### User Atul, 14 different physics problems (optics, termodinamics, electricity)

Set 1 questions

1. 0.097 mol

2. 3.5 W

3. 5.34 *10^7 nm

4. 58.7 degree

5. the potential energy is decreasing and it is moving to a region with a lower potential

W = P*(V2-V1)

P*V1= niu*R*T1

P = niu*R*T1/V1 =0.1*8.31*(300+273)/2*10^-3 =238081.5 N/m^2

Work =238081.5*(2*10^-3 -10^-3) = 238.0815 J

At constant temperature

P1*V1 = P2*V1 (=constant)

infinitesimal work is dW= P*dV dV is infinitesimal change in volume

W = integral (from v1 to V2) P*dV = integral (V1 to V2) constant*dV/V = constant *ln(V2/V1) =

= niu*R*T1 * ln(V2/V1) = 0.1*8.31* 573*ln(2000/100) = 1426.45 J

3.

strong reflection is when the both reflected waves (from both the surfaces) have a difference of path of an integer of wavelength (or lambda)

Since the propagation medium for the wave reflected from the water is oil the speed of light inside the oil is V = C/n(oil) (n is the refractive index, C is the speed of light) and hence

d = lambda(air)*n(oil) =500*10^-9*1.25 = 625*10^-9 m =625 nm

4.

The angle of the refraction of light at the surface of the water from the water to the air (angle of light inside the water with the normal to the surface) is

r = 90-50 =40 degree

If i is the angle of the incidence of light from the air to the water (angle of light above the water with the normal to surface) then

sin(i)/sin(r) = n where n is the refractive index of water.

sin(i)=n*sin(r) = 1.33*sin(40) = 0.85

i = 58.74 degree.

Hence the angle of light with the horizontal is

theta =90-58.74 = 31.25 degree

5.

E1x = k*q1/R1^2 =9*10^9*5*10^-9/(0.02)^2 = 112500 V/m to the right

E3y = k**q3/R3^2 = -5*9/(0.04)^2 = 28125 V/m upwards

E2 = k*q2/R2^2

R2 = sqrt(2*2+4*4) =4.47 cm

E2 = 9*10/(0.047)^2 *= 40742 V/m

tan(theta) = 4/2 =2

theta= 63.43 degree angle that E2 is making with x axis

E2x = E2*cos(theta) = 40742*cos(63.43) =18220 V/m to the right

E2y = E2*sin(theta) = 36439 V/m downwards

Ex = E1x +E2x =18220 + 112500 = 130720 V/m

Ey = E2y - E3y = 36439 -28125 = 8314 V/m

Magnitude is

E = sqrt(Ex^2+ Ey^2) = 130984 V/m

tan (alpha) = Ey/Ex = 8314/130720

alpha = 3.64 degree is the angle made by E with the positive x

6.

F1 = B2*I1*L - B3*I1*L =I1*L*(B2-B3) = I1*L*miu*I/2/pi * (1/R2- 1/R1) = 10*0.5*4*pi*10^-7/2/pi*(1/2-1/4)*100 = 2.5 *10^-5 N

F2 = B1*I2*l + B3*I2*L = I2*L (B1+ B3) = I2*L*miu*I/2/pi * (1/R1 +1/R3) =10*0.5*4*pi*10^-7/2/pi *(1/2+1/2) *100 = 10^-4 N

F3 = F1 from the simetry of the problem

7.

h*c/lambda = Ec +W

niu is the photon frequency, Ec the kinetic energy, W the work function, c is the speed of light

h*c/lambda = 2.8 eV + W

h*c/(lamba1+0.5*lambda1) =1.1 +W

h*c/(1.5*lambda1) = 1.1 eV+ W

lambda = h*c/(2.8 eV+ W)

1.5*lambda = h*c/(1.1 eV +W)

1.5 = (2.8 + W)/(1.1 +W)

1.65 +1.5*W =2.8 +W

0.5*W = 1.15

W = 2.3 eV

h*c/lambda = (2.8+2.3)*1.6*10^-19

lambda = 6.62*10^-34*3*10^8/5.1*1.6*10^-19 =243.4 *10^-9 m =243.4 nm

1. How many moles of an ideal gas are there in a container with a pressure of

115,049 Pa, a temperature of 284 K, and a volume of 2.0 L? The universal gas

constant is 8.314 J/K � mol. (Points : 5) 97 mol 13,000 mol 6,700 mol 0.097 mol 2. (TCO 2) A person consumes a meal containing 18 kcal. What is the power this meal

produces if it is to be "burned off" due to exercise in 6 hours? (Points : 5) 0.8 W 0.0035 W 3.5 W 12,552 W 3. (TCO 3) A double slit illuminated with light of wavelength 498 nm forms a

diffraction pattern on a screen 45 cm away. The slit separation is d = 4196 nm.

