A large electromagnet draws 300A at 270V. The coils of the electromagnet are cooled by a flow of mystery liquid passing over them. The liquid enters the electromagnet at a temperature of 15 deg C, absorbs the Joule heat, and leaves at a higher temperature. If the liquid is to leave at a temperature no higher than 81 deg C, and the maximum flow rate of liquid through the electromagnet is 0.307kg/s. What is the specific heat of the mystery liquid? Pick the closest value. The solution should include the correct units.

Answer:

B(max) =  3.7971 × [tex]10^{-12}[/tex]  T

Explanation:

given data

radius R = 26 mm

plate separation d = 4.0 mm

potential difference Vm = 220 V

frequency f = 76 Hz

V = (220 V) sin[2π(76 Hz)t]

solution

we know that E will be

E = V ÷ d     ............1

put here value

E = [tex]\frac{220 \times sin(2\pi 76\times t)}{d}[/tex]  

and here we take as given r = R

so A = π R²    .................2

and

ФE  = E × A

ФE = [tex]\frac{\pi R^2 \times 220 \times sin(2\pi 76 \times t)}{d}[/tex]  .....................3

so use use here now  Ampere's Law that is

∫ B ds = [tex]\mu_o \times \epsilon_o \times \frac{d\Phi E}{dt} + \mu_o \times I_{encl}[/tex]      .....................4

and

here [tex]I_{encl}[/tex]  is = 0  and r = R

so

[tex]2B \times \pi \times R = \mu_o \times \epsilon_o \times \frac{d\Phi E}{dt}[/tex]      .....................5

and put here value we get

B =  [tex]\frac{\mu_o \times \epsilon_o \times \pi \times f \times R \times V_m cos(2\pi f t)}{d}[/tex]        .....................6

put here value  for B maximum cos(2πft) = 1

and we get B (max)

B(max) = [tex]\frac{\mu_o\times \epsilon_o\times \pi \times f\times R\times V_m}{d}[/tex]     ....................7

put here all value

B(max) = [tex]\frac{4\pi \times10^{-7} \times 8.85\times 10^{-12}\times \pi \times 76 \times 0.026\times 220 }{4\times 10^{-3}}[/tex]      

solve it we get

B(max) =  3.7971 × [tex]10^{-12}[/tex]  T

Answer:

Explanation:

Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli

= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴

=  13216 x 10⁻⁹ weber .

b ) When the coil becomes parallel to magnetic field  , flux through it will become zero.

c ) e m f induced = change in flux / time

= 13216 x 10⁻⁹ / 4.9 x 10⁻²

= 2697.14 x 10⁻⁷ V

= 269.7 x10⁻⁶

269.7 μV.

Answer:

Explanation:

The star is revolving the black hole like earth revolves around the sun .so time period of rotation  T is given by the following relation

T² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex] , R is distance between black hole and star , M is mass of black hole

Given T = 4.8 hours

4.8² =  [tex]\frac{4\pi^2\times R^3}{GM }[/tex]

Using the same equation for earth sun system

24² =  [tex]\frac{4\pi^2\times (50R)^3}{GM_s }[/tex]  , Ms is mass of the sun and 50R is distance between the sun and the earth .

Dividing the equation

[tex](\frac{4.8}{24})^2[/tex] = [tex]\frac{M_s}{M}\times\frac{1^3}{50^3}[/tex]

[tex]\frac{M}{M_s}[/tex] = 2x 10⁻⁴

The mass of the black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.

Given information:

An X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours (T).

The star is orbiting around the black hole.

The distance between the centers of the star and the black hole is one-fiftieth of the distance between the centers of the Earth and our Sun.

Let R be the center to center distance between the black hole and the sun.

So, the time period T of the sun around the black hole will be,

[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\4.8^2=\dfrac{2\pi^2R^3}{GM}[/tex]

where M is the mass of the black hole.

The distance between the earth and our sun will be 50R.

So, the time period of the earth will be,

[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}[/tex]

where [tex]M_s[/tex] is the mass of our sun.

Now, compare the above two relations to get the mass of black hole in terms of mass of our sun as,

[tex]4.8^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}\\\dfrac{24^2}{4.8^2}=\dfrac{50^3}{1}\times \dfrac{M}{M_s}\\\dfrac{M}{M_s}=0.0002\\\dfrac{M}{M_s}=2\times 10^{-4}[/tex]

Therefore, the mass of black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.

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Answer:

The final temperature of the ice is [tex]-40^{\circ}\ C[/tex].

Explanation:

It is given that,

Energy, Q = 10 cal

Mass of ice, m = 2 g

Initial temperature, [tex]T_1=-30^{\circ}\ C[/tex]

We need to find the final temperature of the ice. We know that the specific heat of ice is [tex]0.5\ cal\ g^{-1} ^{\circ} C^{-1}[/tex]

The heat added in terms of specific heat is given by :

[tex]Q=mc(T_1-T_2)[/tex]

[tex]T_2[/tex] final temperature of the ice

c is specific heat of ice

[tex]T_2=T_1-\dfrac{Q}{mc}\\\\T_2=(-30)-\dfrac{10}{2\times 0.5}\\\\T_2=-40^{\circ}[/tex]

So, the final temperature of the ice is [tex]-40^{\circ}\ C[/tex].

Answer: the correct answer is -20

Explanation:

The correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.

Explanation

The moon is the most popular and well-known natural satellite on planet earth, it is a satellite widely studied by humans, they have even walked on the moon. However, from Earth, at night when you see the moon you always see the same face of the moon, which has caused people to wonder why this phenomenon. The answer to this phenomenon is that we always see the same face of the moon because it takes the same time to rotate once on itself as it does to go around the Earth (about 27 days). The result is that the same part of the moon always points towards the earth. According to the above, the correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.

