If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g

Answers

Answer 1

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current [tex]I_1[/tex] = 100 μA

current [tex]I_2[/tex] = 1 mA

forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode  if the bias current is increased to 1mA.

Using the formula:

[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]

[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]

where;

[tex]V_r[/tex] = 0.7

[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]

[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]

[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]

[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]

Suppose n = 1

[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]

Then;

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]

[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]

[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]

[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]

[tex]V_r'=29.37 \times nV_T[/tex]

[tex]V_r'=29.37 \times 25.86[/tex]

[tex]V_r'=759.5 \ mV[/tex]

[tex]Vr' \simeq[/tex] 760 mV

Thus, the voltage drop across this same diode will be 760 mV


Related Questions

Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction

Answers

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

(a)  the final speed of the water after contraction is 15 m/s.

(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.

The given parameters;

initial pressure, P₁ = 3.5 x 10⁵ Painitial speed, v₁ = 5 m/sdensity of water, ρ = 1000 kg/m³

Let the initial area of the pipe = A₁

Apply the continuity equation to determine the final speed of the water after contraction as follows;

[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]

The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;

[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]

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The ancient Greek Eratosthenes found that the Sun casts different lengths of shadow at different points on Earth. There were no shadows at midday in Aswan as the Sun was directly overhead. 800 kilometers north, in Alexandria, shadow lengths were found to show the Sun at 7.2 degrees from overhead at midday. Use these measurements to calculate the radius of Earth.

Answers

Answer:

The  radius of the earth is [tex]r = 6365.4 \ km[/tex]

Explanation:

From the question we are told that

     The distance at  Alexandria is  [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]

      The angle of the sun is  [tex]\theta = 7.2 ^o[/tex]

So we want to first obtain the circumference of the earth

   So let assume that the earth is  circular ([tex]360 ^o[/tex])

  Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many  [tex](7.2 ^o)[/tex]  are in [tex]360^o[/tex]

 i.e    [tex]N = \frac{360}{7.2}[/tex]

=>      [tex]N = 50[/tex]

     With this  value we can evaluate the circumference as

             [tex]c = 50 * 800[/tex]

              [tex]c = 40000 \ km[/tex]

Generally circumference is mathematically represented as

        [tex]c = 2\pi r[/tex]

         [tex]40000 = 2 * 3.142 * r[/tex]

=>        [tex]r = 6365.4 \ km[/tex]

Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction

Answers

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?1. Increasing the wavelength of the laser.2. Increasing the distance to the screen.3. Increasing the frequency of the laser.4. Increasing the number of lines per length.

Answers

Answer:

Increase in frequency of the laser

Explanation:

Because An increase in frequency will result in more lines per centimeter and a smaller distance between each consecutive line. And a decrease in distance between each gratin

A double-slit experiment is performed with light of wavelength 620 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?

Answers

Answer:

1.34 mm

Explanation:

A double slit experiment is conducted with a light which has a wavelength of 620 nm

The fringes are separated 2.3 mm apart

The light is changed to a wavelength length of 360 nm

Let x represent the fringe spacing as a result of the change in wavelength

Therefore,the fringe spacing can be calculated as follows

2.3mm/x= 620nm/360nm

Multiply both sides

x × 620= 2.3×360

620x= 828

x= 828/620

x= 1.34 mm

How are electricity and magnets connected

Answers

Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?

Answers

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?

Answers

Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.

Explanation:

Answer:

Light amplification by stimulated emission of radiation

front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire

Answers

Answer:

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

Explanation:

This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.

Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.

Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

y=k/x, x is halved.
what happens to the value of y

Answers

Answer:

y is doubled

Explanation:

If x is halved, that means the value is doubled. Here is an exmaple:

y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.

Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors

Answers

Answer:

it is going to D. all of these are resistors

A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.

In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?

The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.

Answers

Answer:

a)   Δt = 24.96 s , b)  τ = 0.078 N m

Explanation:

This is a rotational kinematics exercise

        θ = w₀ t - ½ α t²

Let's reduce the magnitudes the SI system

       θ = 60 rev (2π rad / 1 rev) = 376.99 rad

       w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s

       

      α = (w₀ t - θ) 2 / t²

let's calculate the annular acceleration

      α = (43.98 10 - 376.99) 2/10²

      α = 1,258 rad / s²

Let's find the time it takes to reach zero angular velocity (w = 0)

        w = w₀ - alf t

         t = (w₀ - 0) / α

         t = 43.98 / 1.258

         t = 34.96 s

this is the total time, the time remaining is

         Δt = t-10

         Δt = 24.96 s

To find the braking torque, we use Newton's law for angular motion

        τ = I α

the moment of inertia of a circular ring is

       I = M r²

we substitute

         τ = M r² α

we calculate

        τ = 0.625  0.315²  1.258

        τ = 0.078 N m

The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Given data:

The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex]   (rps means rotation per second).

