A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid

Answers

Answer 1

Answer:

The refractive index of fluid 2 is 1.78

Explanation:

Refractive index , n = real depth/apparent depth

For the first fluid, n = 1.37 and apparent depth = 9.00 cm.

The real depth of the container is thus

real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm

To find the refractive index of fluid index of fluid 2, we use the relation  

Refractive index , n = real depth/apparent depth.

Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.

So, n = 12.33 cm/6.86 cm = 1.78

So the refractive index of fluid 2 is 1.78

Answer 2

Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8

Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,

Then,

Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]

Real depth = Index of refraction x apparent depth

Since the same container is used, we can make an assumption that;

real depth of fluid 1 = real depth of fluid 2

That is,

1.37 x 9 = n x 6.86

Where n = Index of refraction for the second fluid.

make n the subject of formula

n = 12.33 / 6.86

n = 1.79

Therefore, the index of refraction of the second fluid is 1.8 approximately.

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Related Questions

An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answers

Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material

Answers

Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]

Explanation:

Given: Speed of red light = 700 nm

= [tex]700\times10^{-9}[/tex] m

[tex]= 7\times10^{-7}[/tex] m

Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]

Speed of light = [tex]3\times10^8[/tex] m

Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]

[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]

Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]

The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].

what is transmission of heat?​

Answers

Answer:

Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are

Conduction ConvectionRadiation.

Example of Conduction:

Heating a metal

Example of Convection:

Sea Breeze

Example of Radiation:

Sun

Hope this helps ;) ❤❤❤

Answer:

Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.

There are three(3) different ways heat can transfer and they are:

a) Conduction (through direct contact).

b) Convection (through fluid movement).

c) Radiation (through electromagnetic waves).

Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).

2. Baking a pie in an oven(radiation).

Hope it helps!!Please mark me as the brainliest!!!

Thanks!!!!

what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates

Answers

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

             

the electric force is

             F_{e} = q E

   

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

         

               m = 2e E / g

     

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

 

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.

Answers

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = [tex]\frac{W}{F}[/tex]

d = [tex]\frac{3}{9}[/tex]

d = 0.33 m

The distance d the block ice moved is 33 cm.

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).

Answers

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases

Answers

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?

Answers

Answer:

72J

Explanation:

distance moved is equal to 3m.then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

The answer is 72J.

Distance moved is equal to 3m.

Then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

Is there any definition of force?

A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

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Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?

Answers

Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.

Required:
a. How much work is done by the tension?
b. How much work is done by gravity?

Answers

B)is pills everything to the surface of the earth not sure about A

The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

The given parameters;

mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 N

The work done by the tension is calculated as follows;

W = Fd

W = 15 x 0.15

W = 2.25 J

The work done by gravity is calculated as;

W = (25 x 9.8) x 0.15

W = 36.75 J

Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

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if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c​

Answers

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/s

The angle between the axes of two polarizing filters is 41.0°. By how much does the second filter reduce the intensity of the light coming through the first?

Answers

Answer:

The  amount by which the second filter reduces the intensity of light emerging from the first filter is

     z =  0.60

Explanation:

From the question we are told that

    The angle between the axes is  [tex]\theta = 41^o[/tex]

The intensity of polarized light that emerges from the second filter is  mathematically represented as

         [tex]I= I_o cos^2 \theta[/tex]

 Where [tex]I_o[/tex] is the intensity of light emerging  from the first filter

        [tex]I = I_o [cos(41.0)]^2[/tex]

      [tex]I =0.60 I_o[/tex]

This means that the second filter reduced the intensity by z =  0.60

           

A ​46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this​ causeway?

Answers

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

Angle and Force

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), [tex]F_t[/tex] = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

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You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2

Answers

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 u s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 u s  = 75 n s

TL = T - Th = 100 ns - 75 n s = 25 n s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?

Answers

Answer:

0.46N

Explanation:

See attached file

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:

Answers

Answer:

5.19 x 10³Hz

Explanation:

The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;

[tex]X_C = \frac{1}{2\pi fC }[/tex]

Where;

f = frequency of the signal through the capacitor

C = capacitance of the capacitor.

Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;

V = I x [tex]X_{C}[/tex]             [Substitute the value of

=> V = I x [tex]\frac{1}{2\pi fC}[/tex]      [Make f the subject of the formula]

=> f = [tex]\frac{I}{2\pi VC}[/tex]                    ---------------------(i)

From the question;

I = 3.33mA = 0.00333A

C = 8.50nF = 8.50 x 10⁻⁹F

V = 12.0V

Substitute these values into equation (i) as follows;

f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex]            [Taking [tex]\pi[/tex] = 3.142]

f = 5.19 x 10³Hz

Therefore, the frequency is closest to f = 5.19 x 10³Hz

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum

Answers

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

[tex]m[/tex] - Mass of the ball, measured in kilograms.

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?

Answers

Answer:

1.3515x10^5pa

Explanation:

Plss see attached file

To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface

Answers

Answer:

Explanation:

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A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?

a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.

Answers

Answer:

the greatest intensity is obtained from   c

Explanation:

An electromagnetic wave stagnant by the expression

           E = E₀ sin (kx -wt)

when two waves meet their electric fields add up

           E_total = E₁ + E₂

the intensity is

           I = E_total . E_total

           I = E₁² + E₂² + 2E₁ E₂ cos θ

where θ  is the phase angle between the two rays

       

Let's examine the two waves

in this case E₁ = E₂ = E₀

          I = Eo2 + Eo2 + 2 E₀ E₀ coasts

         I = E₀² (2 + 2 cos θ )

         I = 2 I₀ (1 + cos θ )

     let's apply this expression to different cases

a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀

b) cos π = -1     this implies that     I_total = 0

c) the cosine is  1,

         I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ

         I = E₀² (5 +4 cos θ)

         I = E₀² 9

         I = 9 Io

d) in this case the cos pi = -1

          I = E₀² (5 -4)

          I = I₀

e) we rewrite the equation

         I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ

         I = Eo2 (10 + 6 cos θ)

         cos π = -1

         I = E₀² (10-6)

         I = 4 I₀

the greatest intensity is obtained from   c

The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.

What is an amplitude?

An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.

In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.

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The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine

Answers

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?

Answers

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration

Answers

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC

Answers

Answer:

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Explanation:

The electric field is given by the following formula:

E = F/q

E= W/q

E = mg/q

where,

E = magnitude of electric field = ?

m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

= charge = 3 nC = 3 x 10⁻⁹ C

Therefore,

E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)

E = 17.64 x 10⁶ N/C = 17.64 MN/C

A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t

Answers

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?

Answers

Answer:

The glider’s speed after the skydiver lets go is 26 m/s

Explanation:

To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum

Mathematically;

mv = mv + mv

so 680 * 26 = (680-60)v + 60 * 26

17680 = 620v + 1560

17680-1560 = 620v

16120 = 620v

v = 16120/620

v = 26 m/s

You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.

Answers

Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]

we can solve for N1 by substituting into the equation above

[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]

the secondary voltage will be 360 volts.

(a) In the primary coil, you have "60 turns".

(b) The emf across the secondary coil would be "360 volts".

Transformer and Voltage

According to the question,

Primary voltage, V₁ = 120 volts

Secondary voltage, V₂ = 8 volts

Secondary turns, N₂ = 40 turns

(a) By applying transformer formula,

→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]

or,

   N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]

By substituting the values,

        = [tex]\frac{40\times 120}{8}[/tex]

        = [tex]\frac{4800}{8}[/tex]

        = 60

(2) Again by using the above formula,

→ V₂ = [tex]\frac{60\times 240}{40}[/tex]

       = [tex]\frac{14400}{40}[/tex]

       = 360 volts.

Thus the above approach is correct.  

Find out more information about voltage here:

https://brainly.com/question/4389563

Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.

Answers

Explanation:

It is given that,

The separation between two parallel wires, r = 11 cm = 0.11 m

Current in wire 1, [tex]q_1=54\ A[/tex]

Current in wire 2, [tex]q_2=45\ A[/tex]

Length of wires, l = 4.3 m

We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]

So, the magnetic force on a 4.3 m length of the wire  on both of currents is F=0.0189 N.

What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c

Answers

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

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