The magnitude of the magnetic field at point P for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at P? (c = 3.0 × 108 m/s)

Answers

Answer 1

Answer:

The electric field is  [tex]E = 636 \ V/m[/tex]

Explanation:

From the question we are told that

     The magnitude of magnetic field is [tex]B = 2.12 \mu T = 2.12*10^{-6} \ T[/tex]

      The value for speed of light is  [tex]c = 3.0 *10^8 \ m/s[/tex]

Generally the magnitude of the electric field at point P is

        [tex]E = B * c[/tex]

substituting values

         [tex]E = 2.12 *10^{-6} * 3.0 *10^{8}[/tex]

         [tex]E = 636 \ V/m[/tex]

Answer 2

The magnitude of electric field for the wave at point P is 636 V/m.

Given data:

The strength of magnetic field at point P is, [tex]B = 2.12 \;\rm \mu T=2.12 \times 10^{-6} \;\rm T[/tex].

The speed of light is, [tex]c = 3.0 \times 10^{8} \;\rm m/s[/tex].

The given problem is based on the concept of electric field and magnetic field. The electromagnetic wave works on the principle of oscillating magnetic field and electric field at the same region. We can find any of the two using the expression,

[tex]E = B \times c[/tex]

here,

E is the strength of electric field.

Solving as,

[tex]E = (2.12 \times 10^{-6}) \times (3 \times 10^{8})\\\\E = 636 \;\rm V/m[/tex]

Thus, we can conclude that the magnitude of electric field for the wave at point P is 636 V/m.

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Related Questions

The magnetic field at the center of a 1 cm diameter loop is 2.5 mT. If a long straight wire carries the same current as the loop of wire, at what distance from the wire is the magnetic field 2.5 mT? A. 0.10 m B. 1.6x10-3 m C. 0.01 m D. 20 m

Answers

Answer:

B. 1.6 x 10⁻³ m

Explanation:

The magnetic field at the center of the loop is given by;

[tex]B = \frac{\mu_o I }{2R}[/tex]

Where;

μ₀ is the permeability of free space

I is the current in the loop

R is the radius of the circular loop

B is the magnetic field

Given;

diameter of the loop = 1cm

radius of the loop, r = 0.5 cm = 0.005 m

magnetic field, B = 2.5mT = 2.5 x 10⁻³ T

The current in the loop is calculated as;

[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]

The magnetic at a distance from the long straight wire is calculated as;

[tex]B = \frac{\mu_o I}{2\pi d}[/tex]

where;

d is the distance from the wire;

[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]

Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.

B. 1.6 x 10⁻³ m

This question involves the concepts of the magnetic field due to a loop and a  current-carrying wire and current.

A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".

The magnetic field at the center of a loop of wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2r}[/tex]

where,

B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = ?

r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m

Therefore,

[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]

I = 19.9 A

Now, the magnetic field at a distance from the straight wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2\pi R}[/tex]

where,

R = distance from wire = ?

Therefore,

[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]

R = 1.6 x 10⁻³ m

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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

At what minimum angle will you get total internal reflection of light traveling in diamond and reflected from ethanol? °

Answers

Answer:

34°

Using the relation

θᶜ = sin^-1(n₂/n₁),

where n1= the refractive index of light is propagating from a medium

And n2 = refractive index of medium into which light is entering

So we know that

refractive index of diamond at 589nm = 2.41= n₁

refractive index of ethanol at 589nm and 20°C = 1.36= n₂

Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°

Explanation:

What is the impedance of an AC series circuit that is constructed of a 10.0-W resistor along with 12.0 W inductive reactance and 7.0 W capacitive reactance

Answers

Answer:

11.2 Ω

Explanation:

The impedance of a circuit is given by;

Z= √R^2 +(XL-XC)^2

Since

Resistance R= 10 Ω

Inductive reactance XL= 12 Ω

Capacitive reactance XC= 7 Ω

Z= √10^2 + (12-7)^2

Z= √100 + 25

Z= √125

Z= 11.2 Ω

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

Mention the importance of writing the physical quantities as vectors.​

Answers

Answer:

Knowledge of vectors is important because many quantities used in physics are vectors. If you try to add together vector quantities without taking into account their direction you'll get results that are incorrect.

Explanation:

An example of the importance of vector addition could be the following:

Two cars are involved in a collision. At the time of the collision car A was travelling at 40 mph, car B was travelling at 60 mph. Until I tell you in which directions the cars were travelling you don't know how serious the collision was.

