A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. Assuming that the mass is at the equilibrium posiiton at t = 0, what is its displacement at t = 1.0 s?

Answers

Answer 1

Answer:

[tex]d =3.7*10^{-3} m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.2kg[/tex]

Spring constant [tex]\mu=22Nm^{-1}[/tex]

Amplitude [tex]A=5cm=0.05m[/tex]

Generally the equation for displacement d is mathematically given by

 [tex]d = Asin(\omega t)[/tex]

Where

 [tex]\omega=angular\ velocity[/tex]

 [tex]\omega=\sqrt{k/m}[/tex]

 [tex]\omega=\sqrt{22/1.2}[/tex]

 [tex]\omega=4.2817rads^{-1}[/tex]

Therefore  

 [tex]d = 0.05*sin(4.2817*1)[/tex]  

 [tex]d =3.7*10^{-3} m[/tex]


Related Questions

Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons

Answers

Answer:

No of proton is 13 and nucleus is 13

a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an angle of 45 and is 51.0 m long/ What will its kinetic energy be wehn it reaches the bottom

Answers

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is [tex]m=230\ kg[/tex]

It reaches the steepest hill with speed of [tex]u=6.2\ km/h\ or \ 1.72\ m/s[/tex]

Hill to bottom is 51 m long with inclination of [tex]45^{\circ}[/tex]

Height of the hill is [tex]h=51\sin 45^{\circ}=36.06\ m[/tex]

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

[tex]\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ[/tex]

Which of the following best describes the upper respiratory tract?
O It takes air in from outside the body.
O It is where oxygen and carbon dioxide are exchanged.
O It is located inside the thorax.
O It is not directly involved in respiration.

Answers

A). It takes air in from outside the body.

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

     

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

           

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

A wave has a frequency of 87.00 Hz and has a wavelength of 74.62 m. What is its
velocity?

Answers

Answer:

v = 6491.94 m/s

Explanation:

We are given;

Frequency; f = 87 Hz

Wavelength;λ = 74.62 m

Formula for velocity(v) of waves from the wave equation is;

v = fλ

Thus;

v = 87 × 74.62

v = 6491.94 m/s

what is the suitable way of using social media​

Answers

Answer:

not using it too much and getting addicted

Explanation:

Don’t used everyday or you will be addicted to it, just used when it’s necessary

A small object with mass 0.200 kg moves with constant speed in a vertical circle of radius 0.500 m. It takes the object 0.500 s to complete one revolution. (a) What is the translational speed of the object

Answers

Answer:

6.28 m/s.

Explanation:

Given that,

The mass of the object, m = 0.2 kg

The radius of the circle, r = 0.5 m

It takes the object 0.500 s to complete one revolution.

We need to find the translational speed of the object. Let it is v. We know that,

[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]

So, the transalational speed of the object is 6.28 m/s.

Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.
-increasing use of hybrid crops
-altering genes in DNA to create new plants
-increasing human population
-increasing climate change
-developing disease or pest resistant crops

Answers

Answer:

- increasing use of hybrid crops

- altering genes in DNA to create new plants

- developing disease or pest resistant crops

Explanation:

The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.

For example,

- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.

- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.

- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.

A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.

Determine the fan's airflow in m3/s.

Answers

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration

Answers

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instead of a discussion board requiring posts in a forum, this assignment has been modified to accept your response to the following questions in this assignment. Be sure to clearly address each of the points below and show all of your work • What is the difference between a vector and a scalar quantity? • List two examples each of vector and scalar quantities • Write the due date of this assignment: Month and Day (For example, July 15 would be Month - 7. Day = 15) • For a building having a height equal to the quantity you have recorded for Day in meters in our example 15 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s) • How fast was the ball traveling when it hit the ground? Submit the kinetics Assignment by 11:59 p.m. (ET) on Monday,

Answers

Answer:

A) vectors: veloicty, force

scalar:  speed, work

B)  t = 1.75 s,  C)   v = - 17 2 m / s

Explanation:

We answer each part separately

A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude

vector quantities: the speed of a car number is the magnitude and direction is where it goes

Force, the number is the magnitude and above that applies gives direction

Scalar magnitude: how quickly the number of the speedometer of the car

Temperature, work

B) I = 15 m height to the soil and get to calculate time = 0

        y = y₀ + v₀ t - ½ g t²

as the ball is loose its initial velocity is zero

       0 = 0 +0  - ½ g t²

       t = [tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]

       t = 1.75 s

C) the velocity to the reach the floor

      v = vo - g t

      v = 0 - g t

      v = - 9.8 1.74

       v = - 17 2 m / s

The negative signt iindicates that the speed goes down

Study the position-time graph for a bicycle. Which statement is supported by the graph? Position vs Time O The bicycle has speed but not velocity. O The bicycle is moving at a constant velocity. O The bicycle has a displacement of 3 m. O The bicycle is not in motion. 3 Position (m) 0 1 2 3 4 5 Time (s) Next Submit Save and Exit Mark this and return tViewers/AssessmentViewer/Activit. 0 M M​

Answers

Answer:

D) The bicycle is not in motion.

