PHYSICS COLLEGE

Early in the morning, when the temperature is 5.5 °C, gasoline is pumped into a car’s 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 °C. Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out in this case?

Answers

Answer 1

Answer:

Answer:

Volume of gasoline spills out is 0.943 L.

Explanation:

Volumetric expansion of both gasoline and steel tank is :

$\beta_(gas)=9.5 *10^(-4)/K\n\beta_(steel \ gas)=3.6 * 10^(-5)/K.$ { source Internet}

We know expansion due to temperature change is :

$\Delta V=\beta*\Delta T* V$

For gasoline:

$\Delta V_g=0.98 \ L.\n$

Similarly for Steel tank:

$\Delta V_(steel \ gas)=0.037\ L$ .

Now, volume of gasoline spills out is equal to difference between expansion in volume.

$\Delta V_(gas)-\Delta V_(Steel \ gas)=0.98-0.037\ L=0.943\ L.$

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A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

Answers

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal = 30 degree

tension in the string = 4.5 N

WE KNOW THAT

Work = force * distance

horizontal force = $Tcos\theta = 4.5*cos30 = 3.89 N$

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

A local ice hockey pond is at 4°C when the air temperature falls, causing the water temperature to drop to 0°C with 10.9 cm of ice forming at the surface at a temperature of 0°C. How much heat was lost if the unfrozen pond is 1 m deep and 50 m on each side?

Answers

Answer: 114.4 GJ

Explanation:

Heat loss Q=U×A×ΔT

Heat loss of size A is determined by the U value of materials and the difference in temperature.

From 10.9cm from the ice

50m= 5000cm

A= 5000×5000

Q== (10.9) (5000) (5000)(4.184)(1×4 + 80)

Q = 95,771,760,000J

Q≈ 95.8 GJ

Linear gradient from the bottom of the pond to the ice:

Q = (89.1)(5000)(5000)(4.184)(1*2)

Q = 18,639,720,000J

Q ≈ 18.6 GJ

Total heat loss:

Q= 95.8GJ + 18.6GJ

Q= 114.4 GJ

n Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is 125 Ω when its temperature is 20.0°C. The wire is then immersed in boiling chlorine, and the resistance drops to 99.6 Ω. The temperature coefficient of resistivity of platinum is α = 3.72 × 10−3(C°)−1. What is the temperature of the boiling chlorine?

Answers

Answer:

- 48.55 degree

Explanation:

Resistance increases when temperature increases. This increase is linear and which is defined by the following relation between resistance and temperature

$R_t=R_0(1-\alpha* t)$

$R_t$ is resistance at temperature t , R₀ is resistance at temperature t₀ , t is rise in temperature and α is temperature coefficient of resistance .

Putting the values we get

125 = 99.6 ( 1 + 3.72x 10⁻³ x t )

t = 68.55

Therefore temperature of boiling chlorine is 20-68.55 = - 48.55 degree celsius .

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????

Answers

Answer:

The number of available energy states per unit volume is $4.01*10^(48)$

Explanation:

Given that,

Average energy $E=2*10^(-4)\ eV$

Photon = $4*10^(-5)\ eV$

We need to calculate the number of available energy states per unit volume

Using formula of energy

$g(\epsilon)d\epsilon=(8\pi E^2dE)/((hc)^3)$

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

$g(\epsilon)d\epsilon=(8*\pi*2*10^(-4)*4*10^(-5)*1.6*10^(-19))/((6.67*10^(-34)*3*10^(8))^3)$

$g(\epsilon)d\epsilon=4.01*10^(48)$

Hence, The number of available energy states per unit volume is $4.01*10^(48)$

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

1 microgram equals how many milligrams?

Answers

Answer: 1 microgram is equal to 0.001 miligrams

Explanation: The factor micro is equal 10^-3 while the factor mili is equal to 10^-3 so to converte the micro to mile we have to multiply by 0.001.

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