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RideAnS [48]
3 years ago
9

(b) Cable of a microwave oven has three wires inside it which have insulation of different colours black, green and red. Mention

the significance of the three colours and potential difference between red and black one.
Physics
1 answer:
MakcuM [25]3 years ago
8 0
The green is ground.  Ideally, no current travels in this one.  The red and black are the power and neutral wires but which colors they are depends on a convention.  In the US, you will actually have a black (power) and a white (neutral)  Here it's red and black and usually in a red/black system the red is the power.  Either way there is a potential of 120V rms between them.
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What percentage of the original kinetic energy is convertible to internal energy?
schepotkina [342]
There would be very less percentage loss<span> of the kinetic energy during </span>the conversion<span> to internal energy, assuming that there is less air in the </span>surroundings<span>. Also, the friction will contribute to the conversion where if it is, the percentage loses is negligible.</span>
7 0
3 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will
nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg

So, the mass that is attached to the spring is 1 kg.

6 0
2 years ago
Whqt do you mean by geo-stationery satellite? Explain.​
Ratling [72]

Answer:

Description: When a geosynchronous satellite is placed directly above the Equator with a circular orbit and angular velocity identical to that of the Earth, the satellite is known as a geostationary satellite

Explanation:

These satellites appear to be stationary above a particular point which is due to the synchronization. This type of satellite provides a distinct benefit of fixing the receiving antennas at one place, hence making them more economical than regular tracking antennas. Over the years, these satellites have helped in revolutionizing the global communications, weather forecasting and television broadcasting. When the orbit of a geosynchronous satellite is not aligned directly over the equator, the orbit is known as an inclined orbit.

8 0
3 years ago
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
A particle moves along a straight line with an acceleration of a = 5&gt;(3s 1&gt;3 + s 5&gt;2) m&gt;s2, where s is in meters. De
Len [333]

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation  a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

\int\limits^a_b {x} \, dx

<em>a=2</em>

<em>b=1</em>

<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>

<em>dx=dv</em>

Integrate the left side the standard method.

\int\limits^a_b {x} \, dx

<em>a=v</em>

<em>b=0</em>

<em>dx=dv</em>

<em>Integrating</em>

=v^2/2

Use Simpson's rule for the right site.

\int\limits^a_b {x} \, dx

<em>a=b</em>

<em>b=a</em>

<em>x=f(x)</em>

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

        =0.8376

Solve for v.  

      v=1.295

5 0
3 years ago
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