Explanation:

Study the position-time graph for a bicycle.

Which statement is supported by the graph?

A) The bicycle has speed but not velocity.

B) The bicycle is moving at a constant velocity.

C) The bicycle has a displacement of 3 m.

D) The bicycle is not in motion.

Solution:

Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.

Speed is the time rate of change of distance. It is the ratio of distance to time taken.

From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.

From the position-time data provided, it can concluded that the bicycle is not in motion.

Motion

Motion of a body involves a change in the position of that body with time.

A body in motion is constantly changing position or orientation as time passes.

The body may move with constant velocity/speed or changes in its velocity.

A position-time graph provides information about the motion of a body.

From the data provided:

At time 0, the bicycle is at position 3At time 1, the bicycle is at position 3At time 2, the bicycle is at position 3At time 3, the bicycle is at position 3At time 4, the bicycle is at position 3At time 5, the bicycle is at position 3

The position of the bicycle remains the same for all time intervals.

Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.

Learn more about motion and position-time graph at: https://brainly.com/question/2356782


find the resistance of wire of
0.65m Radius 0.25
and
resistivity 3x10-6 OHM

Answers

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

[tex] Resistance = P \frac {L}{A} [/tex]

Where;

P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]

[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]

Resistance = 9.95 Ohms

Which phenomenon occurs when one wave is superimposed on another?
A. Interference
B. Refraction
C. Diffraction
D. Polarization

Answers

Answer:Alternativa A.   Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.

According to ____________ , the randomness of the universe is constantly increasing.

a. The first law of thermodynamics
b. The zeroth law of thermodynamics
c. The second law of thermodynamics

Answers

Answer:

According to " The second law of thermodynamics", the randomness of the universe is constantly increasing?

Explanation:

So answer option C. Have a great summer.

A Michelson interferometer operating at a 600nm wavelength has a 2.02-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

19

Explanation:

Given that:

wavelength = 600 nm

Distance (d) = 2.02 cm = 2.02 × 10⁻² m

refraction index of air (n) = 1.00028

Pressure = 1.00 atm

The number of bright-dark-bright fringe shifts can be determined by using the formula:

[tex]\Delta m = \dfrac{2d}{\lambda} (n -1 ) \\ \\ \Delta m = \dfrac{2\times2.02 \times 10^{-2}}{600\times 10^{-9}} (1.00028 -1 ) \\ \\ \Delta m = 67333.33 \times 10^{-5}(1.00028 -1) \\ \\ \Delta m = 67333.33 \times 10^{-5}(2.8\times 10^{-4}) \\ \\ \Delta m = 18.853 \\ \\ \mathbf{\Delta m = 19}[/tex]

what is simple definition of democracy​

Answers

it's a form of government where people elect their representatives

Answer:

The word democracy itself means rule by the people.

20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.​

Answers

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

PLEASE HELP MEE THIS IS DUE IN 45 MINS

Answers

Answer:

The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]

Explanation:

This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:

[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)

Where:

[tex]m[/tex] - Mass of the car, in kilogram.

[tex]v[/tex] - Initial velocity, in meters per second.

[tex]\mu[/tex] - Coefficient of friction, no unit.

[tex]s[/tex] - Travelled distance, in meters.

Then we derive an expression for the distance travelled by the vehicle:

[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]

[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]

As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]

How does gravity affect your ability to live on a planet?

Answers

If we didn’t have it, we might just float into space. Or it would be hard to live with everything floating around
If it were too light, we would float off, too heavy, and then our bones would have to be extremely dense lest we want to get crushed by just existing

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible

Answers

Answer:

By the Pythagorean Theorem the distances from the speakers os

5 and 5.5 (rounding) meters   - let y be the wavelength in the solution

n y = 5      n is number of wavelengths from speaker

(n + m) y = 5.5      m must be integral for constructive interference

m y = .5       subtracting equations

m = 2      and y = ,25   for the above conditions

(n + 2) y = 5.5     substituting for m

n = 5.5 / .25 - 2 = 20

f = v / y     using frequency of sound

f = 340 / .25 = 1360 / sec    for lowest frequency

Check: D1 = y n  = ,25 * 20 = 5

and D2 = .25 * 22 = 5.5     for the distances traveled

A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constant is 105 N/m, how fast will the mass be moving when it reaches the equilibrium position? Hint: You cannot ignore the change in gravitational potential energy in this problem. Please give your answer in units of m/s, however, do not explicitly include units when typing your answer into the answer box.

