A diffraction grating 19.2 mm wide has 6010 rulings. Light of wavelength 337 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of θ at which maxima appear on a distant viewing screen?

Answers

Answer 1

Answer:

(a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .

Explanation:

Given that,

Width of diffraction grating [tex]w= 19.2\ mm[/tex]

Number of rulings[tex]N=6010[/tex]

Wavelength = 337 nm

We need to calculate the distance between adjacent rulings

Using formula of distance

[tex]d=\dfrac{w}{N}[/tex]

Put the value into the formula

[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]

[tex]d=3.19\times10^{-6}\ m[/tex]

We need to calculate the value of m

Using formula of constructive interference

[tex]d \sin\theta=m\lambda[/tex]

[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]

Here, m = 0,1,2,3,4......

[tex]\lambda[/tex]=wavelength

For largest value of  θ

[tex]\dfrac{m\lambda}{d}>1[/tex]

[tex]m>\dfrac{d}{\lambda}[/tex]

Put the value into the formula

[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]

[tex]m>9.46[/tex]

[tex]m = 9[/tex]

(a). We need to calculate the largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=71.9^{\circ}[/tex]

(b). We need to calculate the second largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=57.7^{\circ}[/tex]

(c). We need to calculate the third largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=47.7^{\circ}[/tex]

Hence, (a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .


Related Questions

What physical feature of a wave is related to the depth of the wave base? What is the difference between the wave base and still water level?

Answers

Answer:

physical feature of a wave is related to the depth of the wave base is The circular orbital motion

B. The wave base is the depth, and the still water level is the horizontal level

"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit

Answers

Answer:

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Explanation:

Given that:

the time period T = 88.5 min = 88.5 × 60  = 5310 sec

The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg

if  the radius of orbit is r,

Then,

[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e }{r}[/tex]

[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Similarly :

[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]

where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Then:

[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]

[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]

[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]

[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]

[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]

[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]

[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]

[tex]r= 2565.38^2[/tex]

r = 6579225 m

The  circumference of the satellites  orbit can now be determined by using the formula:

 circumference = 2π r

 circumference = 2π  × 6579225 m

 circumference = 41338489.85 m

 circumference of the satellite orbit  = 4.13 × 10⁷ m

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 .

21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fricción μr =0.55. ¿Cuánto desciende antes de detenerse?

Answers

Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

             

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

         

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

A nozzle with a radius of 0.22 cm is attached to a garden hose with a radius of 0.89 cm that is pointed straight up. The flow rate through hose and nozzle is 0.55 L/s.
Randomized Variables
rn = 0.22 cm
rh = 0.94 cm
Q = 0.55
1. Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m.
2. Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed assuming the same flow rate.

Answers

Answer:

1. 0.2m

1. 66m

Explanation:

See attached file

The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;

1) The maximum height when the water leaves the hose is: Δy = 0.20 m

2) The maximum height of the water leaves the nozzle is: Δy = 68.6m

Given parameters

The flow rate  Q = 0.55 L/s = 0.55 10⁻³ m³ / s Nozzle radius r₁ = 0.22 cm = 0.22 10⁻² m Hose radius r₂ = 0.94 cm = 0.94 10⁻² m

To find

   1. Maximum height of water in hose

  2. Maximum height of water at the nozzle

Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:

The continuity equation. It is an expression of the conservation of mass in fluids.

           A₁v₁ = A₂.v₂

Bernoulli's equation. Establishes the relationship between work and the energy conservation in fluids.

          P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂

Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.

1, Let's find the exit velocity of the water in the hose.

Let's use subscript 1 for the nozzle and subscript 2 for the hose.

The continuity equation of the flow value that must be constant throughout the system.

      Q = A₁ v₁

      v₁ = [tex]\frac{Q}{A_1 }[/tex]  

The area of ​​a circle is:

     A = π r²

Let's calculate the velocity in the hose.

    A₁ = π (0.94 10⁻²) ²

    A₁ = 2.78 10⁻⁴ m²

    v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]

    v₁ = 1.98 m / s

Let's use Bernoulli's equation.

