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Sloan [31]
2 years ago
14

n Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wi

re. 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?
Physics
1 answer:
Evgen [1.6K]2 years ago
6 0

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\times 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 .

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nika2105 [10]

Answer:

0.000003782 m

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\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
2 years ago
What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl
trasher [3.6K]

Net flux through the cylindrical surface is given as

\phi = \frac{q}{epsilon_0}

here q = enclosed charge in the surface

so here in order to find the value of q

q = \lambda* L

so now we have

\phi = \frac{\lambda * L}{\epsilon_0}

so this is the total flux

now by Gauss's law we can find the electric field

\int E.dA = \phi

\int E.dA = \frac{\lambda * L}{\epsilon_0}

E* 2\pi rL = \frac{\lambda * L}{epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

<em>by above expression we can find the electric field at required position</em>

8 0
1 year ago
A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end
Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

and linear mass density= \frac{8.25 g}{65 cm}

Linear mass density= \frac{8.25*10^-3}{65*10^-2}

Linear mass density= 0.0127 kg/m

Velocity= 2*\frac{l}{t}

Velocity= 2 * \frac{65*10^-2}{7.84}

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So putting all these values in equation we get

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

165.8= \sqrt{\frac{m*9.81}{0.0127} }

Solving we get

m= 35.58 kg

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3 0
2 years ago
Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi
Doss [256]

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

5 0
2 years ago
If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th
cluponka [151]

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

7 0
2 years ago
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