What is the distance ?x between orders m = 17 and m = 16? (Points : 5) 10.68 x 107 nm 379.16 x 107 nm 5.34 x 107 nm 2.67 x 107 nm 4. (TCO 4) The index of refraction of the core of a piece of fiber optic cable is

1.72. If the index of the surrounding cladding is 1.47, what is the critical angle

for total internal reflection for a light ray in the core incident on the

core-cladding interface? (Points : 5) 52.8� 47.0� 64.6� 58.7� 5. (TCO 5) A parallel plate capacitor contains a positively charged plate on the

left and a negatively charged plate on the right. An electron in between the plates

is moving to the right. Which statement is correct? (Points : 5) The potential energy of the electron is increasing and it is moving to a region having a lower potential. The potential energy of the electron is decreasing and it is moving to a region having a lower potential. The potential energy of the electron is decreasing and it is moving to a region having a higher potential. The potential energy of the electron is increasing and it is moving to a region having a higher potential. 6. (TCO 6) A straight bar magnet is initially 4 cm long, with the north pole on the

right and the south pole on the left. If you cut the magnet in half, the right

half will (Points : 5) still contain a north pole on the right and a new south pole on the left. only contain a south pole. only contain a north pole. be unmagnetized. 7. (TCO 8) Monochromatic light is incident on a metal surface and electrons are

ejected. If the intensity of the light is increased, what will happen to the

ejection rate and maximum energy of the electrons? (Points : 5) Same rate; greater maximum energy. Greater rate; greater maximum energy. Greater rate; same maximum energy. Same rate; same maximum energy.

Set 2 questions

This questions requires a numeric or algebraic answer. In order to receive full

credit for this problems you must show all your work. To show your work for this

problems you may scan neatly handwritten work or type up your work in MS Word using

the equation editor and submit it using the Final Exam Dropbox. You should make

your submissions after you complete the exam.

1. (TCO 1) A copper block is removed from a 300� C oven and dropped into 1.00 liter of water

at 20� C. The water quickly reaches 25.5� C and stays at that temperature. What is

the mass of the copper block? (Points : 20) 2. (TCO 2) A 2,000 cm3 container holds 0.10 mol of helium gas at 300� C. How much work must be

done to compress the gas to 1,000 cm3 at constant pressure? How much work to

compress the gas to 100 cm3 at constant temperature? (Points : 20) 3. (TCO 3) A very thin oil film (n = 1.25) floats on water (n = 1.33). What is the thinnest

film that produces a strong reflection for green light with a wavelength of 500 nm?

(Points : 20) 4. (TCO 4) An underwater diver sees the sun 50 degrees above horizontal. How high is the sun

above the horizon to a fisherman in a boat directly above the diver? (Points : 20) 5. (TCO 5)

What are the strength and direction of the electric field at the position indicated

by the dot in the figure? Give your answer in component form and as a magnitude and

an angle referred to the +x-axis. (Points : 20) 6. (TCO 6) What is the net force (magnitude and direction) on each wire in the figure?

(Points : 20) 7. (TCO 7) The maximum kinetic energy of photoelectrons is 2.8 eV. When the wavelength of the

light is increased by 50%, the maximum energy decreases to 1.1 eV. What are the

work function of the cathode and the initial wavelength of the light? (Points : 20)

AnswersSet 1 questions

1. 0.097 mol

2. 3.5 W

3. 5.34 *10^7 nm

4. 58.7 degree

5. the potential energy is decreasing and it is moving to a region with a lower potential

6. still contain a north pole on the right and a new south pole on the left.

