A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)

For the first question

Answers:

b.) All these technologies use radio waves, including high-frequency microwaves

d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.

e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.

f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.

For second question

Answer:

B. The earth absorbs visible light and emits radiation with a longer wavelength

C. The earth absorbs visible light and emits radiation with a lower frequency.

Both options are correct

Explanation:

In the EM spectrum, wavelenght reduces with increasing energy and frequency increases with increasing energy. The order of increasing energy is radio wave, microwave, infrared rays, visible rays, ultraviolet rays, xrays, gamma rays.

If the earth absorbs visible rays of higher energy, and radiates infrared rays of lower energy, then this means that the radiated radiation will have a longer wavenght, and a lower frequency.

1) The statements that correctly describe the various applications given in the question are;

B. All these technologies use radio waves, including high-frequency microwaves

D. Microwave ovens emit in the same frequency band as some wireless Internet devices.

E. The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies.

F. All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m.

2) The correct statements on the EM spectrum in the question are;

B. The earth absorbs visible light and emits radiation with a longer wavelength.

C. The earth absorbs visible light and emits radiation with a lower frequency.

1) We are told about different waves being used by different technologies such as;

- Radio and TV emitting radio waves

- microwave oven using microwaves to cook food

- Dentists checking teeth with x-rays

All these waves are electromagnetic waves.

Looking at the options, options B, D, E and F are true because;

2) The correct options are Options B and C because;

In the EM radiation spectrum, when the energy increases, the wavelength usually reduces and the frequency usually increases. The order of  radio waves according to increase in energy is;

radio waves < infrared radiation < visible rays < ultraviolet rays <  xrays < gamma rays.

Now, if we say that the earth absorbs visible rays of higher energy, and radiation of infrared rays with lower energy, then from our definition, we can conclude that radiated waves will have a longer wavelength, and a lower frequency.

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Answer:

The answer is "[tex]1.94 \ m^2[/tex]".

Explanation:

Formula:

[tex]F= \frac{1}{2} \times C_D \times density \times area \times velocity^2\\\\Power (P) = F \times velocity \\\\P = \frac{C_D}{2} \times density \times area \times velocity^3\\[/tex]

Given value:

[tex]\ P = 10.25 \ W \\\\\ density = 1.2 \ kg/m^3 \\\\\ velocity= 2.5 \\\\[/tex]

[tex]10.25 = \frac{C_D A}{2} \times 1.2 \times 2.5^3\\\\C_D A= \frac{10.25 \times 2 }{1.2 \times 2.5^3}\\\\C_D A = 1.94 m^2\\[/tex]

Answer:

The total angular momentum is 292.59 kg.m/s

Explanation:

Given that :

Rotation of the horizontal circular platform [tex]\omega[/tex] = 0.919 rad/s

mass of the platform (m) = 90.7 kg

radius (R) = 1.91 m

mass of the poodle [tex]m_p[/tex] = 20.5 kg

Your mass  [tex]m'[/tex] = 73.5 kg

speed v = 1.05 m/s with respect to the platform

[tex]V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}[/tex]

r = 0.955

Mass of the mutt [tex]m_m[/tex] = 18.5 kg

[tex]r' = \frac{3}{4} \ R[/tex]

Your angular momentum is calculated as:

Your angular velocity relative to the platform is [tex]\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s[/tex]

[tex]Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s[/tex]

[tex]I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2[/tex]

[tex]L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s[/tex]

For poodle :

[tex]Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s[/tex]

Actual [tex]\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s[/tex]

[tex]I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2[/tex]

[tex]L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s[/tex]

[tex]I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2[/tex]

[tex]L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s[/tex]

Disk [tex]I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2[/tex]

[tex]L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s[/tex]

Total angular momentum of system is:

L = [tex]L_D +L_Y+L_P+L_M[/tex]

= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s

= 292.59 kg.m/s

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

To calculate the horsepower of the engine required for 30 skiers to be pulled up a hill, we need to consider the work done against friction and gravity. By plugging in the given values into the appropriate equations, we can find the required horsepower of the engine.

To calculate the horsepower of the engine required for 30 skiers to be pulled up a hill, we need to consider the work done against friction and gravity. First, we find the net force of the skier parallel to the incline by subtracting the force due to friction from the component of the skier's weight parallel to the incline. Then, we calculate the work done against friction and gravity using the formula W = Fd. Finally, we convert the work done to horsepower using the conversion factor 1 hp = 746 W.