When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2

      ½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂

      y₂-y₁ = ½  [tex]\frac{v_i^2 - v_2^2}{g}[/tex]  

At the highest point of the trajectory the velocity must be zero.

     y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]

Let's calculate

     y₂-y₁ =  [tex]\frac{1.98^2}{2 \ 9.8}[/tex]  

     Δy = 0.2 m

 

2.  Let's find the exit velocity of the water at the nozzle

          A₁ = π r²

          A₁ = π (0.22 10⁻²) ²

          A₁ = 0.152 10⁻⁴ m / s

With the continuity and flow equation.

           Q = A v

            v₁ = [tex]\frac{Q}{A}[/tex]  

             v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]  

             v₁ = 36.67 m / s

Using Bernoulli's equation, where the speed of the water at the highest point is zero.

           y₂- y₁ =  [tex]\frac{v^1^2}{g}[/tex]  

Let's calculate.

           Δy =  [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]  

           Δy = 68.6m

In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:

      1) The maximum height when the water leaves the hose is:

          Δy = 0.20 m

      2) The maximum height of the water when it leaves the nozzle is:

          Δy = 68.6 m

Learn more here:  https://brainly.com/question/4629227

7. A sound wave begins traveling through a thin metal rod at one end with a speed that is 15 times the speed of sound in air. If an observer at the other end of the rod hears the sound twice, one from the sound traveling through the rod and one from the sound traveling through the air, with a time delay of 0.12 s, how long is the rod? The speed of sound in air is 343 m/s.

Answers

Answer:

   L = 44,096 m

Explanation:

The speed of the sound wave is constant therefore we can use the relations of uniform kinematics

             v = x / t

the speed of the wave in the bar is

            v = 15 v or

            v = 15 343

             v = 5145 m / s

The sound at the bar goes the distance

             L = v t

Sound in the air travels the same distance

             L = v_air (t + 0.12)

as the two recognize the same dissonance,

             v t = v_air (t +0.12)

             t (v- v_air) = 0.12 v_air

              t = 0.12 v_air / (v -v_air)

l

et's calculate

             t = 0.12 343 / (5145 - 343)

             t = 8.57 10-3 s

The length of the bar is

              L = 5145 8.57 10-3

              L = 44,096 m

A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.

Answers

Answer:

Explanation:

When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .

Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .

hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .

A rod on a compressed spring exerts 12 N of force on a 0.05-kg steel ball. The
rod pushes the ball 0.03 m. How much work does the spring do on the ball?
A) 36
B) 36 N
C) 60 N
D)1.00

Answers

Answer:

Work = 0.36N

Explanation:

Given

Force = 12N

Distance = 0.03m

Weight = 0.05kg

Required

Determine the work done

Workdone is calculated as thus;

Work = Force * Distance

Substitute 12N for Force and 0.03m for Distance

Work = 12N * 0.03m

Work = 0.36Nm

Using proper S.I units

Work = 0.36N

Hence, work done by the spring on the ball is 0.36N

At the B end of the recessed horizontal bar, forces F1 and F2 of magnitudes 3 KN and 2KN respectively are applied and oriented as shown in the figure. Determine the magnitude of the resulting force and its orientation with respect to the horizontal.

Answers

Answer:

2.98 kN, 69.1° below the horizontal

Explanation:

Sum of forces in the x direction:

Fₓ = 3 kN cos 30° − 2 kN cos 40°

Fₓ ≈ 1.07 kN

Sum of forces in the y direction:

Fᵧ = -3 kN sin 30° − 2 kN sin 40°

Fᵧ = -2.79 kN

Magnitude of the resultant force is:

F² = Fₓ² + Fᵧ²

F = 2.98 kN

The direction of the resultant force is:

tan θ = Fᵧ / Fₓ

θ = -69.1°

An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum

Answers

Explanation:

The De-Broglie wavelength is given by :

[tex]\lambda=\dfrac{h}{p}[/tex]

h is Planck's constant

p is momentum

In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.

Only momentum is the factor that remains same for both particles i.e. momentum.

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).