Answers

Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)  

Apply the principle of conservation of mechanical energy;

ΔK.E  +  ΔP.E = 0

¹/₂m(v₂² - v₁²)  +  mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0)  +  mg(0 - h₁) = 0

¹/₂mv₂²  -  mgh₁  =  0

¹/₂mv₂²  = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

A 7.5-kg rock and a 8.9 × 10-4-kg pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.

Answers

Answer:

F' = 73.7 N

F = 8.749×10⁻³ N

a' = a =  9.83 m/s²

Explanation:

(a)

For the rock

Applying

F' = Gm'm/r²................... Equation 1

Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.

From the question,

Given: m' = 7.5 kg

Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m

Substitute these values into equation 1

F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²

F' = 7.37×10¹ N

F' = 73.7 N

Also, For the pebble,

F = GMm/r².............. Equation 2

Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth

Given: M = 8.9×10⁻⁴ kg,

Substitute into equation 2

F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²

F = 8.749×10⁻³ N

(b)

For the rock,

a' = F'/m'

Where a' = magnitude of the acceleration of the rock

a' = 73.7/7.5

a' = 9.83 m/s²

For the pebble,

a = F/M

Where a = acceleration of the pebble

a = (8.749×10⁻³)/(8.9×10⁻⁴)

a = 9.83 m/s²

You are using a constant force to speed up a toy car from an initial speed of 6.5 m/s
to a final speed of 22.9 m/s. If the toy car has a mass of 340 g, what is the work
needed to speed this car up?

Answers

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W82 J

In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?

Answers

Answer:

W = 16.4 kJ

Explanation:

Given that,

There are 135 steps from the ground floor to the sixth floor.

Each step is 16.6 cm tall.

The mass of a person, m = 73.5 kg

We need to find the work done by the person. We know that,

Work done = Fd

Where

d is the displacement, d = 135 × 0.166 = 22.41

So,

W = 73.5 × 10 × 22.41

= 16471.35  J

or

W = 16.4 kJ

So, 16.4 kJ is the work done by the person.

Need an answer in hurry u can make the pic big

Answers

answer: C

hope this helps! please give me brainliest :)

Identify each statement as an example of melting or sublimation,
lodide produces fumes when heated.
Melting
Sublimation
An iceberg turns to ocean water.
Candle wax turns to liquid when hot.
"Fog" is created from dry ice.

Answers

1st is sublimation

2and is melting

3red is melting

4th is sublimation

sublimation is just "skipping" the liquid phase / state

(c) Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0= 1.68×10-21 J and R0= 3.82×10-10 m. Find the frequency of small oscillations of one Ar atom about its equilibrium position.

Answers

Answer:

Explanation:

Answer:

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Explanation:

The formula for calculating the elastic potential energy is:

[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]

By rearrangement and using (K) as the subject;

[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]

[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]

k = 2.3 × 10⁻² N/m

Now; the formula used to calculate the frequency of the small oscillation is:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

where;

m = mass of each atom

assuming

m = 1.66 × 10⁻²⁶ kg

Then:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Which of the statements below are TRUE! Group of answer choices The carbon rod in batteries react to form a carbon cation. A good car battery gives you a little bit of power for a long period of time. A good car battery gives you a lot of power in a short period of time. The carbon rod in batteries is used as an inert electrode.

Answers

Answer:

The carbon rod in batteries is used as an inert electrode

Explanation:

A battery is considered as a power source that consists of one or more electrochemical cells having an external connections to provide power to the electrical devices such as the lights, bulbs, fans, mobile phones, etc.

It contains a positive terminal and a negative terminal.

The carbon rod in the battery does not help in the electrochemical reactions. It acts as an inert electrode and helps to flow the electrons only.

Thus the true statement is :

The carbon rod in batteries is used as an inert electrode.

The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?

Answers

Answer:

[tex]I_2=0.50 w/m^2[/tex]

Explanation:

From the question we are told that:

initial Intensity [tex]I_1=0.020 w/m^2[/tex]

Final Electric field [tex]E_2=5E[/tex]

Generally the equation for Relation ship between intensity and Electric field is mathematically given by

 [tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]

Therefore

 [tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]

 [tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]

 [tex]I_2=0.50 w/m^2[/tex]

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