The work done against friction and gravity is:

W = F_net * d

The power required is:

P = W / t

And finally, we convert the power to horsepower:

Power (hp) = P / 746

By plugging the given values into these equations and multiplying the final power by 30 (for 30 skiers), we can find the required horsepower of the engine.

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The mutual inductance of a Tesla coil system with two solenoids will double if the length and number of turns of the primary solenoid are both doubled, assuming all other factors remain constant.

When we are given a Tesla coil configuration with a solenoid of length l and cross-sectional area A, closely wound with N1 turns of wire, and another coil with N2 turns surrounding it at its center, we can calculate the mutual inductance based on the properties of the solenoids. If a new solenoid is introduced that has twice as many turns and is twice as long, the mutual inductance M of the system will be affected.

To understand how the mutual inductance changes, let's remember that for a closely wound solenoid, the mutual inductance can be calculated by a formula incorporating the number of turns, permeability of the core material, cross-sectional area, and the length of the solenoid. The mutual inductance is directly proportional to the product of the number of turns of each coil, the magnetic permeability of the core, and the area of the cross-section, and inversely proportional to the length of the solenoid. Therefore, if we double the length l and the number of turns N1 while keeping all other factors constant, the mutual inductance will also double.

Answer:

Explanation:

The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.

Applying Ampere's law :-

∫ B dl = μ₀ I      ; I is current passing through ampere's loop

B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ T.

Answer:

l = 10.16 m

Explanation:

In this case, we have the period of oscillation of the pendulum is 6.4 s. It is required to find the height of the tower.

We know that the pendulum executes SHM. Let l is the height of the tower. The time period of simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

g is acceleration due to gravity

We need to rearrange the above equation such that,

[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(6.4)^2\times 9.8}{4\pi ^2}\\\\l=10.16\ m[/tex]

So, the height of the tower is 10.16 m.

Answer:

The diameter of the second pipe is [tex]D_2 = 1.095 D[/tex]

Explanation:

From the question we are told that

       The length of the connected pipe is  [tex]d = 2L[/tex]

        The pressure drop for the first pipe is  [tex]\Delta p __{1}} = 1.44* \Delta p__{2}}[/tex]

        The diameter of the pipe is  [tex]D[/tex]

The rate at which the fluid flows for laminar flow is mathematically represented as

          [tex]\r m = \frac{\pi D^4 \Delta p}{128 \mu L}[/tex]

Where L is the length of the pipe

           [tex]\mu[/tex] is the dynamic viscosity

          [tex]\Delta p[/tex] is the difference in pressure

          [tex]\r m[/tex] is the flow rate of the fluid

From the equation of continuity

         [tex]\r m_ 1 = \r m_2[/tex]

Where [tex]\r m_1[/tex] is the flow rate in pipe one

           [tex]\r m_2[/tex] is the flow rate in pipe two

So

        [tex]\frac{\pi D^4 \Delta p_1}{128 \mu L} = \frac{\pi D^4_2 \Delta p}{128 \mu L}[/tex]

Where [tex]D_2[/tex] is the diameter of the second pipe

=>     [tex]\frac{\pi D^4 (1.44 \Delta p_2)}{128 \mu L} = \frac{\pi D^4_2 \Delta p}{128 \mu L}[/tex]

=>   [tex]1.44 D^4 = D_2 ^4[/tex]

       [tex]D_2 =\sqrt[4]{ \frac{D ^4 }{1.44 } }[/tex]

      [tex]D_2 = 1.095 D[/tex]

The equation modeling the exponential decay of the Beryllium-11 isotope is 50 = 800(1/2)^(t/14). This lets you find out the time that has passed since there were initially 800 atoms.

The situation described is an example of exponential decay, modeled by the formula N = N0(1/2)^(t/h). In this case, N0 is the original number of atoms (800), N is the remaining number of atoms (50), t is the time that has passed, and h is the half-life of the isotope (14 seconds for Beryllium-11).

So the equation modelling this scenario is 50 = 800(1/2)^(t/14). This equation will let you solve for the time t (in seconds) that has passed since the sample initially contained 800 atoms of Beryllium-11.

