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3 years ago

7

When it is brought near a neutral object, a charged object can _________ a charge in a neutral object.A) induce, B) cancel,C) ne

utralize, D) add or subtract

1 answer:

Bess [88]3 years ago

3 0

The answer is a induce.

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What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?

lisabon 2012 [21]

<h3>Answer:</h3>

E≈2,5 eV

<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

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$\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\; eV\approx \boldsymbol{2,5\; eV}$

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1 year ago

It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Ass

Vinvika [58]

The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

<h3>Acceleration of the car </h3>

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

<h3>Distance traveled when the speed is 32 m/s</h3>

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

Learn more about distance here: brainly.com/question/4931057

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1 year ago

How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity $(K.E.=\frac{1}{2}mv^2)$ . If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height $(P.E.=mgh)$ . If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.

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3 years ago

What type of energy waves are used in Solar Cells, and what is it’s position on the electromagnetic spectrum?

Artyom0805 [142]

Answer: Electromagnetic waves (Ultraviolet light, between 100 nm and 380 nm)

Explanation:

Solar cells work by the photoelectric effect, which consists of the emission of electrons (electric current) when light (electromagnetic waves) falls on a metal surface under certain conditions.

In this sense, the portion of the electromagnetic spectrum this cells use is Ultraviolet light (UV) from the Sun, whose wavelength is approximately between 100 nm and 380 nm.

It is important to note, this is a type of electromagnetic radiation that is not visible to the human eye.

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3 years ago

A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio

lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

$W = F\times d$

$KE = 0.5\times m\times v^2$

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

$W = F \times d = 2.62 N \times 100 m$

$W = 261.6 N\times m$

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

$0.5 m\times v2^2 = 0.5 m\ v1^2 - W$

Now solve for v2

$v2 = \sqrt{v1^2 - {\frac{2W}{M}}}$

$= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}$

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

$Ff = Us\times N$

Now solve for Us

$= \frac{Ff}{N}$

$= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}$

= 0.00494

4 0

2 years ago