, etc.

Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.
QL
Determine:
i. the tension in the string,
ii. the acceleration of each mass, and
iii. the distance each mass moves in the first second of motion if they start from rest​

Answers

i. [tex]T = 36.8\:\text{N}[/tex]

ii. [tex]a = 2.45\:\text{m/s}^2[/tex]

iii. [tex]x = 1.23\:\text{m}[/tex]

Explanation:

Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]

Forces acting on m1:

[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]

Forces acting on m2:

[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]

Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us

[tex]m_2g - m_1g = m_2a + m_1a[/tex]

or

[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]

Solving for a,

[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]

[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]

We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).

[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]

Assuming that the two objects start from rest, the distance that they travel after one second is given by

[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]

train starts from rest and accelerates at 1m/ s²
for 10 seconds how far does it move​

Answers

Answer:

s=50m

Explanation:

you can use the formula

s=ut+1/2at²

s=0t+1/2(1)10²

=1/2(100)

=50

I hope this helps

An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton

Answers

Answer:

the  speed of the electron at the given position is 106.2 m/s

Explanation:

Given;

initial position of the electron, r = 9 cm = 0.09 m

final position of the electron, r₂ = 3 cm = 0.03 m

let the speed of the electron at the given position = v

The initial potential energy of the electron is calculated as;

[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]

When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;

[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]

Apply the principle of conservation of energy;

ΔK.E = ΔU

[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]

Therefore, the  speed of the electron at the given position is 106.2 m/s

Action and Reaction are equal in magnitude and opposite in direction.Then,Why do they not balance each other?​

Answers

Answer:

Newton's third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel each other out.

Accommodation of the eye refers to its ability to __________. see on both the brightest days and in the dimmest light see both in air and while under water move in the eye socket to look in different directions focus on both nearby and distant objects

Answers

Answer:

to adjust from distant to the near objects

Explanation:

The process of accommodation is achieved by changing in the shape and position of the eye ball. Just like adjusting the lens of the camera.

Answer:

The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.

Explanation:

The human eye is the optical instrument which works on the refraction of light.

The ability of eye lens to change its focal length is called accommodation power of eye.

The focal length of eye lens is changed by the action of ciliary muscles.

When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.

Required:
What is the lowest possible frequency of sound for which this is possible?

Answers

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?

Answers

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

state the laws of reflection​

Answers

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface  called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.

why did Rita's hands get hot when she rubbed them ?

Answers

Answer:

due to production of heat through friction

Explanation:

because of the friction produce between her hands

De que esta hecho el sol? plisss ayuda.no necesito un texto de 100 reglones, puede ser resumido en solo 2 renglones

Answers

El sol es una bola de gas así como las estrellas:)

what aspect of the US justice system has its roots in Jewish scripture?​

Answers

The aspect of the US justice system that has its roots in Jewish scripture is:

the idea that all people are subject to the same rules and laws.

It is the doctrine of "equality before the law."  Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc.  Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism.  It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.

Answer:

answer is C

the idea that all people are subject to the same rules and laws

Explanation:

hope this helps!

Question 5 of 10
What must be the same for two resistors that are connected in parallel?

Answers

Answer:

in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .

Explanation:

In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?

a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time

Answers

Answer:

b. The heavier one reaches the bottom first.

Answer:

B

Explanation:

The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and we can model the situation to see why. Assume that the man has a mass of 90 kg, with a center of gravity 1.0 m above the ground. The action of the wind on his torso, which we approximate as a cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground, produces a force that tries to tip him over backward. To keep from falling over, he must lean forward.
A. What is the magnitude of the torque provided by the wind force? Take the pivot point at his feet. Assume that he is standing vertically. Assume that the air is at standard temperature and pressure.
B. At what angle to the vertical must the man lean to provide a gravitational torque that is equal to this torque due to the wind force?

Answers

Answer:

a)  [tex]t=195.948N.m[/tex]

b)  [tex]\phi=13.6 \textdegree[/tex]

Explanation:

From the question we are told that:

Density [tex]\rho=1.225kg/m^2[/tex]

Velocity of wind [tex]v=14m/s[/tex]

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient [tex]\mu=2.05[/tex]

a)

Generally the equation for Force is mathematically given by

[tex]F=\frac{1}{2}\muA\rhov^2[/tex]

[tex]F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2[/tex]

[tex]F=163.29[/tex]

Therefore Torque

[tex]t=F*r*sin\theta[/tex]

[tex]t=163.29*1.2*sin90[/tex]

[tex]t=195.948N.m[/tex]

b)

Generally the equation for torque due to weight is mathematically given by

[tex]t=d*Mg*sin90[/tex]

Where

[tex]d=sin \phi[/tex]

Therefore

[tex]t=sin \phi*Mg*sin90[/tex]

[tex]195.948=833sin \phi[/tex]

[tex]\phi=sin^{-1}\frac{195.948}{833}[/tex]

[tex]\phi=13.6 \textdegree[/tex]

A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?