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Answer:3m/s

Explanation:

mass(m)=50kg

Kinetic energy =225 joules

Kinetic energy=(m x (velocity)^2)/2

225=(50 x (velocity)^2)/2

Cross multiplying we get

225x2=50x(velocity)^2

450=50x(velocity)^2

Divide both sides by 50

450/50=(50x(velocity)^2)/50

9=(velocity)^2

Taking the square root of both sides we get

√9=velocity

3=velocity

Velocity=3m/s

Answer:

The 500 watts motor

Explanation:

Power is the workdone per unit time, it can be expressed as;

P = VI

P = I^2 × R

Where;

P = power

V = potential difference

I = current

R = resistance

Making current the subject of formula;

I = P/V

I^2 = P/R

Since they are both connected to the same standard home outlet. The potential difference across them is thesame( V is constant). Therefore the higher the power the higher the current.

I ∝ P

So the electrical motors with higher power 500 watts draws greater current.

The slope of the line is calculated by dividing the change in distance over the change in time.

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop [tex]A=\pi r^2[/tex]

[tex]A=3.14\times 0.045^2=0.00635m^2[/tex]

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field [tex]dB=250-350=100mT[/tex]

Emf induced is given by

[tex]e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}[/tex]

[tex]e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V[/tex]

Magnitude of induced emf is equal to 0.00635 V

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The minimum mass of [tex]M_1 = 90\ kg[/tex] correct option is  E

Explanation:

 Free body diagram of the set up  in the question is shown on the third uploaded image

  The mass of board is  [tex]M = 60kg[/tex]

   The length of the board is [tex]L = 6 \ m[/tex]

    The length extending over the edge is [tex]L_e = 4 \ m[/tex]

    The second mass is  [tex]M_2 = 30kg[/tex]

Now to obtain [tex]M_1[/tex] we take moment about the edge of the platform

               [tex]M_1 g L_1 = Mg \frac{L}{2} + M_2 g L_2[/tex]

              [tex]M_1 L_1 = M \frac{L}{2} + M_2 L_2[/tex]

  Substituting value  

               [tex]M_1 (2) = (60)(1) + (30)(4)[/tex]

               [tex]M_1 = 90 \ kg[/tex]

The minimum value of M1 needed to keep the board from falling off the platform is 90 kg.

From the information given, we are to find:

Given that:

Consider the center of gravity in the board that lies at the length of the board midpoint.

Then, the distance (D) of the gravity center from the platform end = 3 - 2

= 1 m

∴

Considering the moment about the platform end, the mass (M1) placed on the left side edge of the board can be computed as:

[tex]\mathbf{M_1gL_1 = MgD + M_2gL_2} \\ \\ \mathbf{M_1L_1 = MD + M_2gL_2} \\ \\ \mathbf{M_1(2) = 60 \ kg \times 1 + 30 \ kg \times (4)} \\ \\ \mathbf{ M_1 =\dfrac{60 \ kg + 120 kg }{2} } \\ \\ \mathbf{ M_1 =\dfrac{180 \ kg}{2} } \\ \\ \mathbf{ M_1 =90 \ kg }[/tex]

Therefore, we can conclude that the minimum value M1 needed to keep the board from falling off the platform is 90 kg.

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Answer: Vertical displacement = -27.6m

And takes 2.375 s

A pumpkin thrown at a horizontal speed of 4.0 m/s and travels 9.5 m horizontally before hitting the ground will take 2.375 seconds to hit the ground.

Given the following information:

Horizontal speed of the pumpkin = 4.0 m/s

Horizontal distance traveled before hitting the ground = 9.5 m

Ignoring air resistance

Use the following formula to calculate the time it takes the pumpkin to hit the ground:

time = horizontal distance / horizontal speed

time = 9.5 m / 4.0 m/s = 2.375 seconds

Therefore, it takes the pumpkin 2.375 seconds to hit the ground.

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Answer:

Decrease the voltage,and do not change the resistance,the current will also decrease

Explanation:

Decrease the voltage,and do not change the resistance, the current will also decrease, because voltage is directly proportional to current

Answer:

B.If we decrease the voltage, and do not change the resistance, then the current will also decrease.

Explanation:

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Answer:

a) 37.8 W

b) 2 Nm

Explanation:

180 g = 0.18 kg

We can also convert 180 revolution per minute to standard angular velocity unit knowing that each revolution is 2π and 1 minute equals to 60 seconds

180 rpm = 180*2π/60 = 18.85 rad/s

We can use the heat specific equation to find the rate of heat exchange of the steel drill and block:

[tex]\dot{E} = mc\Delta \dot{t} = 0.18*420*0.5 = 37.8 J/s[/tex]

Since the entire mechanical work is used up in producing heat, we can conclude that the rate of work is also 37.8 J/s, or 37.8 W

The torque T required to drill can be calculated using the work equation

[tex]E = T\theta[/tex]

[tex]\dot{E} = T\dot{\theta} = T\omega[/tex]

[tex]T = \frac{\dot{E}}{\omega} = \frac{37.8}{18.85} = 2 Nm[/tex]

Calculating the rate of working of a drill involves using the specific heat capacity, mass of the block, and rate of temperature increase to find the power. The torque can be derived from this power and the angular velocity of the drill.

