In a physics lab, light with a wavelength of 490 nm travels in air from a laser to a photocell in a time of 17.5 ns . When a slab of glass with a thickness of 0.800 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.What is the wavelength of the light in the glass? Use 3.00×108 m/s for the speed of light in a vacuum. Express your answer using two significant figures.

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

Answer:

4.38 m/s

Explanation:

Answer:

The induced emf  is  [tex]\epsilon = 0.1041 \ V[/tex]  

Explanation:

From the question we are told that

   The  radius of the circular loop is  [tex]r = 9.50 \ cm = 0.095 \ m[/tex]

     The  intensity of the wave is  [tex]I = 0.0215 \ W/m^2[/tex]

      The wavelength is  [tex]\lambda = 6.90\ m[/tex]

Generally the intensity is mathematically represented as

         [tex]I = \frac{ c * B^2 }{ 2 * \mu_o }[/tex]

Here  [tex]\mu_o[/tex] is the permeability of free space with value  

         [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]

B is the magnetic field which can be mathematically represented from the equation as

          [tex]B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }[/tex]

substituting values

          [tex]B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }[/tex]

          [tex]B = 1.342 *10^{-8} \ T[/tex]

The  area is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

       [tex]A = 3.142 * (0.095)^2[/tex]

       [tex]A = 0.0284[/tex]

The angular velocity is mathematically represented as

        [tex]w = 2 * \pi * \frac{c}{\lambda }[/tex]

substituting values          

       [tex]w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }[/tex]  

        [tex]w = 2.732 *10^{8} rad \ s^{-1}[/tex]  

Generally the induced emf is mathematically represented as

        [tex]\epsilon = N * B * A * w * sin (wt )[/tex]

At maximum induced emf  [tex]sin (wt) = 1[/tex]

    So

         [tex]\epsilon = N * B * A * w[/tex]

substituting values

         [tex]\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}[/tex]  

         [tex]\epsilon = 0.1041 \ V[/tex]  

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]

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

.

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;

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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Answer:

a) 3.7 x 10^-3 s

b) 230.41 rad/s

Explanation:

The angular speed N = 2200 rpm (revolution per minute)

==> 2200/60 revolutions per sec = 36.67 rps

The total angle turned in one second = 36.67 x 360° = 13201.2°

if it takes 1 sec to revolve 13201.2°

then it will take t sec to rotate 49.0°

time t = 49/13201.2 = 3.7 x 10^-3 s

conversion to rad/s = 2πN/60 = (2 x 3.142 x 2200)/60 = 230.41 rad/s

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field  = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E  x B is along k direction or z  - axis so wave is propagating along z - axis .

The direction of motion of electromagnetic wave will be +z-direction.

Electromagnetic waves are waves that consist of the electric field and magnetic field.

The electric and magnetic fields are perpendicular to each other and the wave propagates in the direction perpendicular to both the fields.

Now, the direction of wave motion can be estimated by taking the cross-product of directional unit vectors of the electric and magnetic fields.

The electric field is in the +y direction and the magnetic field is in the +x-direction.

So, the direction of the wave will be,

[tex]i\times j=k[/tex]

Therefore, the direction of motion of electromagnetic wave will be +z-direction.

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Answer:

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

Explanation:

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.

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.

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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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

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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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

Answer:

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

Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

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

Answer:

v = 345.6m/s

Explanation:

v = 384 x 0.9 = 345.6

v = 345.6m/s

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.

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

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

The question is incomplete. Here is the complete question.

The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.

(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu  left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.

(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.

Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]

              (b) 91 years.

Explanation:

(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:

Q(t) = [tex]0.5Q_{0}[/tex]

[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]

[tex]0.5=e^{-87.7k}[/tex]

[tex]ln(0.5)=ln(e^{-87.7k})[/tex]

[tex]ln(0.5)=-87.7k[/tex]

[tex]k = \frac{ln(0.5)}{-87.7}[/tex]

k = 0.0079

Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]

(b) Using function:

[tex]Q(t)=3.5e^{-0.0079t}[/tex]

[tex]1.7=3.5e^{-0.0079t}[/tex]

[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]

[tex]e^{-0.0079t}=0.4857[/tex]

[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]

[tex]-0.0079t=-0.7221[/tex]

[tex]t = \frac{-0.7221}{-0.0079}[/tex]

t = 91.41

t ≈ 91 years

Scientists will be able to receive data for approximately 91 years.

Answer:

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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.

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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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

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.