Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

Answers

Answer 1

Answer:

The radiation pressure of the light is 3.33 x 10⁻ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]

Therefore, the radiation pressure of the light is 3.33 x 10⁻ Pa.


Related Questions

What is an understood decimal

Answers

The decimal place accuracy of a number is the number of digits to the right of the decimal point. ... If there is no decimal point, it is understood to be after the last digit on the right and there is no place (zero place) accuracy. The significant digits of a number are those digits that are most accurate.

Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.

Answers

Answer:

I know the answer

Explanation:

We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.

You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.

Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.

So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.

Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).

An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for the blood in an artery during heart surgery). Such a device is illustrated. The conducting fluid moves with velocity v in a tube of diameter d perpendicular to which is a magnetic field B. A voltage V is induced between opposite sides of the tube. Given B = 0.120 T, d = 1.2 cm., and a measured voltage of 2.88 mV, determine the speed of the blood.

Answers

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

[tex]E = Blv[/tex]

where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

[tex]v[/tex] is the velocity of the fluid through the field = ?

[tex]B[/tex] is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]

2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]

[tex]v[/tex] = 2.88/1.44 = 2 m/s

The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle

Answers

Answer:

4.38 m/s

Explanation:

The answer is 4.38 m/s

Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?

Answers

Each electron winds up with kinetic energy of

(270 keV)

plus

(whatever KE it had when it started accelerating).

Ayudaaa :(
Calcula la resistencia total del siguiente circuito eléctrico.

Answers

I believe it is 17 hope it helps!

A solid metal sphere of radius 3 m carries a total charge of -5.5 uc. What is the magnitude of the
electric field at a distance from the sphere's center of (a) 2.9 m and (b) 8 m? How would the answers
differ if the sphere was (c) a thin shell.
IN​

Answers

Answer:

2.9::: 5.87*10*3 N/C

8: 7.73 × 10 ^2  N/C

Explanation: https://study.com/academy/answer/a-solid-metal-sphere-of-radius-3-00-m-carries-a-total-charge-of-5-50-muc-what-is-the-magnitude-of-the-electric-field-at-each-of-the-following-distances-from-the-sphere-s-center-a-3-10-m-b-8-00-m.html

At TTT = 14 ∘C∘C, how long must an open organ pipe be to have a fundamental frequency of 262 HzHz ? The speed of sound in air is v≈(331+0.60T)m/sv≈(331+0.60T)m/s, where TT is the temperature in ∘C∘C.

Answers

Answer:

Length of pipe organ(L) = 0647 m (Approx)

Explanation:

Given:

Temperature (T) = 14°C

Fundamental frequency (F) = 262 Hz

Speed of sound (v) = 331 + 0.60(T) m/s

Find:

Length of pipe organ(L)

Computation:

Speed of sound (v) = 331 + 0.60(14) m/s

Speed of sound (v) = 339.4

Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)

Length of pipe organ(L) = 339.4 / 2 (262)

Length of pipe organ(L) = 0647 m (Approx)

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed

Answers

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

if you jog at a speed of 1.5m/s for 20 seconds how far di you travel

Answers

Answer: 30m

Explanation:

Given:

Speed: 1.5m/s

Time: 20 seconds

Distance = speed × time

Distance = 1.5 × 20

= 30m

Therefore you will travel 30m

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If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​

Answers

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?

Answers

Answer:

3.067MHz

Explanation:

The formula for calculating the voltage across an inductor is expressed as

[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor [tex]V_l[/tex] = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]

Hence, the frequency required is 3.067MHz

A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?

Answers

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{1}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{1}=0.01021\ Wb[/tex]

We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{2}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{2}=0.00127\ Wb[/tex]

We need to calculate the magnitude of the average induced emf

Using formula of emf

[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]

Put the value into t5he formula

[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]

[tex]\epsilon=0.016867\ V[/tex]

[tex]\epsilon=16.87\ mV[/tex]

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

[tex]I=\dfrac{\epsilon}{R}[/tex]

Put the value into the formula

[tex]I=\dfrac{0.016867}{2.5}[/tex]

[tex]I=0.00675\ A[/tex]

[tex]I=6.75\ mA[/tex]

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .

Answers

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"

Answers

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

A student wants to create a 6.0V DC battery from a 1.5V DC battery. Can this be done using a transformer alone

Answers

Answer:

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

Explanation:

A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.

Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.

So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

A 384 Hz tuning fork produces standing waves with a wavelength of 0.90 m inside a resonance tube. The speed of sound at experimental conditions is

Answers

Answer:

v = 345.6m/s

Explanation:

v = 384 x 0.9 = 345.6

v = 345.6m/s

Which of the following statements is true vibrations ?

Answers

Answer:

C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.

Explanation:

The complete question is

Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.

Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.

Answer:

Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."

Explanation:

My options were:

A)  Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.

B)  Resonance occurs as a result of sympathetic vibrations.

C)  A non-vibrating object can begin to vibrate as a result of forced vibrations.

D)  Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.

A is correct

.

A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=5.0rad/s. A second cylinder with rotational inertia I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω2=8.0rad/s. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?