The question involves calculating the rate of working of the drill in Watts and the torque required to drive the drill based on the information that all mechanical work is converted to heat that raises the temperature of the steel block and drill at 0.5 0C/s. Given the mass of the steel block and drill is 180 g and the specific heat capacity of steel is 420 J/(kgK), we can use the formula power (P) = mcigtriangleup T/igtriangleup t, where m is the mass, c is the specific heat capacity,  is the temperature change, and  is the change in time. To find the torque, we can use the power-torque relationship P = au imes w, where P is power in watts, au is the torque in Newton-meters, and w is the angular velocity in radians per second. As the drill makes 180 revolutions per minute (rpm), we'll convert that to radians per second to use in the power-torque equation.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The gauge pressure that water has at the House A  [tex]P_A = 257020.68 Pa[/tex]

The gauge pressure that water has at the House B  [tex]P_B = 188454 \ Pa[/tex]

Explanation:

From the question we are told that

    The mass of water when full is  [tex]m_f = 7.41* 10^{5} kg[/tex]

Generally the volume of water in this tank is mathematically represented as

              [tex]V = \frac{m }{\rho}[/tex]

Where  [tex]\rho[/tex] is the density of water with a value of with a value of [tex]\rho = 1000 kg /m^3[/tex]

   substituting values

                  [tex]V = \frac{7.41 *10^5}{10^3}[/tex]

                  [tex]V = 741 m^3[/tex]

This volume is the volume of a sphere since the tank is spherical so

            [tex]V = \frac{4 \pi ^3}{3}[/tex]

  making r the subject of the formula

           [tex]r =\sqrt[3]{ \frac{741 *3 }{4\pi} }[/tex]              

        [tex]r = 5.6134 m[/tex]

Now we can use this parameter to obtain the diameter

  So

        [tex]d = 2 * r[/tex]

substituting values

        [tex]d = 2 * 5.6134[/tex]

        [tex]d = 11.23m[/tex]

The pressure  the water has at  faucet in House A is mathematically evaluated as

        [tex]P_A = \rho g h_A[/tex]

This height is obtained as follows

                       [tex]h_A = d+ 15[/tex]

The value 15 is gotten from the diagram

  so

          [tex]h_A = 15 + 11.23[/tex]

          [tex]h_A = 26.22 m[/tex]

Now substituting values

         [tex]P_A = 26.23 * 9.8 * 1000[/tex]

         [tex]P_A = 257020.68 Pa[/tex]

      The pressure  the water has at  faucet in House B is mathematically evaluated as

        [tex]P_B = \rho g h_B[/tex]

This height is obtained as follows

                       [tex]h_B = d+ 15[/tex]

The value 15 is gotten from the diagram

  so

          [tex]h_B = d + 15 -h[/tex]

substituting values

         [tex]h_B =11.23 + 15 -7[/tex]

          [tex]h_A = 19.23 m[/tex]

Now substituting values

         [tex]P_B = 19.23 * 9.8 * 1000[/tex]

         [tex]P_B = 188454 \ Pa[/tex]

Answer:

The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ   = 3472.

Explanation:

Refractive index of Glass (given) = 1.5

For the case of a constructive interference,

2nt = (m + 1/2) λ

For case 1,

2nt = (m + 1/2) 496 nm

For case 2,

2nt = (m +1+ 1/2) 386 nm

2nt = (m+3/2) * 386 nm

(m + 1/2) 496 nm = (m+3/2) * 386 nm

m = 3

Inserting the value of m in 1.

