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2 years ago

Materials with resistivities between those of good conductors and those of insulators are called:

Physics

2 answers:

bogdanovich [222]2 years ago

8 0

semiconductors ... silicon, germanium etc ...

nevsk [136]2 years ago

8 0

Answer:

Semiconductors

Explanation:

The resistivity of conductors is low and the resistivity of insulators is high. The materials having resistivity between conductors and insulators are called semiconductors.

For example, silicon, gemanium, etc.

The elements having valency +4 are the semiconductors.

By increasing the temperature os a semiconductor, the conductivity increases or by the process of dopping the conductivity of semiconductor also increases.

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A constant magnetic flux through a closed loop of wire induces an emf in that loop. True or false?

miskamm [114]

Answer: False

Explanation: In order to explain this problem we have to use the Faraday law, which say

dФm/dt=-ε it means that the variation of the magnetic field flux with time is equal to the emf ( electromotive force). In our case the magnetic flux is constant then there is not a emf induced in a wire closed loop.

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3 years ago

A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars

Sergeeva-Olga [200]

Answer:

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2 years ago

Read 2 more answers

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

8 0

2 years ago

An object of mass 100kg is raised 2m above the ground using an Inclined Plane of length 10m calculate the effort parallel to inc

MArishka [77]

Answer:

Slope = 2 m / 10 m = 1/5

For every 5 m of effort the object will be raised 1 m

W = work done on object = M g h increase in PE of object

E S = W where E is effort and S the distance thru which the effort acts

E S = M g H

E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N

Check: total work = 2 * 9.8 * 100 = 1960 J

Force Needed = 1960 J / 2 m = 980 Newtons

Mechanical advantage = 980 / 196 = 5 as one would expect since the object is raised 1 m for every 5 m of force input

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2 years ago

What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300

Vinvika [58]

The efficiency of an ideal Carnot heat engine can be written as:
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the temperature of the cold region
$T_{hot}$ is the temperature of the hot region

For the engine in our problem, we have $T_{cold}=300 K$ and $T_{hot}=400 K$ , so the efficiency is
$\eta= 1 - \frac{300 K}{400 K}=0.25$

4 0

2 years ago