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Semmy [17]
3 years ago
14

If the resistivity of copper is less than that of gold at room temperature, which of the following statements must be true? Gold

has a higher resistance than copper.The sample of gold is thinner than the sample of copper. Electrons in gold are more likely to be scattered than electrons in copper at room temperature when they are accelerated by the same electric field. There is more current flowing in the gold than in the copper.
Physics
1 answer:
KiRa [710]3 years ago
5 0

Answer:

Gold Has A Higher Resistance Than Copper. The Sample Of Gold Is Thinner Than The Sample Of Copper. Electrons In Gold Are More Likely To Be Scattered Than Electrons In Copper At Room Temperature When they are exelerated by the same electric field.

Explanation:

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The net force does not depend on the mass.

We have 12N to the right, and 19N to the left.
The net force is (19.0-12.0)N=7.0N to the left.

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A laser pointer used in a lecture hall emits light at 650 nm what is the frequency
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Don Walsh and Jacques Piccard:
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A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be​
zhuklara [117]

The question is incomplete, the complete question is;

A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?

Answer:

See explanation

Explanation:

From Newton's law of cooling;

θ1 - θ2/t = K(θ1 + θ2/2 - θo]

Where;

θ1 and θ2 are initial and final temperatures

θo is the temperature of the surroundings

K is the constant

t is the time taken

Hence;

100 - 60/5 = K(100 + 60/2 - θo)

100 - 40/10 = K(100 + 40/2 - θo)

8= (80 - θo)K -----(1)

6= (70 - θo)K -----(2)

Diving (1) by (2)

8/6 = (80 - θo)/(70 - θo)

8(70 - θo) = 6(80 - θo)

560 - 8θo = 480 - θo

560 - 480 = -θo + 8θo

80 = 7θo

θo = 11.4°

Again from Newton's law of cooling;

θ = θo + Ce^-kt

Where;

t= 0, θ = 60° and θo = 11.4°

60 = 11.4 + C e^-K(0)

60 - 11.4 = C

C= 48.6°

To obtain K

40 = 11.4 + 48.6e^-10k

40 -11.4 = 48.6e^-10k

28.6/48.6 = e^-10k

0.5585 = e^-10k

-10k = ln0.5585

k= ln0.5585/-10

K= 0.0583

Hence, the temperature in 15 minutes;

θ= 11.4 + 48.6e^(-0.0583 × 15)

θ= 31.7°

4 0
2 years ago
In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required
konstantin123 [22]

Answer:

2.46 eV

Explanation:

It is given that,

The energy of light that fall on the metal = 3.56 eV

The stopping potential of the metal = 1.1 V

We need to find the work function of the metal. It is given by the relation as follows :

W = E-KE ...(1)

Where KE is the kinetic energy of the ejected electron and it is given by :

KE = V×e

= 1.1 eV

Put all the values in formula (1)

W = 3.56 eV - 1.1 eV

= 2.46 eV

Hence, the work function of the metal is 2.46 eV. Hence, the correct option is (c).

5 0
2 years ago
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