Answers

Answer:

V = V0 + a t

V = 75 - 9.8 * 3 = 45.6 m/s

The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.

What is the equation of motion?

The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.

Given, the initial velocity of the projectile, u = 75 m/s

The time taken by the projectile, t = 3 sec

The acceleration due to gravity upward, g = - 9.8 m/s²

From the first equation of motion we can calculate the final velocity of the projectile:

v = u + at

v = u - gt

v = 75 - 9.8 ×(3)

v = 75 - 29.4

v = 45.6 m/s

Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.

Learn more about the equation of motion, here:

brainly.com/question/16982759

#SPJ2

Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?

a.) a loudspeaker

b.) a headphone

c.) a light bulb

d.) a microphone

I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure

Answers

Answer:

I THINK C

Explanation:

BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.

Use the DC Construction kit to build a simple circuit to perform the following task:

You are asked to use a single resistor and a 110 V DC battery for the purpose of boiling a litter of water (4,184 Joule/Kg*degree Celsius), with a starting temperature of 20 C, in exactly 4 minutes.

Answers

Answer:

The resistance is 8.7 ohm.

Explanation:

Voltage, V = 110 V

mass, m = 1 kg

change in temperature, T = 100 - 20 = 80 C

time, t = 4 min = 4 x 60 = 240 s

specific heat, c = 4184 J/kg C

let the resistance is R.

The heat generated by the heater is used to the heat the water.

[tex]\frac{V^2}{R} t = m c T \\\\\frac{110^2}{R}\times 240 = 1\times 4184\times 80\\\\R = 8.7 ohm[/tex]

Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por
3.00 s a lo largo de su línea inicial de movimiento con una aceleración de 30.0 m/s2
. En este momento, sus motores fallan,
y el cohete procede a moverse como un proyectil. Determine: (a) la altitud máxima que alcanza el cohete, (b) su tiempo
total de vuelo y (c) su alcance horizontal

Answers

Answer:

Explanation:

v = u + at

v₃ = 100 +30.0(3.00) = 190 m/s

s = vt + ½at²

y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m

x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m

a) v² = u² + uas  

s = (v² - u²) / 2a

ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m

b) t₁ = 3.00 s

   t₂ = (190sin53) / 9.80 = 15.5 s

   t₃ = √(2(1522) / 9.80) = 17.6 s

t = 3.00 + 15.5 + 17.6 = 36.1 s

c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m

The standard unit of brightness is called the candela.
True
False

Answers

Answer:

TRUE

Explanation:

CHEGG Over the course of a multi-stage 4820-km bicycle race, the front wheel of an athlete's bicycle makes 2.40x106 revolutions. How many revolutions would the wheel have made during the race if its radius had been 1.4 cm larger?

Answers

Answer:

θ' =  14.44 × [tex]10^{6}[/tex]

Explanation:

given data

total distance is d = 4820

radius = 1.4 cm

solution

we get here total angle by which the wheel rotates traveling is express as

⇒  [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex]     ................1

and

total angle (θ)  and the total distance (d) express as

⇒ d = r × θ       ...............2

here r is radius

and here rotated through some other angle θ' so put value in given equation and find revolutions    

⇒  d = (r+r)θ'      ........3

here  r = d/θ

so

⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]    

so put value and get θ'

⇒  θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]  

⇒  θ' =  14.44 × [tex]10^{6}[/tex] rev

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________

Answers

Answer:  

Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].

Explanation:  

E inside [tex]= KQr_{1} /R^{3}[/tex]  

so if r1 will be the same then  

E  [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

Answer:

The electric field is E/8.

Explanation:

The electric field due to a solid sphere of uniform charge density inside it is given by

[tex]E =\frac{\rho r}{3}[/tex]

where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.

For case I:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]

So, electric field at a distance r is

[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]

Case II:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]

So, the electric field at a distance r is

[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]

what do we mean by thrust?

Answers

Answer:

the answer is push example: she thrust her hand into her pocket

In a single-slit diffraction pattern, the central fringe is 360 times as wide as the slit. The screen is 14,000 times farther from the slit than the slit is wide. What is the ratio /W, where is the wavelength of the light shining through the slit and W is the width of the slit

Answers

Answer:

0.01286

Explanation:

In a given single-slit, the central fringe (Y) is 360 times as wide as the slit (a). Then

2Y₁ = 360a

Y₁ = 360a/2

= 180a

The distance D = 14000a

In a given single-slit diffraction, the ratio = [tex]\dfrac{\lambda }{W}[/tex]

and since the angle is infinitesimally small;

sin θ ≅ tan θ = [tex]\dfrac{Y}{D}[/tex]

For the first dark fringe;

Suppose:  [tex]\dfrac{a}{2}sin \theta = \dfrac{\lambda }{2}[/tex]

then,

[tex]\dfrac{a}{2} \ \dfrac{Y_1}{D} = \dfrac{\lambda }{2}[/tex]

[tex]aY_1 = \lambda D[/tex]

[tex]\dfrac{\lambda }{a} = \dfrac{Y_1}{D}\\ \\ \\ \implies \dfrac{180\ a}{14000 \ a} \\ \\ \mathbf{\dfrac{\lambda }{a} = 0.01286}[/tex]

An airplane propeller is 2.16 m in length (from tip to tip) with mass 100 kg and is rotating at 2900 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy? Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

Answers

a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.

b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.