7. Greater rate; same maximum energy.

7. Greater rate; same maximum energy.

Set 2questions

1.

m(copper)*c*(t2(copper)-t1(copper)) = m(water)*cwater*(t2(water)-t1(water))

m is the mass of the block, t1 and t2 the initial and final temperatures

c (copper) =123 J/kg/k

c(water) =4186 J/kg/k

Q =m(water)*cwater*(t2(water)-t1(water)) =1*4186*(25.5-20) = 23023 J

m(copper) = Q/c(copper)/(t2-t1) =23023/123/(300-25.5) =0.681 kg copper

m(copper)*c*(t2(copper)-t1(copper)) = m(water)*cwater*(t2(water)-t1(water))

m is the mass of the block, t1 and t2 the initial and final temperatures

c (copper) =123 J/kg/k

c(water) =4186 J/kg/k

Q =m(water)*cwater*(t2(water)-t1(water)) =1*4186*(25.5-20) = 23023 J

m(copper) = Q/c(copper)/(t2-t1) =23023/123/(300-25.5) =0.681 kg copper

2.At constant pressure the work is

W = P*(V2-V1)

P*V1= niu*R*T1

P = niu*R*T1/V1 =0.1*8.31*(300+273)/2*10^-3 =238081.5 N/m^2

Work =238081.5*(2*10^-3 -10^-3) = 238.0815 J

At constant temperature

P1*V1 = P2*V1 (=constant)

infinitesimal work is dW= P*dV dV is infinitesimal change in volume

W = integral (from v1 to V2) P*dV = integral (V1 to V2) constant*dV/V = constant *ln(V2/V1) =

= niu*R*T1 * ln(V2/V1) = 0.1*8.31* 573*ln(2000/100) = 1426.45 J

3.

strong reflection is when the both reflected waves (from both the surfaces) have a difference of path of an integer of wavelength (or lambda)

Since the propagation medium for the wave reflected from the water is oil the speed of light inside the oil is V = C/n(oil) (n is the refractive index, C is the speed of light) and hence

d = lambda(air)*n(oil) =500*10^-9*1.25 = 625*10^-9 m =625 nm

4.

The angle of the refraction of light at the surface of the water from the water to the air (angle of light inside the water with the normal to the surface) is

r = 90-50 =40 degree

If i is the angle of the incidence of light from the air to the water (angle of light above the water with the normal to surface) then

sin(i)/sin(r) = n where n is the refractive index of water.

sin(i)=n*sin(r) = 1.33*sin(40) = 0.85

i = 58.74 degree.

Hence the angle of light with the horizontal is

theta =90-58.74 = 31.25 degree

5.

E1x = k*q1/R1^2 =9*10^9*5*10^-9/(0.02)^2 = 112500 V/m to the right

E3y = k**q3/R3^2 = -5*9/(0.04)^2 = 28125 V/m upwards

E2 = k*q2/R2^2

R2 = sqrt(2*2+4*4) =4.47 cm

E2 = 9*10/(0.047)^2 *= 40742 V/m

tan(theta) = 4/2 =2

theta= 63.43 degree angle that E2 is making with x axis

E2x = E2*cos(theta) = 40742*cos(63.43) =18220 V/m to the right

E2y = E2*sin(theta) = 36439 V/m downwards

Ex = E1x +E2x =18220 + 112500 = 130720 V/m

Ey = E2y - E3y = 36439 -28125 = 8314 V/m

Magnitude is

E = sqrt(Ex^2+ Ey^2) = 130984 V/m

tan (alpha) = Ey/Ex = 8314/130720

alpha = 3.64 degree is the angle made by E with the positive x

6.

F1 = B2*I1*L - B3*I1*L =I1*L*(B2-B3) = I1*L*miu*I/2/pi * (1/R2- 1/R1) = 10*0.5*4*pi*10^-7/2/pi*(1/2-1/4)*100 = 2.5 *10^-5 N

F2 = B1*I2*l + B3*I2*L = I2*L (B1+ B3) = I2*L*miu*I/2/pi * (1/R1 +1/R3) =10*0.5*4*pi*10^-7/2/pi *(1/2+1/2) *100 = 10^-4 N

F3 = F1 from the simetry of the problem

7.

h*c/lambda = Ec +W

niu is the photon frequency, Ec the kinetic energy, W the work function, c is the speed of light

h*c/lambda = 2.8 eV + W

h*c/(lamba1+0.5*lambda1) =1.1 +W

h*c/(1.5*lambda1) = 1.1 eV+ W

lambda = h*c/(2.8 eV+ W)

1.5*lambda = h*c/(1.1 eV +W)

1.5 = (2.8 + W)/(1.1 +W)

1.65 +1.5*W =2.8 +W

0.5*W = 1.15

W = 2.3 eV

h*c/lambda = (2.8+2.3)*1.6*10^-19

lambda = 6.62*10^-34*3*10^8/5.1*1.6*10^-19 =243.4 *10^-9 m =243.4 nm

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