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1 year ago

Does everything in a black hole contain a singularity?

2 answers:

8 0

In the real universe, no black holes contain singularities. In general, singularities are the non-physical mathematical result of a flawed physical theory.

melamori03 [73]1 year ago

4 0

Answer:

All black holes contain singularities, however not all singularities involve black holes. A neutron star may be dense, matter the size of a pinhead can weigh as much as the earth, but there seems to be a mathematical cut-off point beyond which a black hole is formed.

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In what part of the scientific method do you record data?

densk [106]

C..............................

6 0

3 years ago

When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

Nata [24]

Answer:

The energy absorbed by the atomic electrons in the mercury atom is $7.84 \times 10^{-19}$ J

Explanation:

Given:

Potential $V = 4.9$ V

According to the conservation law,

Loss in kinetic energy = Gain in potential energy

Here, energy absorbed by the atomic electrons is given by,

$E = eV$

Where $e = 1.6 \times 10^{-19} C$ ( charge of electron )

$E = 1.6 \times 10^{-19 } \times 4.9$

$E = 7.84 \times 10^{-19}$ J

Therefore, the energy absorbed by the atomic electrons in the mercury atom is $7.84 \times 10^{-19}$ J

5 0

2 years ago

How do paved parking lots and roads with concrete or asphalt affect water flow on the land?

harina [27]

On the regular ground, water usually seeps in to the soil, which has a number of benefits for the wildlife in that area. However, with asphalt and concrete, there is no soft and absorbent soil to take in the water, so it just keeps flowing down to the lowest part of land it can. Additionally, some water runoff can carry fertilizer and other harmful chemicals with it in to the oceans and lakes it’s dumped in, which harms the ecosystem in them as well.

8 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.

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3 years ago

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What forms when two air masses meet and create weather?

Minchanka [31]

When two different air masses meet, a boundary is formed. the boundary between two air masses is called a front. weather at a front is usually cloudy and stormy. there at four different fronts: cold, warm, stationary, and occluded

4 0

3 years ago

Read 2 more answers