The time interval is, t' = 10.0 s.

The number of rotations made by wheel is, n = 60.0.

The mass of bike wheel is, m = 0.625 kg.

The radius of wheel is, r = 0.315 m.

The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,

[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]

Here, [tex]\theta[/tex] is the angular displacement, and its value is,

[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]

And, angular speed is,

[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]

Solving as,

[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]

Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.

[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]

Then total time is,

T = t - t'

T = 35.18 - 10

T = 25.18 s

Now, use the standard formula to obtain the value of braking torque as,

[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]

Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

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1.) When the acceleration is zero, what can you say about the velocity of an object?

Answers

Answer:

it is either constant or zero

Explanation:

If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)

Answers

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K,is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K in the body?

Answers

Answer:

0.03143 Gy

Explanation:

Mass of the human body = 70 kg

Mass of potassium in the human body = 140 g

chemical atomic mass of potassium = 39.1

From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms

Thus,

140g of potassium will contain;

(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms

We are told that the natural abundance of one of the 40K isotopes is 0.012%.

Thus;

Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms

Formula for activity of K-40 is given as;

Activity = (0.693 × number of K-40 atoms)/half life

Activity = (0.693 × 7.2276 × 10^(19))/1300000000

Activity = 3.85 × 10^(10)

We are told that each decay deposits 1.0 MeV of energy into the body.

Thus;

Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV

Now, 1 MeV = 1.602 × 10^(-13) joules

Thus;

Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J

1 Gy = 1 J/kg

Thus;

Yearly dose = 2.25/70 = 0.03143 Gy

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?

Answers

Answer:

6000 counts per second

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample

Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <

Answers

Answer:

1. Energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared >> visible > ultraviolet

Explanation:

Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.

Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:

1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.

i.e energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.

i.e frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.

i.e  wavelength: infrared >> visible > ultraviolet

In terms of lowest to the highest energy,frequency, and wavelength is;

Energy: infrared > visible light > ultraviolet

Frequency: infrared > visible light > ultraviolet

Wavelength: ultraviolet > visible light > infrared

The electromagnetic spectrum:

The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.

The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.

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A baseball (m=145g) traveling 35 m/s moves a fielder's glove backward 23 cm when the ball is caught. What was the average force exerted by the ball on the glove?

Answers

Answer:

386.13 N

Explanation:

The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).

Therefore, KE of the ball

[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]

Now, workdone in moving the glove

W= Fd

where F = Force applied, d = displacement of the glove= 0.23 cm.

88.81 = F×0.23

F= 88.81/0.23 = 386.13 N

An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?

Answers

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

The amount of charge after 5 seconds is:

5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?

Answers

Answer:

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

PLEASE HELP ANSWER FAST As the vibration of molecules decreases, the _____ of the substance decreases. 1.temperature 2.internal energy 3.kinetic energy 4.all of the above

Answers

I think 1 and 3 is absolutely right but im not sure about number 2.
I think the answer is 4 all of the above because as the vibration decrease automatically the kinetic energy decrease and the temperature is decrease because when the vibration of molecules decrease thats mean the substances is slightly become a solid and you can get a solid cube of liquid if you freeze them

A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light

Answers

Answer:

b

Explanation:

A resistance heater having 20.7 kW power is used to heat a room having 16 m X 16.5 m X 12.3 m size from 13.5 to 21 oC at sea level. The room is sealed once the heater is turned on. Calculate the amount of time needed for this heating to occur in min. (Write your answer in 3 significant digits. Assume constant specific heats at room temperature.)

Answers

Answer:

t = 23.6 min

Explanation:

First we need to find the mass of air in the room:

m = ρV

where,

m = mass of air in the room = ?

ρ = density of air at room temperature = 1.2041 kg/m³

V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³

Therefore,

m = (1.2041 kg/m³)(3247.2 m³)

m = 3909.95 kg

Now, we find the amount of energy consumed to heat the room:

E = m C ΔT

where,

E = Energy consumed = ?

C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C

ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C

Therefore,

E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)

E = 29324.62 KJ

Now, the time period can be calculated as:

P = E/t

t = E/P

where,

t = Time needed = ?

P = Power of heater = 20.7 KW

Therefore,

t = 29324.62 KJ/20.7 KW

t = (1416.65 s)(1 min/60 s)

t = 23.6 min

Test Bank, Question 18.83 Inside a room at a uniform comfortable temperature, metallic objects generally feel cooler to the touch than wooden objects do. This is because: a given mass of wood contains more heat than the same mass of metal the human body, being organic, resembles wood more closely than it resembles metal metal conducts heat better than wood heat tends to flow from metal to wood

Answers

Answer:

metal conducts heat better than wood.

