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fenix001 [56]
3 years ago
6

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.400 m and

the length of the copper section is 0.800 m . Each segment has cross-sectional area 0.00700 m2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice-water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings.
(a) What is the temperature of the point where the brass and copper segments are joined?
(b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?
Physics
1 answer:
vlabodo [156]3 years ago
7 0

Answer:

a) 36°

b) 0.109 kg

Explanation:

Heat flows from brass to copper with the brass having its temperature

Length of brass = 0.4

Length of copper = 0.8

Temperature of = 36.15

See attachment for calculation

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

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So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr
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Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

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F = 2.16×10⁻⁶ N

5 0
2 years ago
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