Answers

Answer:

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia [tex]I[/tex] = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia [tex]I[/tex] = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = [tex]I[/tex]ω

where [tex]I[/tex] is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = [tex]I[/tex]ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = [tex]I[/tex]ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia [tex]I_{t}[/tex] = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = [tex]I_{t}[/tex][tex]w_{f}[/tex]

where [tex]I_{t}[/tex] is their total rotational inertia

[tex]w_{f}[/tex] = their final angular speed together

Final angular momentum = 3 x [tex]w_{f}[/tex] = 3[tex]w_{f}[/tex]

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

this means that

2 =  3[tex]w_{f}[/tex]

[tex]w_{f}[/tex] = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy = [tex]\frac{1}{2} Iw^{2}[/tex]

where [tex]I[/tex] is the rotational inertia

[tex]w[/tex] is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = [tex]\frac{1}{2} Iw^{2}[/tex] = [tex]\frac{1}{2}*2*5^{2}[/tex] = 25 J

for second cylinder = [tex]\frac{1}{2}*1*8^{2}[/tex] = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = [tex]\frac{1}{2}I_{t}w^{2} _{f}[/tex]

where

where [tex]I_{t}[/tex] is the total rotational inertia of the cylinders

[tex]w_{f}[/tex] is final total angular speed of the coupled cylinders

Final kinetic energy =  [tex]\frac{1}{2}*3*0.67^{2}[/tex] = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s

B) The percentage of the original kinetic energy lost to friction is;

percentage energy lost = 98.82%

We are given;

Rotational Inertia for first cylinder; I₁ = 2 kg.m²

Angular speed of first cylinder; ω₁ = 5 rad/s

Angular speed of second cylinder; ω₂ = 8 rad/s

Rotational Inertia for second cylinder; I₂ = 1 kg.m²

From conservation of angular momentum, we know that;

Initial angular Momentum([tex]L_{i}[/tex]) = Final angular Momentum([tex]L_{f}[/tex])

Thus;

I₁ω₁ + I₂ω₂ = I₃ω₃

Where;

ω₃ is the angular speed when the two cylinders are combined

I₃ = I₁ + I₂

I₃ = 2 + 1

I₃ = 3 kg.m²

Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;

(2 * 5) + (1 * -8) = 3ω₃

10 - 8 = 3ω₃

3ω₃ = 2

ω₃ = 2/3

ω₃ = 0.67 rad/s

B) Let us find initial kinetic energy;

E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²

E_i = ¹/₂((2 * 5²) + (1 * 8²)

E_i = 57 J

Final kinetic energy is;

E_f = ¹/₂I₃ω₃²

E_f = ¹/₂ * 3 * 0.67²

E_f = 0.67335 J

Energy lost = 57 - 0.67335 = 56.32665 J

percentage energy lost = (56.32665/57) * 100%

percentage energy lost = 98.82%

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If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL

Answers

Answer:

B. mineral oil – density of 0.910 g/mL.

Explanation:

Hello,

In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.

In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.

Best regards.

A stonecutter's chisel has an edge area of 0.7 cm2. If the chisel is struck with a force of 42 N, what is the pressure exerted on the stone

Answers

Answer:

The pressure is [tex]P = 583333 \ N/m^2[/tex]

Explanation:

From the question we are told that

  The area of the edge is  [tex]A = 0.72 cm^2 = 0.72 *10^{-4}\ m[/tex]

    The  force is [tex]F = 42 \ N[/tex]

The pressure is mathematically represented as

            [tex]P = \frac{F}{A}[/tex]

substituting values

           [tex]P = \frac{42}{0.72*10^{-4}}[/tex]

           [tex]P = 583333 \ N/m^2[/tex]

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Answers

Answer:

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Explanation:

2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s

Answers

Answer:

9 Brainly hahaha ............huh

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.


Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?

Answers

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

please help !!!!!!!!!!

Answers

Answer:

Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.

Explanation:

As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.

A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg
crate by a light rope. The light rope remains taut. Compared to the 6.00-kg crate,
the lighter 4.00-kg crate

Please explain why any of these multiple choices is correct!

Answers

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

The net force on both crates is the same and the acceleration of both crates is the same.

The given parameters;

mass of the crate, m = 6 kgmass of the second crate, = 4 kg

The force on the 4kg crate is calculated as follows;

[tex]F_{4kg } = T + F[/tex]

The force on the 6kg crate is calculated as follows;

[tex]F_{6 kg} = -T + F[/tex]

The net force on both crates is calculated as follows;

[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]

Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.

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30 POINTS HELP PLEASE
A galvanometer consists of a loop of wire in a magnetic field.
Question 19 options:
True
False

Answers

Answer:

A galvanometer consists of a loop of wire in a magnetic field.

TRUE

The correct answer is true

A disk between vertebrae in the spine is subjected to a shearing force of 375 N. Find its shear deformation, taking it to have a shear modulus of 1.60×109 N/m2. The disk is equivalent to a solid cylinder 0.750 cm high and 6.50 cm in diameter.

Answers

Answer:

5.29×10^-7

Explanation:

shear stress τ = F/ A

shear deformation δ = (VL)/ (AG)

= (τL)/ G

V=shear force

L=height of disk=6.50×10^-2

A=cross sectional area

G= shear modulus= (1.60x10^9N/m^2)

A=πd^2/4

Then substitute the values we have

4×(375N)(0.00750m)

________________ = δ

(π*0.00650^2)(1.60x10^9N/m^2)

= 5.29×10^-7

You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?

Answers

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.

Answers

Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer

Explanation:

Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.

What is friction?

Friction is the force that prevents one hard material from scooting or rolling over the other.

Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.

We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.

Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.

In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.

Thus, the correct option is B.

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