How to determine the angular speed of a airplane propeller

Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:

K = 0.5 · I · ω²  (1)

Where:

I - Moment of inertia of the airplane propeller, in kilogram-square meters.ω - Angular speed, in radians per second

In addition, the moment of inertia of a slender rod rotating around its center is:

I = 0.0833 · M · L²   (2)

Where:

M - Mass of the propeller, in kilogramsL - Length of the propeller, in meters

a) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:

K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²

K = 1792152.287 J

The rotational kinetic energy of the airplane propeller is 1792152.287 joules. [tex]\blacksquare[/tex]

b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:

[tex]M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}[/tex]

[tex]\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}[/tex]   (3)

If we know that [tex]\omega_{o} = 2900\,\frac{rev}{min}[/tex], [tex]M_{o} = 100\,kg[/tex] and [tex]M_{f} = 75\,kg[/tex], then the angular speed of the airplane propeller is:

[tex]\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }[/tex]

[tex]\omega_{f} \approx 3348.631\, \frac{rev}{min}[/tex]

The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. [tex]\blacksquare[/tex]

To learn more on rotational kinetic energy, we kindly invite to check this verified question: https://brainly.com/question/20261989

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).

Answers

Answer:

1.6031 miles

Explanation:

Given the following data;

Speed = 344 m/s

Time = 7.5 seconds

To find how many miles away was the lighting strike;

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 344 * 7.5

Distance = 2580 meters

Next, we would have to convert the value of the distance travelled in meters to miles;

Conversion:

1609.344 metres = 1 mile

2580 meters = X mile

Cross-multiplying, we have;

X * 1609.344 = 2580

X = 2580/1609.344

X = 1.6031 miles

A 40 kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate without friction. The boy tries to walk around the platform in a counterclockwise direction. As he does:

a. the platform doesn't rotate.
b. the platform rotates in a clockwise direction just fast enough so that the boy remains stationary relative to the ground.
c. the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.
d. both go around with equal angular velocities but in opposite directions

Answers

Answer:

the correct one is C

Explanation:

To find the answer, let's propose the solution of the problem

We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.

Initial instant. Before starting to walk

          L₀ = 0

Final moment. After the child is walking

          L_f = I₁ w₁ + m r v₂

where index 1 is used for the platform and index 2 for the child

linear and angular velocity are related

          v₂ = w₂ r

           

angular momentum is conserved

          0 = I₁ w₁ + m r (w₂ r)

          w₁ =  [tex]- \frac{m r^2}{I1} \ w_2[/tex]

the moment of inertia of the platform bringing it closer to a disk or cylinder

         I₁ = [tex]\frac{1}{2}[/tex] M r²  

sustitute

          w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]

          W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]

from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds

when examining the answers the correct one is C

Answer:

Option C (the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground)

Explanation:

relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system

so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero

so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong

option B is wrong because relative to ground their angular momentum should be equal and opposite

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) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2,5][−2,5].

Answers

Answer:

89.87m/s

Explanation:

Given the velocity function

v(t)=−t²+3t−2

In order to get the displacement function, we will integrate the velocity function as shown:

[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]

at t = 5

[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]

at t = -2

[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]

Required displacement = d(5) - d(-2)

Required displacement = 89.2 - (-0.67)

Required displacement = 89.2 + 0.67

Required displacement = 89.87m/s

What is the de Broglie wavelength of a red blood cell with a mass of 1.00 * 10-11 g that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?

Answers

Answer:

The wavelength is "[tex]=16.5675\times 10^{-18} \ m[/tex]".

Explanation:

Given:

Mass,

m = [tex]1\times 10^{-11} \ g[/tex]

Speed,

V = [tex]0.400 \ cm/s[/tex]

or,

  = [tex]0.4\times 10^{-2}[/tex]

According to De Broglie,

The wavelength will be:

⇒ [tex]\lambda = \frac{h}{mV}[/tex]

      [tex]=\frac{6.627\times 10^{-34}}{1\times 10^{-11}\times 10^{-3}\times 0.4\times 10^{-2}}[/tex]

      [tex]=16.5675\times 10^{-18} \ m[/tex]

So, blood cells move these wavelength.

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