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nadya68 [22]
2 years ago
14

The metal wire in an incandescent lightbulb glows when the lights is switch on and stops glowing when switched off. This simple

process is which kind of a change?
Physics
2 answers:
Radda [10]2 years ago
7 0

Opening or closing a circuit.

lutik1710 [3]2 years ago
4 0

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.  

Explanation:

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A small glass bead has been charged to 8.0 nc. what is the magnitude of the electric field 2.0 cm from the center of the bead?
astraxan [27]
<span>Charge of the glass bead Q = 8.0 x 10^-9 C Distance d = 2.0 cm = 0.02 m Coulombs constant K = 8.99 x 10^9 Nm^2/C^2 Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2 E = 71.92 / 0.0004 = 17.98 x 10^4 The electric field is 1.8 x 10^5 N/C</span>
7 0
3 years ago
A wind turbine takes in energy from wind with the goal of converting it into electrical energy. Much of the wind energy is also
Shalnov [3]
German physicist  Albert Betz  (in 1919) demonstrated  that the highest efficiency you can achieve with a wind turbine is around 59%

We would have to analyze the design of an specific turbine to determine its efficiency, however it is unlikely to achieve 50% , as todays turbines have an average efficiency in the 20-35%

The answer would be around 25%

6 0
3 years ago
Read 2 more answers
Can you have zero displacement and nonzero average velocity? Zero displacement and nonzero velocity? Illustrate your answers on
Svetlanka [38]

a) Not possible

b) Yes, it's possible (see graph in attachment)

Explanation:

a)

The average velocity of a body is defined as the ratio between the displacement and the time elapsed:

v=\frac{\Delta x}{\Delta t}

where

\Delta x is the displacement

\Delta t is the time elapsed

In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,

\Delta x = 0

And therefore as a consequence,

v=0

which means that the average velocity is zero.

B)

Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking  about average velocity, but we are talking about (instantaneous) velocity.

On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.

In all of this, we notice that the total displacement of the object is zero:

\Delta x = 0

However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

3 0
3 years ago
We have a 10kg ball and a 15kg ball. We drop them from the top of the science building while objects are at rest and there is no
Mashcka [7]
They will hit the ground at the same time, as mass is negligible when calculating the acceleration of gravity when there is no air resistance
3 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid
bixtya [17]

Answer:

u_{s}=0.56

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

F_{s.max}=\frac{mv^2}{r} \\

Where the r is the radius of merry-go-round and v is the tangential speed

but

F_{s.max}=u_{s}F_{N}=u_{s}mg

So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

Substitute the given values

So

u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56

6 0
3 years ago
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