2nt = (m + 1/2) 496 nm

2*1.5t = (3 + 1/2) * 496 nm

t = ((3 + 1/2) * 496 nm)/ 3

t = 578.6 nm

The thickness of the glass, t = 578.6 nm

b)

It is generally known that for constructive interference,

2nt = (m + 1/2) λ

λ = 2nt / ((m + 1/2))

For Longest Wavelength, m = 0

λ = 2*1.5*578.6/ (1/2)

λ = 3472 nm

Answer:

The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]

Explanation:

From the question we are told that

     The distance of the point source  from the screen is  [tex]d = 1.0 m[/tex]

      The length of a side of the  first square hole is  [tex]L_1 = 0.020 \ m[/tex]

      The distance of the cardboard from the point source is [tex]D_1 = 0.50\ m[/tex]

   The distance of the second cardboard from the point source is [tex]D_2 = 0.25 \ m[/tex]

Let take the  [tex]\alpha_{max }[/tex] as  the angle at which the light is passing through the edges of the cardboards square hole

     Since the bright square casted on the screen by both  square holes on the   individual cardboards are then it means that

              [tex]\alpha_{max} __{1}} = \alpha_{max} __{2}}[/tex]

This implies that

             [tex]tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})[/tex]      

Looking at this from the SOHCAHTOA concept

               [tex]tan (\alpha_{max} __{1}}) = \frac{opposite}{Adjacent}[/tex]

     Here opposite is  the length of the side of the  first cardboard square hole

     and    

      Adjacent is  the  distance of the from the  first cardboard square hole to the point source

And for  

            [tex]tan (\alpha_{max} __{2}}) = \frac{opposite}{Adjacent}[/tex]

    Here opposite is  the length of the side of the  second  cardboards square hole (let denote it with [tex]L_2[/tex])

and

Adjacent is the distance of the from the  second  cardboards square hole to the point source

         So

                 [tex]tan (\alpha_{max} __{1}}) = \frac{0.020}{0.50}[/tex]

         And  

                [tex]tan (\alpha_{max} __{2}}) = \frac{L_2}{0.25}[/tex]

Substituting this into the above equation

                 [tex]\frac{0.020}{0.50} = \frac{L_2}{0.25}[/tex]

Making [tex]L_2[/tex] the subject

                   [tex]L_2 = \frac{0.25 *0.020}{0.50}[/tex]

                 [tex]L_2 = 0.01m[/tex]

Since it is a square hole the sides are the same hence

The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]

Answer:

L = 182.4 m      

Explanation:

Given:-

- The number of turns of the coil, N = 50

- The shape of the coil = square

- The angle between the coil and magnetic field, θ = 30°

- The change in magnetic field, ΔB = ( 700 - 250 ) μT

- The time duration in which magnetic field changes, Δt = 0.3 s

- The induced emf, E = 60.0 mV

Solution:-

- The problem at hand is an application of Faraday's law. The law states that the induced emf ( E ) is proportional to the negative rate of change of magnetic flux ( ΔФ / Δt ) and number of turns of the coil ( N ).

- The Faraday's law is mathematically expressed as:

                    E =  - N* ( ΔФ / Δt )

Where,

- The flux ( Ф ) through a current carrying with an cross-sectional area ( A ) at a normal angle ( θ ) to the direction of magnetic field ( B ) is given by the following relationship.

                    Ф = B*A*cos ( θ )

- We need the rate of change of magnetic flux ( ΔФ / Δt ) for the Faraday's law. I.e the induced emf ( E ) is proportional to rate of change in magnetic field ( ΔB / Δt ), rate of change of angle between the coil and magnetic field ( Δθ / Δt ) or rate of change of cross-sectional area of the coil under the influence of magnetic field.

- To determine the exact relationship. We will derive the multi-variable function of flux ( Ф ) with respect to time "t":

                     Ф ( B , A , θ ) = B*A*cos ( θ )

- The first derivative would be ( Use chain and product rules )

    ( ΔФ / Δt ) = ΔB / Δt*A*cos ( θ ) + B*ΔA/Δt*cos ( θ ) - B*A*sin ( θ )*Δθ/Δt

- For the given problem the only dependent parameter that is changing is magnetic field ( B ) with respect to time "t". Hence, ( ΔA/Δt = Δθ/Δt = 0 ):

                        ΔФ / Δt  = (ΔB/Δt)*A*cos ( θ )

- Substitute the rate of change of magnetic flux  ( ΔФ / Δt ) into the expression for Faraday's Law initially stated:

                        E =  - N*(ΔB/Δt)*A*cos ( θ )

- Plug in the values and evaluate the Area of the square coil:

                       A =  - E / ( N*(ΔB/Δt)*cos ( θ ) )

                       A = - 0.06 / ( 50*[ (250-700)*10^-6/0.3 ] *cos ( 30° ) )