Complete Question

Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55 min if she is not going to arrive late. Her exit is 43 mi away. How much time would it take at the posted 60 mph speed?

Answer:

The time it would take at the given speed is  [tex]x = 43.00 \ minutes[/tex]

Explanation:

From the question we are told that

      The time taken to exist the highway is  [tex]t = 55 min[/tex]

      The distance to the exist is  [tex]d = 43\ mi[/tex]

       Alberta speed is [tex]v = 60 mph[/tex]

The time it would take travelling at the given speed is mathematically represented as

        [tex]t_z = \frac{d}{v}[/tex]

  substituting values

        [tex]t_z = \frac{43}{60}[/tex]

        [tex]t_z = 0.71667\ hrs[/tex]

Converting to minutes

         1  hour =  60 minutes

So      0.71667 hours = x minutes

   Therefore

               [tex]x = 0.71667 * 60[/tex]

                [tex]x = 43.00 \ minutes[/tex]

At a speed of 60 mph, it would take Alberta approximately 43 minutes to travel the 43 miles to her grandmother's house, which is within the 55 minutes time frame she has to avoid being late.

To determine how long it will take Alberta to reach her grandmother's house if she travels at a constant speed of 60 mph, we need to use the formula for time which is time = distance ÷ speed. Alberta's exit is 43 miles away and the speed limit is 60 mph.

First, we calculate the time it would take her to travel 43 miles at 60 mph:

Time = Distance ÷ Speed
= 43 miles ÷ 60 mph
= 0.7167 hours

Since time in hours is not always intuitive, let's convert it to minutes by multiplying by 60 (since there are 60 minutes in one hour):

Time in minutes = 0.7167 hours × 60 minutes/hour
= 43 minutes

Thus, it will take Alberta approximately 43 minutes to reach her exit at the posted speed of 60 mph.

Answer:764.4N

Explanation:

Mass=78kg

Acceleration=9.8m/s^2

force=mass x acceleration

Force=78 x 9.8

Force=764.4

Force=764.4N

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

[tex]\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}[/tex]     ..............3

put here value and we get  

[tex]\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}[/tex]  

[tex]\frac{m1+0.5}{m2+0.5}[/tex]   = 1.4

[tex]\frac{m1+0.5}{m2+0.5} = \frac{7}{5}[/tex]   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm

Using the formula for thin film interference, the minimum thickness of the plastic film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.

The question is asking for the minimum thickness of the film for which maximum light is reflected for the provided wavelengths. This is a classic example of thin film interference.

For constructive interference (maximum reflection), the thickness of the film (t) is given by the formula: t = mλ/2n. Here 'm' is the order of the bright fringe, 'λ' is wavelength, and 'n' is the index of refraction.

Considering the first-order maximum, we find the thickness for each wavelength:
t1 = (1)(477.1 x 10^-9 m)/2(1.58) = 1.51 x 10^-7 m (or 151 nm for 477.1 nm light)
t2 = (1)(668.0 x 10^-9 m)/2(1.58) = 2.12 x 10^-7 m (or 212 nm for 668.0 nm light)

Therefore, the minimum thickness of the film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.

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Answer:

Option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.

Explanation:

Regarding Rayleigh's criterion, the angular resolution is given as below:

θ = 1.22λ/D

From this expression, it is observed that the larger the aperture size, the smaller will be the value of the angular resolution, and the better the device will be.

This signifies that at very high angular differences, the precision for distinguishing two points is high.

Therefore, option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.

Answer:

867755.73 V/m

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

20m

Explanation:

Pressure = pgh

p = density of water 1000

kg/m^3

g = acceleration due to gravity 9.81 m/s^2

h is the depth of water

Pressure = 201 kPa = 201 x 10^3 Pa

201 x 10^3 = 1000 x 9.81 x h

201 x 10^3 = 9810h

h = 20.49 m

Approximately 20 m

Answer:

This is because of the inertia of the object.

Explanation:

This is because of the inertia of the object.

When you push a static object you must overcome the static friction force of the object. Once you have overcome the static friction the inertia law demands that the object tend to conserve its motion. That is the reason why you need less force when the object is already in motion. In other words, the inertia gained by the object with the initial force, "helps" you with the work of moving the object. This is also the reason why the kinetic friction of an object in motion over a surface is lower than the static friction.

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

Answer:they repel

Explanation:

Like charges repels, unlike charges attracts

Answer:oxygen

Explanation:algae are plants that lives in aquatic habitat,with a few of them occurring in land. Algae have chlorophyll and as a result are autotrophic in nutrition. Algae uses carbon dioxide as a raw material for photosynthesis which is the process where they produce food. They also give off oxygen which results from the splitting of water by light.

This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.

Final answer:

Carbon dioxide is a metabolic waste of photosynthesis in algae that can be recycled back into the ecosystem through cellular respiration.

Explanation:

Among the metabolic wastes produced by algae during photosynthesis, carbon dioxide (CO₂) is a byproduct that can be recycled and used in cellular respiration. Photosynthesis is the process in which algae absorb light energy to create carbohydrates within chloroplasts, releasing oxygen as a byproduct. Alternatively, cellular respiration involves using oxygen to break down carbohydrates, primarily in the cytoplasm and mitochondria, releasing ATP and carbon dioxide. These two processes are synchronously linked in a biological cycle that allows organisms to harness energy from the sun, guide by electron transport chains that drive cellular reactions.