Explanation:

Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.

Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm

Answers

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]

  = [tex]\frac{10+7155}{2650}[/tex]

[tex]\frac{1}{f}[/tex]  = [tex]\frac{7165}{2650}[/tex]

⇒ f = [tex]\frac{2650}{7165}[/tex]

      = 0.37

The focal length of the eye is 0.37 cm.

If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L

Answers

Answer:

The highest percentage of change corresponds to the thinnest rod, the correct answer is a

Explanation:

For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity

               F / A = Y ΔL/L

where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.

In this case the bars are made of the same material by which Young's modulus is the same for all

              ΔL / L = (F / A) / Y

the area of ​​the bar is the area of ​​a circle

               A = π r² = π d² / 4

               A = π / 4 d²

we substitute

              ΔL / L = (F / Y) 4 /πd²

changing length

               ΔL = (F / Y 4 /π) L / d²

The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change

a) values ​​given d and 3L

               ΔL = cte 3L / d²

               ΔL = cte L /d²  3

To find the percentage, we must divide the change in magnitude by its value and multiply by 100.

                ΔL/L % = [(F /Y  4/π 1/d²) 3L ] / 3L 100

                ΔL/L  % = cte 100%

 

b) 3d and L value, we repeat the same process as in part a

               ΔL = cte L / 9d²

               ΔL = cte L / d² 1/9

               ΔL / L% = cte 100/9

               ΔL / L% = cte 11%

   

c) 2d and 2L value

               ΔL = (cte L / d ½ )/ 2L

               ΔL/L% = cte 100/4

               ΔL/L% = cte 25%

d) value 4d and L

               ΔL = cte L / d² 1/16

                ΔL/L % = cte 100/16

                ΔL/L % = cte 6.25%

   

The highest percentage of change corresponds to the thinnest rod, the correct answer is a

The temperature of the hot spots caused by the impact of transferred matter onto the surface of a pulsar can be 108 K. What is the peak wavelength in the blackbody spectrum of such a spot, and in what range of the electromagnetic spectrum does it occur

Answers

Given that,

Temperature = 10⁸ K

We need to calculate the peak wavelength in the blackbody spectrum

Using formula of peak wavelength

[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{T}[/tex]

Where, T= temperature

Put the value into the formula

[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{10^{8}}[/tex]

[tex]peak\ wavelength = 2.90\times10^{-11}\ m[/tex]

[tex]peak\ wavelength = 290\ nm[/tex]

This range of wavelength is ultraviolet.

Hence, The peak wavelength in the blackbody spectrum is 290 nm and the range of wavelength is ultraviolet electromagnetic spectrum .

A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.

Required:
Calculate the electric energy dissipated in this process.

Answers

Answer:

Explanation:

change in magnetic flux = .16 x .16 x .510 - 0

= .013056 weber .

rate of change of flux = change in flux / time

= .013056 / 40.5 x 10⁻³

= .32237

voltage induced = .32237 V

electrical energy dissipated = v² / R where v is voltage , R is resistance

= .32237² / 6.35

= 16.36 x 10⁻³ J .

15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis​

Answers

Answer: Photosynthesis

Explanation: every food chain starts with plant life, therefore photosynthesis comes first.

The Curiosity rover now on Mars analyzed rocks and found magnesium to have the following isotopic composition.
79.70% Mg-24 (23.9872 amu), 10.13% Mg-25 (24.9886 amu), and 10.17% Mg-26 (25.9846 amu).
A. How many neutrons are in Mg-25?
B. What is the average atomic mass of magnesium in these rocks?
C. Is the magnesium composition on Mars the same as that on Earth? Explain.

Answers

Answer:

A.   number of neutrons of Magnesium Mg = 13

B.   The average mass of Mg = 22.29 amu

C.   the magnesium composition on Mars is not the same as that on Earth.

Explanation:

Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.

The atomic number of Magnesium Mg = 12

The atomic number of an element is the number of protons present in the atomic nucleus of the element

i.e Atomic number = number of protons = 12

The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.

Mass number = number of protons + number of neutrons

Given that the mass number of Mg = 25

Then;

25 = 12 + number of neutrons

25 - 12 = number of neutrons

13 = number of neutrons

number of neutrons of Magnesium Mg = 13

B. What is the average atomic mass of magnesium in these rocks?

The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Therefore the average atomic mass of magnesium can be calculated as:

= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]

= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]

= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]

The average mass of Mg = 22.29 amu

C. Is the magnesium composition on Mars the same as that on Earth? Explain.

The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.

There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.

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