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3 years ago

14

What is the sharpness of the light and the amount of voltage when you put the magnet through the lower number of coils first vs

the higher number of coils

1 answer:

fomenos3 years ago

4 0

Answer:hjvuihi

Explanation:lk

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Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0

3 years ago

Does research lessen its value for its being cyclical? Enumerate Two Possible advantages of a cyclical research

Tanya [424]

Answer:

Research usually is a cyclical process because it starts with a problem and ends with a problem.

This is bad or lessens the value of the research?

No, because there are problems that in nature are cyclical, (for example the related ones to action research).

And it also may be a good thing, solving one "problem" leads to another problem, but in the process of solving the first one we may win a lot of knowledge, and the same happens with the next one, and so on.

Two possible advantages of cyclical research are:

Knowing beforehand that the research will be cyclical, will allow us to estimate better the amount of time and money needed because we already know (more or less) where to aim.

It also may lead to a better end product, as we already know that we must focus in solving one thing and then we can focus in the next one.

Another possible advantage may be that we know that after the work, there will be a new thing to research, and it is fun, so if you are curious enough this may be a good thing (especially in scientific areas, physics, chemistry, etc).

7 0

3 years ago

The force required to stretch a Hooke’s-law

Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K

Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J

5 0

3 years ago

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

So

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

So

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

So

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

So

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.

4 0

3 years ago

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

Free_Kalibri [48]

Answer:

$\beta = 41.68°$

Explanation:

according to snell's law

$\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }$

refractive index of water n_w is 1.33

refractive index of glass n_g is 1.5

$sin\alpha = \frac{n_w}{n_g}* sin30$

$sin\alpha = 0.443$

now applying snell's law between air and glass, so we have

$\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}$

$sin\beta = \frac{n_g}{n_a} sin\alpha$

$\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]$

we know that $sin\alpha = 0.443$

$\beta = 41.68°$

7 0

3 years ago