An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µk) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving?

Answer:

since the bed is a cuboid, we find the volume by L×W×H

4.50 × 2.50 × 1.50 = 16.875m³

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Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]

So, its initial position was 471 m.

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

[tex]37.8=\frac{1}{2}(3)v^2[/tex]

[tex]v^2=\frac{37.8\times 2}{3}[/tex]

[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]

v=5.02 m/s

c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]

[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/tex]

Explanation:

If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :

[tex]K=\dfrac{1}{2}mv^2[/tex]

m is mass of the object

We can rearrange the above equation such that,

[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]

Hence, this is the velocity at the end of a rollercoaster ride.

Answer:

Option B - displacement can be specified by a magnitude and a direction.

Explanation:

A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c

Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.

Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.

Looking at the options, the only one that truly justifies this definition is option B.

Explanation:

Momentum = mass × velocity

p = mv

p = (1000 kg) (14 m/s)

p = 14000 kg m/s

The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

Given the data in the question

Momentum is the product of the mass of a particle and its velocity.

Momentum = Mass × Velocity

[tex]P = m \ * \ v[/tex]

We substitute our given values into the equation

[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]

Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

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Answer:  arrows

Explanation:

A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.

What is the vector quantity unit?

The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.

The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.

Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).

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Answer:

The nature of the wave formed is a transverse  progressive wave.

Explanation:

A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.

A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.

The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.

Answer:

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Explanation:

The Free Body Diagram associated with the experiment is presented as attachment included below.

Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.

Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:

[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]

[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]

Where:

[tex]W[/tex] - Weight of the eraser, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force, measured in newtons.

[tex]\theta[/tex] - Angle of the incline, measured in degrees.

The maximum allowable static friction force is:

[tex]f = \mu_{s} \cdot N[/tex]

Where:

[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

Likewise, the kinetic friction force is described by the following model:

[tex]f = \mu_{k} \cdot N[/tex]

Where:

[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])

In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:

[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]

[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]

But [tex]N = m\cdot g \cos \theta[/tex], so that:

[tex]\mu = \tan \theta[/tex]

Now, coefficients of static and kinetic friction are, respectively:

[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]

[tex]\mu_{s} \approx 0.716[/tex]

[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]

[tex]\mu_{k} \approx 0.596[/tex]

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Answer:

16kg m/s

Explanation:

P=mv

8 times 2=16kg m/s

Answer:

The momentum of moving body is calculated by

p= mv

In this question m= 8kg

v= 2m/s

so p = 8*2 = 16 kg m/s.

"Mass in motion" can be used to describe momentum. Mass exists in all things. Therefore, if an object is moving, it has momentum—its mass is moving. There are two factors that determine an object's momentum level: how much and how quickly the objects are moving.

Mass and velocity are two variables that affect momentum. An object's momentum can be expressed mathematically as the product of its mass multiply by its velocity.

The equation above can be rewritten as p = m • v, where m is the mass and v is the velocity, since momentum is represented by the lower case p in physics. The equation demonstrates that an object's momentum is directly proportional to its mass and velocity.

The quantity momentum is a vector. A vector quantity is a quantity that is fully described by magnitude and direction, as was discussed in a previous unit. Information about the bowling ball's magnitude as well as its direction must be included in order to fully describe the momentum of a 5-kg ball traveling westward at 2 m/s. The ball has a momentum of 10 kg m/s.

Until information about the ball's direction is provided, the ball's momentum cannot be fully described. The direction of the ball's velocity and the direction of the momentum vector are identical. It was mentioned in a previous unit that the velocity vector moves in the same way that an object moves. The bowling ball's momentum can be fully described as 10 kg m/s westward if it is moving westward. The magnitude and direction of an object's momentum can be used to fully describe it as a vector quantity.

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Answer:

Explanation:

fundamental frequency at closed pipe = 40.4 Hz

overtones are odd harmonics in closed pipe

first three overtones are

3 x 40.4 , 5 x 40.4 , 7 x 40.4 Hz

= 121.2 Hz , 202 Hz , 282.8 Hz .

speed of sound given is 337 , fundamental frequency is 233 Hz

wavelength = velocity of sound / frequency

= 337 / 233

= 1.446 m

for fundamental note in open pipe

wavelength /2 = length of tube

length of tube = 1.446 / 2

= .723 m

= 72.30 cm .

first overtone will be two times the fundamental ie 466. In open pipe all the harmonics are found , ie both odd and even .

Answer: Short Answer: NO ( In Most Cases)

Explanation:

If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.