A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of San Francisco.
a) true
b) false

Answers

Answer:

a) true

Explanation:

The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state. This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance

Answers

Answer:

The near point is  [tex]n =44.8 \ cm[/tex]

Explanation:

From the question we are told that

   The power is  [tex]P = 1.50[/tex]

   The  distance from the eye is  [tex]k = 1.8 \ cm[/tex]

    The distance of the book from the eye is [tex]z = -28 \ cm[/tex]

Generally the focal length of the glasses is  

       [tex]f = \frac{1}{P}[/tex]

=>   [tex]f = \frac{1}{1.50 }[/tex]

=>   [tex]f = 0.667 \ m[/tex]

=>   [tex]f = 66.7 \ cm[/tex]

The object distance is evaluated as

     [tex]u = z + k[/tex]

=>   [tex]u = -28 + 1.8[/tex]

=>  [tex]u = -26.2 \ cm[/tex]

The image distance is evaluated from lens formula as

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

=>   [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]

=>   [tex]v=- \frac{1}{0.0232}[/tex]

=>    [tex]v=- 43 \ cm[/tex]

The  near point is evaluated as

      [tex]n = -v + k[/tex]

=>    [tex]n =-(-43) + 1.8[/tex]

=>    [tex]n =44.8 \ cm[/tex]

A homeowner purchases insulation for her attic rated at R-15. She wants the attic insulated to R-30. If the insulation she purchased is 10 cm thick, what thickness does she need to use

Answers

Answer:

she need to use 20 cm thick

Explanation:

given data

wants the attic insulated = R-30

purchased = 10 cm thick

solution

as per given we can say that

10 cm is for the R 15

but she want for R 30

so  

R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]  

R 30 thickness = 20 cm

so she need to use 20 cm thick

Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, which is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C

Answers

Answer

Work done is 57.9KJ

Explanation

First solve the problem according to work done due to variation in temperature

So W= intergral Cu( 1-Tu/T). at Tu and T

So Given that

C = Heat capacity of the Brick

TEPc= Cold Temperature

TEPh = Hot Temperature

W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)

So

W= (1)-(300-150)-300 (1) ln 2

W= -57.9KJ

An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west

Answers

Answer:

e) west

Explanation:

According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.

An electron in the first energy level of the electron cloud has an electron in the third energy level

Answers

Answer:

a lower energy than

Explanation:

sorry im a month late but is lower energy than

In which example is kinetic friction most involved? a sled stuck on a snowy hill a bottle of water wedged in a vending machine an explorer unsuccessfully pushing on a massive stone that is blocking the entrance to a cave a volleyball player sliding across the court while diving for the ball

Answers

Answer:

I believe the answer is A volleyball player sliding across the court while diving for the ball.

Explanation:

Kinetic friction is a body moving on the surface experiences a force in the opposite direction of its movement.

Hope this helps! (づ ̄3 ̄)づ╭❤~

QUESTION 27
The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of the
plane is 400 degrees C and the linear coefficient of expansion for titanium is 5x10-6/C when flying at 3 times the speed of sound, how
much would a 10-meter long (originally at oC) portion of the airplane expand? Write your final answer in centimeters and show all of your
work.

Answers

Answer:

2 cm.

Explanation:

Data obtained from the question include the following:

Original Length (L₁ ) = 10 m

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Linear expansivity (α) = 5×10¯⁶ /°C

Increase in length (ΔL) =..?

Next, we shall determine the temperature rise (ΔT).

This can be obtained as follow:

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Temperature rise (ΔT) =..?

Temperature rise (ΔT) = T₂ – T₁

Temperature rise (ΔT) = 400 – 0

Temperature rise (ΔT) = 400°C

Thus, we can obtain the increase in length of the airplane by using the following formula as illustrated below:

Linear expansivity (α) = increase in length (ΔL) /Original Length (L₁ ) × Temperature rise (ΔT)

α = ΔL/(L₁ × ΔT)

Original Length (L₁ ) = 10 m

Linear expansivity (α) = 5×10¯⁶ /°C

Temperature rise (ΔT) = 400°C

Increase in length (ΔL) =..?

α = ΔL/(L₁ × ΔT)

5×10¯⁶ = ΔL/(10 × 400)

5×10¯⁶ = ΔL/4000

Cross multiply

ΔL = 5×10¯⁶ × 4000

ΔL = 0.02 m

Converting 0.02 m to cm, we have:

1 m = 100 cm

Therefore, 0.02 m = 0.02 × 100 = 2 cm.

Therefore, the length of the plane will increase by 2 cm.

