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3 years ago

Helppppppl!!!!!!!!!!!

Physics

1 answer:

RUDIKE [14]3 years ago

8 0

Answer:

I think deposition of C is the oldest deposit

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Help me please brainlest

crimeas [40]

B . it should be convert energy.

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3 years ago

Read 2 more answers

12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be

Nat2105 [25]

Answer:

$81:256$ .

Explanation:

Let $T$ denote the absolute temperature of this object.

Calculate the value of $T$ before and after heating:

$T(\text{before}) = 27 + 273 = 300\; \rm K$ .

$T(\text{after}) = 127 + 273 = 400\; \rm K$ .

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to $T^4$ .

Ratio between the absolute temperature of this object before and after heating:

$\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}$ .

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

$\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}$ .

4 0

2 years ago

If a pizza delivery guy declares himself the leader over the other pizza delivery people, he probably won't have many people obe

NemiM [27]

Answer:

the last one i think...

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2 years ago

Unlike mechanical waves such as sound or earthquake waves, EM waves can travel through empty space. True or false

Alex777 [14]

Answer:

its true

Explanation:

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2 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

2 years ago

Helppppppl!!!!!!!!!!!​

Helppppppl!!!!!!!!!!!