                       A = - 0.06 / -0.07216

                       A = 0.8314 m^2

- The square coil has equal sides ( x ). The area of a square A is given by:

                      A = x^2

                      x = √0.8314

                      x = 0.912 m

- The perimeter length of a single coil in terms of side length "x" is given as:

                      P = 4x

Whereas for a coil of N turns the total length ( L ) would be:

                      L = N*P

                      L = 4Nx

                      L = 4 * 50 * 0.912

                      L = 182.4 m                 ... Answer

Answer:20 cm

Explanation:

Given

Refractive index of material [tex]n=1.55[/tex]

Outer radius [tex]R_1=1.98\ cm[/tex]

Inner radius [tex]R_2=2.42\ cm[/tex]

using lens maker formula

[tex]\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})[/tex]

[tex]\frac{1}{f}=(1.55-1)(\frac{1}{1.98}-\frac{1}{2.42})[/tex]

[tex]f=\frac{10.895}{0.55}[/tex]

[tex]f=19.81\approx 20\ cm[/tex]

Answer:

towards the object

Explanation:

Final answer:

To balance the forces on the 1 kg object, a force should be applied in the downward direction.

Explanation:

The forces on the 1 kg object can be balanced by applying a force in the opposite direction to the net force acting on the object. In this case, the net force is the sum of the weight of the object and the tension in the string. Since the weight acts downward and the tension in the string acts upward, the force should be applied in the downward direction to balance the forces on the 1 kg object.

Answer:

It is perceived as louder.

Explanation:

Amplitude affects the loudness of sound.

Frequency and wavelength affect the pitch of the sound.

Answer:

Explanation:

Half width - 3.75 m

Rainfall intensity - 90 mm/hr

Longitudinal slope - 0.75%

Cross slope - 2.25%

Allowable limit of gutter flow, Q road = (Design constant x Intensity of rain x Area) / 360

= (0.91×90×(9×L1×10∧-4)/360 = 0.000204 L1

Using the standard design chart, allowable limit of gutter flow =0.018 m3/s

Therefore,0.018 = 0.000204 L1

L1 = 0.018 / 0.000204 = 88.24

Inlet spacing to be adapted is 88m

Answer:

As the Ballon pulls up, the distance between the Ballon and the center of the earth increases, the gravitational pull reduces and the gravity potential energy increases.

Answer:

W_apparent = 93.1 kg

Explanation:

The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.

            W_apparent = W - B

The push is given by the expression of Archimeas

            B = ρ_fluide g V

            ρ_al = m / V

            m = ρ_al V

we substitute

            W_apparent = ρ_al V g - ρ_fluide g V

            W_apparent = g V (ρ_al - ρ_fluide)

we calculate

           W_apparent = 980 50 (2.7 - 0.8)

           W_apparent = 93100 g

            W_apparent = 93.1 kg

The plants that are typically the first to live in soil are known as pioneer plants or pioneer species.

Pioneer plants or pioneer species are the plants that are often the first to live in soil. These plants are distinguished by their propensity to colonize bare or disturbed soil, such as on newly formed land or in the wake of a natural disaster.

The process of primary succession, which is the gradual formation of plant and animal communities in a region devoid of soil or organic matter, depends critically on pioneer plants.

Learn more about plants:https://brainly.com/question/30072240

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The true statement is A. A cylindrical capacitor is a parallel-plate capacitor rolled into a tube, C. The charge on a capacitor increases quickly at first, then much more slowly as the capacitor charges E. One of the principal purposes of a capacitor is to store electric potential energy. F. A capacitor charges rapidly when connected to an RC circuit with a battery and the false statement are B. The dielectric constant indicates that the distance by which the two plates of a capacitor are separated and D. The voltage across a capacitor in an RC circuit increases linearly during charging.

Let's look at each statement one by one to categorize them as true or false.
a. True. A cylindrical capacitor can be thought of as a parallel-plate capacitor with the plates rolled into cylindrical shapes.

b. False. The dielectric constant is a measure of a material's ability to increase the capacitance of a capacitor, not a measure of the distance between the plates.

c. True. The charge on a capacitor follows an exponential curve, increasing rapidly at first and then more slowly as it approaches its maximum charge.

d. False. The voltage increases exponentially, not linearly, when charging a capacitor in an RC circuit.

e. True. Capacitors store electric potential energy in the electric field between their plates.

f. True. Initially, the capacitor in an RC circuit charges quickly, but the rate of charging decreases over time as it gets closer to full charge.