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True [87]
3 years ago
7

Calculate the internal energy (in J) of 86 mg of helium at a temperature of 0°C.

Physics
1 answer:
coldgirl [10]3 years ago
6 0

Answer:

The internal energy is 73.20 J.

Explanation:

Given that,

Weight of helium = 86 mg

Temperature = 0°C

We need to calculate the internal energy

Using formula of internal energy

U =nc_{v}T

Where, c_{v} = specific heat at constant volume

He is mono atomic.

So, The value of c_{v}= \dfrac{3}{2}R

now, 1 mole of Helium = 4 g helium

n =number of mole of the 86 mg of helium

n = \dfrac{86\times10^{-3}}{4}

n =2.15\times10^{-2}\ mole

T = 0°C=273 K

Put the value into the formula

U = 2.15\times10^{-2}\times\dfrac{3}{2}\times8.314\times273

U = 73.20\ J

Hence, The internal energy is 73.20 J.

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A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is
Arada [10]

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

          \frac{1}{p} = \frac{1}{f} - \frac{1}{q}

          1 / p = 1 / 2.2 - 1/3.29

           1 / p = 0.15059

           p = 6.64 cm

therefore the farthest distance from the object is 6.64 c

3 0
3 years ago
A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic
Mazyrski [523]

Answer:

The total resistance of the wire is = 1.917\times10^{-3}

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, R_{Total}

\frac{1}{R_{Total}}  =\frac{1}{R_{cu}  }+\frac{1}{R_{al}}

Parameters given:

Length of wire = 1 m

Cross sectional area of copper A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3}  )^{2} =3.142\times10^{-6} m^{2}

Cross sectional area of aluminium wire  

A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3}  )^{2}-(1\times 10^{-3}  )^{2}] =9.42\times10^{-6} m^{2}\\

Resistivity of copper \rho _{cu}=1.7\times 10^{-8}  \Omega .m

Resistivity of Aluminium \rho _{al}=2.8\times 10^{-8}  \Omega .m

Resistance of copper R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} }  =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega

Resistance of aluminium R_{al}= \frac{\rho_{al} \times l}{A_{al} }  =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega

The total resistance of the wire can be obtained as follows;

\frac{1}{R_{Total}}  =\frac{1}{5.41\times10^{-3}  }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}

R_{Total}= 1.917\times 10^{-3}\Omega

∴ The total resistance of the wire = 1.917\times 10^{-3}\Omega

4 0
3 years ago
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A wild pig attacks a hunter with a velocity of 5 m/s. At the instant when the pig is at a distance of 100 m the hunter shoots an
castortr0y [4]

Answer:

31.55 m/s

Explanation:

Let the initial velocity of the arrow is u metre per second.

Angle of projection, θ = 40 degree

range = 100 m

Use the formula for the range.

R = \frac{u^{2}Sin2\theta }{g}

100 = u^2 Sin(2 x 40) / 9.8

100 x 9.8 = u^2 Sin 80

u^2 = 995.11

u = 31.55 m/s

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alexdok [17]

Answer:

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Explanation:

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