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Bezzdna [24]
2 years ago
7

Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is or

iginally at 24°C, neglecting any changes in dimensions? Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is originally at 24°C, neglecting any changes in dimensions?
Physics
1 answer:
sleet_krkn [62]2 years ago
8 0

Answer:

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

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belka [17]

Answer:

Disruption to electricity power grid

Explanation:

We're looking a a solar flare. This will whip solar particles at high velocity into space and, If they are near earth, will interact with the earth's magnetic field. These magnetic changes will be measurable in the electric grid. Whether they are strong enough to cause "disruption" depends on a huge number of factors such as strength of and angles of the interacting magnetic fields and location of grid infrastructure,

4 0
2 years ago
Read 2 more answers
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
3 years ago
Match the following:
Art [367]

Answer:

1) Tailored software  Library management system

2) Utility software Scan viruses

3) Operating system Used to coordinate the hardware of the computer

4) Packaged software set of programs Microsoft office

Explanation:

1) A tailored software, also known as a custom software, is one that is designed and tailor-made only for a particular organisation

2) A utility software is a computer maintenance and analysis software used to enable proper functioning of the computer by performing restorative and maintenance tasks

3) Operating system software

The operating system software controls the operation of the computer hardware within the system and enables the operation of other programs in the computer

4) Packaged software are a collection of programs that are oriented to perform interrelated tasks that a focused to a particular area, such as Microsoft Office.

8 0
3 years ago
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³
Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, \rho_o = 800 kg/m³

density of water, \rho_w = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0
3 years ago
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
3 years ago
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