Answer:

a) v = -30 - 32 t ,  s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s

c) v (1) = -62 ft / s,  v (3) = -126 ft / s , d) t = 7.13 s , e)  v = -258.16 ft / s

Explanation:

a) For this exercise they give us the function of the position of the ball

          s (t) = s (o) + v_o t - 16 t²

notice that you forgot to write the super index

indicate the initial position of the ball

        s (o) = 600 ft

also indicates initial speed

        v_o = - 30 ft / s

let's substitute in the equation

        s (t) = 600 - 30 t -16 t²

to find the speed we use

       v = ds / dt

       v = v_o - 32 t

       v = -30 - 32 t

b) To find the average speed, look for the speed at the beginning and end of the time interval

t = 1 s

     v (1) = -30 -32 1

     v (1) = - 62 ft / s

t = 3 s

     v (3) = -30 -32 3

     v (3) = -126 ft / s

the average speed is

    v = (v (3) -v (1)) / (3-1)

    v = (-126 +62) / 2

    v = -32 ft / s

c) instantaneous speeds, we already calculated them

    v (1) = -62 ft / s

    v (3) = -126 ft / s

d) the time to reach the ground

in this case s = 0

    0 = 600 - 30 t -16 t²

     t² + 1,875 t - 37.5 = 0

we solve the quadratic equation

     t = [-1,875 ±√ (1,875² + 4 37.5)] / 2

     t = [1,875 ± 12.39] / 2

     t₁ = 7.13 s

     t₂ = negative

Since the time must be positive, the correct answer is t = 7.13 s

e) the speed of the ball on reaching the ground

     v = -30 - 32 t

     v = -30 - 32 7.13

      v = -258.16 ft / s

Answer:

V = 0.0283 m³ = 28300 cm³

T₂ = 1200 K

Explanation:

The volume of the gas can be determined by using General Gas Equation:

PV = nRT

where,

P = Pressure of Gas = (72 cm of Hg)(1333.2239 Pa/cm of Hg) = 95992.12 Pa

V = Volume of Gas = ?

n = no. of moles = mass/molar mass = (35 g)/(32 g/mol) = 1.09 mol

R = General Gas Constant = 8.314 J/ mol.k

T = Temperature of Gas = 27°C + 273 = 300 k

Therefore,

(95992.12 Pa)(V) = (1.09 mol)(8.314 J/mol.k)(300 k)

V = 2718.678 J/95992.12 Pa

V = 0.0283 m³ = 28300 cm³

The Kinetic Energy of gas molecule is given as:

K.E = (3/2)(KT)

Also,

K.E = (1/2)(mv²)

Comparing both equations, we get:

(3/2)(KT) = (1/2)(mv²)

v² = 3KT/m

v = √(3KT/m)

where,

v = r.m.s velocity

K = Boltzamn Constant

T = Absolute Temperature

m = mass of gas molecule

At T₁ = 300 K, v = v₁

v₁ = √(3K*300/m)

v₁ = √(900 K/m)

Now, for v₂ = 2v₁ (double r.m.s velocity), T₂ = ?

v₂ = 2v₁ = √(3KT₂/m)

using value of v₁:

2√(900 K/m) = √(3KT₂/m)

4(900) = 3 T₂

T₂ = 1200 K

Answer:

The ratio is  [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       [tex]F_{T2} = ma[/tex]

 Considering the first car

 The force causing it s movement  is mathematically represented as

      [tex]F = F_{T1} -F_{T2} = ma[/tex]

=>   [tex]F_{T1} -ma = ma[/tex]

=>   [tex]F_{T1} = 2 ma[/tex]

=> [tex]\frac{F_{T1}}{ma} = 2[/tex]

=> [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]

Answer:

d. 1.00 m

Explanation:

In 1905, Einstein proposed special theory of relativity of light.

This theory had a number of consequences or results. One of them is called "Length Contraction".

According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.

But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.

In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:

d. 1.00 m

Answer:

[tex]\large \boxed{42\, \mu \text{C}}$[/tex]

Explanation:

The formula for the force exerted between two charges is

[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

[tex]F=k\dfrac{q^{2}}{r^2}[/tex]

[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex]     (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

[tex]E_p=h\frac{c}{\lambda}[/tex]         (2)

c: speed of light = 3*10^8 m/s

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

[tex]K=E_p-\Phi[/tex]     (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]

You replace in the equation (3):

[tex]K=1.95eV-1.7eV=0.25eV[/tex]

The kinetic energy of the electrons emitted by the metal is 0.25 eV

(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]

(2). The maximum kinetic energy of the electron is 0.37eV.

(1). The power of light is given as,

            [tex]P=1.4*10^{-6}W[/tex]

    Energy is given as,

             [tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]

Number of photons per second is,

                    [tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]

(2). the maximum kinetic energy of the electron is,

              [tex]K.E=E-\phi[/tex]

Where [tex]\phi[/tex] is work function.

         [tex]K.E=2.07-1.7=0.37eV[/tex]

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Answer:

It will lose electrical potential energy.

Explanation:

A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.

Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2  V² = g x 2r + 1/2 v²

 V² = g x 4r +  v²

 V² = 9.8 x 4 +  8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T =   mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T  - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

Answer:

B. revolve around the sun

   rotate on an axis

   generally have moons

Explanation:

edge 2021

Answer:

12.0 V

Explanation:

Data :

Potential difference due to a single charge (+Q), E = 3.0 V

The Electric potential for the system of charges is given as:

[tex]E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}][/tex]

for single charge, E = 3.0 V = [tex]\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}][/tex]  ->eq(1)

And for 4 charges:

[tex]E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}][/tex] -eq(2)

from eq(1) and (2) we have

E = 4 × 3.0 V = 12 V

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

Answer:

11.8 Joules

Explanation:

Given:-

- The height of the target ball, si = 0.860 m

- The mass of target and steel ball, m = 0.012 kg

- The target ball travels a distance ( x ) after being struck = 1.40 m

Find:-

What is the kinetic energy (in joules) of the target ball just after it is struck?

Solution:-

- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.

- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:

                        [tex]vf^2 = vi^2 - 2*g*x[/tex]

Where,

                  vf: The final velocity of target ball at maximum height = 0

                  g: The gravitational acceleration constant = 9.8 m/s^2

- Plug in the required parameters and evaluate the ( vi ) as follows:

                      [tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]

- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:

                       [tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]

Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.

Answer:

Explanation:

Moment of inertia of the rod = 1/12 m L²

m is mass of the rod and L is its length

= 1/2 x 2.3 x 2 x 2

= 4.6 kg m²

Moment of inertia of masses attached with the rod

= m₁ d² + m₂ d²

m₁ and m₂ are masses attached , and d is their distance from the axis of rotation

= 5.3 x 1² + 3.5 x 1²

= 8.8 kg m²

Total moment of inertia = 13.4 kg m²

B )

Rotational kinetic energy = 1/2 I ω²

I is total moment of inertia and ω is angular velocity

= .5 x 13.4 x 2²

= 26.8 J .

C )

when mass of rod is negligible , moment of inertia will be due to masses only

Total moment of inertia of masses

= 8.8 kg m²

D )

kinetic energy of the system

= .5 x 8.8 x 2²

= 17.6 J .

(A) Total moment of inertia is 13.4 kgm²

(B) Total kinetic energy is 26.8J

(C) Moment of inertia is  8.8 kgm²
(D) Kinetic energy is 17.6J

(A) The moment of inertia of the rod is given by:

I = 1/12 mL²

where m is the mass of the rod

and L is the length

So,

I = (1/12) × 2.3 × 2²

I = 4.6 kgm²

Now, the moment of inertia of masses attached to the rod is given by:

I' = m₁ d² + m₂d²

where m₁ and m₂ are masses

and d is their distance from the axis of rotation

I' = 5.3 × 1² + 3.5 × 1²

I' = 8.8 kgm²

The total moment of inertia of the system is given by:

I(tot) = I + I'

I(tot) = 13.4 kgm²

(B) The rotational kinetic energy of an object with a moment of inertia I and angular velocity ω is given by:

KE = 1/2 I(tot)ω²

KE = 0.5 × 13.4 × 2²

KE = 26.8J

(C) If the mass of the rod is negligible, then the moment of inertia of the rod will be zero. So the total moment of inertia will be

I(tot) = I' = 8.8 kgm²

(D) the kinetic energy of the system when the mass of the rod is negligible and the angular speed is 2 rad/s is given by:

KE = 1/2 I'ω²

KE = 0.5 × 8.8 × 2²

KE = 17.6J

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Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.

Since

number of turns, N = 200 turns

cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

the magnitude of magnetic field strength, B = 30 x 10⁻³ T

the maximum value of the induced emf, E = 8 V

Now

Maximum induced emf should be

E = NBAω

here,

ω is angular velocity (ω = 2πf)

Now

E = NBA2πf

here,

f is the minimum frequency

So,

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz.

Learn more about frequency here: https://brainly.com/question/24470698

Answer:

The electric flux is  [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]

Explanation:

From the question we are told that

  The magnitude of the electric point charge  [tex]q = 22.5 nC = 22.5 *10^{-9} \ C[/tex]

   The length of the one side of the cube is [tex]l = 16.3 \ cm = 0.163 \ m[/tex]

   The number of  sides is  [tex]N= 6[/tex]  

The electric flux according to Gauss law is mathematically evaluated as

          [tex]\phi = \frac{q}{\epsilon_o}[/tex]

Where [tex]\epsilon _ o[/tex] is the permitivity of free space with value  [tex]\epsilon_o = 8.85*10^{-12}\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

  substituting values

        [tex]\phi = \frac{22.5 *10^{-9}}{8.85 *10^{-12}}[/tex]

         [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]

Answer:

the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔

Explanation:

i hope this will help you

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

Answer:

0.981 rad/sec^2

Explanation:

mass that pulls on string = 5 kg

weight due to mass = 5 x 9.81 = 49.05 N

radius of rod = 0.1 m

torque produced by this force on the rod = force x radius

torque = 49.05 x 0.1 = 4.905 N-m

mass of disk = 125 kg

radius of disk = 0.2 m

moment of inertia of the disk I = m[tex]r^{2}[/tex]

I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2

from the equation, T = Iα

where T is torque

I is moment of inertia

α is angular acceleration

imputing values,

4.905 = 5α

α = 4.905/5 = 0.981 rad/sec^2