You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C

Answers

Answer:

Option (c) : 20°C

Explanation:

[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]

T(final) = 500* 10 + 100*70/600 = 20°C

A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)

Answers

Answer:

Change in the frequency (in Hz) = 104.96 Hz

Explanation:

Given:

Speed of sound in air (v) = 343 m/s

Speed of car (v1) 36 m/s

Frequency(f) = 500 Hz

Find:

Change in the frequency (in Hz)

Computation:

Frequency hear by the observer(before)(f1) = [f(v+v1)] / v

Frequency hear by the observer(f1) = [500(343+36)] / 343

Frequency hear by the observer(f1) = 552.48 Hz

Frequency hear by the observer(after)(f2) = [f(v-v1)] / v

Frequency hear by the observer(f2) = [500(343-36)] / 343

Frequency hear by the observer(f2) = 447.52 Hz

Change in the frequency (in Hz) = f1 - f2

Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz

Change in the frequency (in Hz) = 104.96 Hz

One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.

Answers

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

ΔP.E = 6.48 x 10⁸ J

ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rises feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder​

Answers

Answer:

692 J/kg/°C

Explanation:

Electric energy added = amount of heat

Power × time = mass × SHC × increase in temperature

Pt = mCΔT

(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)

C = 692 J/kg/°C

The element sodium can emit light at two wavelengths, λ1 = 588.9950 nm and λ2 = 589.5924 nm. Light in sodium is being used in a Michelson interferometer. Through what distance must mirror M 2 be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?

Answers

Answer:

The distance is  [tex]d = 0.00029065 \ m[/tex]

Explanation:

From the question we are told that

    The  first wavelength is  [tex]\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m[/tex]

     The  second wavelength is  [tex]\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m[/tex]

     The  difference in the  fringe pattern is  n =  1.0  

Generally the equation defining the effect of the movement of  the mirror M 2 in a Michelson interferometer is mathematically represented as

          [tex]2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n[/tex]

Here d is the mirror M 2  must be moved

substituting values

         [tex]2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0[/tex]

        [tex]d = 0.00029065 \ m[/tex]

Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit

Answers

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)

d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)

  = 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047

  = 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?

Answers

Answer:

10.75 A

The current is in opposite direction since it causes a repulsion force between the wires

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]

where [tex]F/l[/tex] is the force per unit length on the wires

[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A

[tex]I_{1}[/tex] = current on the first wire = 0.520 A

[tex]I_{2}[/tex] = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] =  [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]

4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]

[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A

The current is in opposite direction since it causes a repulsion force between the wires.

What happens to the deflection of the galvanometer needle (due to moving the magnet) when you increase the number of loops

Answers

Answer:

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force

Explanation:

galvanometer is an instrument that can detect and measure small current in an electrical circuit.

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.

The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.

g suppose he used an alpha particle with an energy of 8.3 MeV, what would be the speed of this alpha particle

Answers

Answer:

speed of the alpha particle is 2 x 10^7 m/s.

Explanation:

energy of alpha particle = 8.3 Mev

1 Mev = 1.602 x 10^-13 J

8.3 Mev = [tex]x[/tex]

solving, [tex]x[/tex] = 8.3 x 1.602 x 10^-13 = 1.329 x 10^-12 J

mass of a alpha particle = 6.645 x 10^−27 kg

The energy of the alpha particle is the kinetic energy KE of the alpha particle

KE = [tex]\frac{1}{2}mv^{2}[/tex]

where m is the mass of the alpha particle

v is  the velocity of the alpha particle

substituting values, we have

1.329 x 10^-12 = [tex]\frac{1}{2}*6.645*10^{-27}*v^{2}[/tex]

[tex]v^{2}[/tex] = 4 x 10^14

[tex]v = \sqrt{4*10^{14} }[/tex] = 2 x 10^7 m/s

Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?

Answers

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]

(b) The angular velocity of disk 2 is :

[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]

(c) The moment of inertia for the two disk system is given by :

[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]

Hence, this is the required solution.

The primary difference between a barometer and a manometer is
A. a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
B. a barometer uses mercury, while a manometer can use any liquid. a barometer is used to measure atmospheric pressure, and a manometer is used to measure absolute pressure.
C a barometer reads in mm, while a manometer reads in Pa.
D a barometer can measure either positivee or negative pressure, while a manometer only
E positive pressure. measures

Answers

Answer:

a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.

Explanation:

A barometer measures air pressure at any locality with sea level as the reference.

However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure

The algebraic sum of these two values gives the absolute pressure.

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