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3 years ago

Why vacuum flask is known as thermos flak?

1 answer:

8 0

The flask was later developed using new materials such as glass and aluminum; however, Dewar refused to patent his invention. ... The name later became a genericized trademark after the term "thermos" became the household name for such a vacuum-insulated container for liquids.

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Imagine that the light you observe from space is exhibiting a redshift. This would mean that the Universe is Choose one: A. trav

docker41 [41]

Answer: C. Expanding

Explanation:

When an object moves away from an observer, the light waves emitted by the object are stretched out making them have move towards the red end of the electromagnetic spectrum, where light has a longer wavelength. This phenomenon is known as the redshift (cosmological redshift). For most astronomical objects, the observed spectral lines are all shifted to longer wavelengths. It is caused solely by the expansion of the Universe and thus, the value of a redshift indicates the recession velocity of the expanding object, or its distance.

Redshift is opposed to the blueshift. Blue light wavelengths tend to be closer together hence blueshift light tends to be emitted from objects that are collapsing.

7 0

3 years ago

A cube that is 20 nanometer on an edge contains 399,500 silicon atoms, and each silicon atom has 14 electrons and 14 protons. In

Sergeu [11.5K]

Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

N-type Impurities: Pentavalent impurities e.g. Phosphorous, Arsenic are added to have an additional electron in the structure. Thus a pentavalent impurity creates 1 additional electron.
P-type Impurities: Trivalent impurities e.g. Boron, Aluminium are added to have a positive "hole" in the structure. Thus a trivalent impurity creates 1 hole.

Now for estimation of extra electrons in the impured structure is as

$N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\$

Now for estimation of "holes" in the impured structure is as

$N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\$

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

$N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3$

So total 3 "holes" are available for conduction of current at 300K.

6 0

2 years ago

What happens to a gas that is enclosed in a rigid container when the temperature of the gas is increased?

kiruha [24]

A. the gas particles move faster and collide more frequently, which causes an increase in pressure.

Increasing temperature increases the energy of the gas, which causes the kinetic energy of the molecules to increase.

4 0

3 years ago

The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an ang

V125BC [204]

Answer:

2954.6 N/C, 46.36 degree from positive axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E = 1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

5 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

3 years ago

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