The cars could have been travelling in the same direction, in which case car B crashed into the back of car A, and the relative velocity between them was 20 mph. Or the cars could have been travelling in opposite directions, in which case it was a head on collision with a relative velocity between the cars of 100 mph!

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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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.

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

How wide is the central diffraction peak on a screen 2.30 m behind a 0.0368-mm-wide slit illuminated by 558-nm light

Answers

Answer:

The  value [tex]y = 0.0349 \ m[/tex]

Explanation:

From the question we are told that

   The  distance of the screen is  [tex]D = 2.30 \ m[/tex]

   The  width of the slit is  [tex]d = 0.0368 \ nm = 0.0368 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 558 \ nm = 558 *10^{-9} \ m[/tex]

The  width of the central diffraction peak is  mathematically represented as

        [tex]k = 2 * y[/tex]

Where  y is the distance from the center to the high peak which  is mathematically represented as

       [tex]y = \frac{\lambda * D }{d }[/tex]

substituting values

      [tex]y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }[/tex]

      [tex]y = 0.0349 \ m[/tex]

A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards

Answers

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

Which one of the following lists gives the correct order of the electromagnetic spectrum from low to high frequencies?
A) radio waves, infrared, microwaves, ultraviolet, visible, x-rays, gamma rays
B) radio waves, ultraviolet, x-rays, microwaves, infrared, visible, gamma rays
C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays
D) radio waves, microwaves, visible, x-rays, infrared, ultraviolet, gamma rays
E) radio waves, infrared, x-rays, microwaves, ultraviolet, visible, gamma rays

Answers

Answer:

C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays

Explanation:

radio waves have lowest  energy , lowest  frequency and highest  wavelength

gamma rays  have highest  energy , highest  frequency and least  wavelength

Answer: C

Explanation:

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and the Sun."

Answers

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

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.

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

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.

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.

To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.68 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.

Answers

Answer:

1.8/0.61 =2.95 ft

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

;)

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 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

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

A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?

Answers

Answer:

The direction of the force will be towards the east

Explanation:

From the question we are told that  

    The direction of the  downward

Generally according to Fleming's right-hand rule(

          Thumb -  direction of force

           Middle finger -  direction of current

           Index finger -  direction of the magnetic field

) and the fact that the earth magnetic field acts  from south to north with respect to the four cardinal points then the direction of the  force will be toward the east with respect to the four cardinal point on the earth

Determine the magnitude and direction of the force between two parallel wires 30 m long and 6.0 cm apart, each carrying 30 A in the same direction.

Answers

Answer:

0.09N, attractive

Explanation:

It can be deducted from the question that the currents are arranged in parallel settings, then it is obvious that the force on each of the wire will be attractive toward the other wire.

the magnitude of force can be determined by using below formula;

F2 = (μ₀/2π)(I₁I₂/d)I₂

μ₀ = constant = 4π × 10^-7 H/m,

I₁, I₂ = currents= 30A

L = the length o the wire=30m

d = distance between these two wires= 0.06m

Since the current are arranged in the same direction, they exhibit attractive force on each other.

Then plugging the values Into the formula above we have

F₂ = (4π × 10^-7 T.m/A)/2π) × ((30A)²/ 0.06m)× 30 m

= 0.09 N, attractive

Therefore, the magnitude and direction of the force is 0.09 N, attractive

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

light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb

Answers

Answer:

121ohms

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

Hence, the resistance of this bulb is 121 ohms

the atomic number of a nucleus increases during which nuclear reactions

Answers

Answer:

Answer A : Fusion followed by beta decay (electron emission)

Explanation:

Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).

Therefore, answer A is the correct one.

Answer:

A : Fusion followed by beta decay (electron emission)

Explanation:

Ap3x

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]

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 .

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]

Learn more here:https://brainly.com/question/19566865

In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in s) of the material

Answers

Answer:

2113 seconds

Explanation:

The general decay equation is given as [tex]N = N_0e^{-\lambda t} \\\\[/tex], then;

[tex]\dfrac{N}{N_0} = e^{-\lambda t} \\[/tex] where;

[tex]N/N_0[/tex] is the fraction of the radioactive substance present = 1/16

[tex]\lambda[/tex] is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

[tex]\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328[/tex]

Half life f the material is expressed as [tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]

[tex]t_{1/2} = \frac{0.693}{0.000328}[/tex]

[tex]t_{1/2} = 2,112.8 secs[/tex]

Hence, the half life of the material is approximately 2113 seconds

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