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4 years ago

Compare lunar and solar eclipses and answer the following questions:

Physics

1 answer:

xeze [42]4 years ago

3 0

So lunar eclips earth between sun and moon
Solar eclips moon between sun and earth.
About the 3th.. im not sure, it depends on if you meen a total solar eclips or not... i think total is more rare then a lunar eclipse..

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What is an example of a non contact force

olchik [2.2K]

Answer: Gravitational force

Explanation:

A non contact force can be described as a force applied to an object by another body that is not in direct contact with it.

For example, an object thrown upwards will return back due to the force of gravity acting on it. So, it means Gravitational force is acting on the body without necessarily being in contact with that body.

8 0

3 years ago

What can make a stronger electric or magnetic field?

rusak2 [61]

Answer:

add more coils

Explanation:

this increases the strength of the magnetic field

5 0

3 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

Physics - Electricity and Magnetism

Orlov [11]

Answer:

Explanation:

because the unit for resistance is in ohms

4 0

3 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago