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SVEN [57.7K]
3 years ago
9

Describe the working of an electric bell​

Physics
1 answer:
denpristay [2]3 years ago
3 0

Answer:

An electric bell contains an electromagnet, consisting of coils of insulated wire wound round iron rods. When an electric current flows through the coils, the rods become magnetic and attract a piece of iron attached to a clapper. The clapper hits the bell and makes it ring.

Explanation:

⠀⠀⠀⠀▕▔╱▔▔▔━▁⠀⠀⠀⠀

⠀⠀⠀▕▔╱╱╱▆⠀╲▂▔▔╲⠀

⠀⠀▕▔╱╱╱╱▔▂▂▂▂▂▂▏

⠀▕▔╱▕▕╱╱╱⠀▽▽▽▽▽⠀

▕▔╱⠀⠀╲╲╲╲▂△△△△⠀⠀

▔╱⠀⠀╱▕╲▂▂▂▂▂▂╱⠀⠀

╱⠀⠀╱┉▕┉┋╲⠀⠀⠀⠀⠀⠀⠀

⠀⠀╱┉┋▕┉┋┉╲

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During the period that the moon’s phases are changing from new to full, the moon is _____. waxing exhibiting retrograde motion w
Sindrei [870]
The answer that best fits the blank is the term WAXING. The moon is waxing whenever it reaches to the period that its phases are transitioning from new to full. The answer is the first option. This is when it is more that half is illuminated. Hope this helps.
3 0
3 years ago
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A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime
Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0
3 years ago
ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces
saul85 [17]

Answer:

The magnitudes of the second force is   Z = 129.9 N

The magnitudes of the  resultant force is   R = 256.047 N

Explanation:

From the question we are told that  

    The force is  F = 187 \ lb

     The angle made with second force \theta_o = 73 ^o 36' =  73 + \frac{36}{60}  =  73.6^o

     The angle between the resultant force and the first force \theta _1  = 29 ^o 1 ' = 29 + \frac{1}{60}  = 29.0167^o

For us to solve problem we are going to assume that

     The magnitude of the second force is  Z N

     The magnitude of the resultant force is R N

According to Sine rule

                \frac{F}{sin (\theta _o - \theta_1 }  = \frac{Z}{\theta _1}

Substituting values

             \frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}  

             267.82 =\frac{Z}{0.4851}  

              Z = 129.9 N

According to cosine rule

       R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }

Substituting values

     R = \sqrt{187^2 + 129.9 ^2  + 2 (187 ) (129.9) cos (73.6)}

     R = 256.047 N

 

3 0
3 years ago
An object weighs 39.36 newtons. What is its mass if a gravitometer indicates that g = 9.83 m/s2?
CaHeK987 [17]
The force applied to an object is said to be a product of its mass and the acceleration. For this case, acceleration is the reading on the gravitometer. We calculate as follows:

F = mg
39.36 N = m(9.83 m/s^2)
m = 4.00 kg

Hope this answers the question. Have a nice day.
6 0
4 years ago
Read 2 more answers
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

M = \frac{gr^2}{G}

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

g = 9.8m/s^2\\G  = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m

Replacing at this equation we have that

M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg

The Volume of a Sphere is equal to

V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3

Therefore using the relation between mass, volume and density we have that

\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3

6 0
3 years ago
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