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2 years ago

What happens when you stroke the prongs with the wire?

1 answer:

8 0

When you stroke the prongs with a wire, a static sound is being produced. This kind of minimum-information noise is called white noise, by analogy with white light which is a uniform mixture of all the different possible colors. Hope this helps.

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1) Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

Anton [14]

Answer:

The order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

Option (A) is correct.

Explanation:

Given:

Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

From the formula of energy in terms of frequency.

$E = hf$

Where $h =$ planck constant, $f =$ frequency of light.

From above formula we can conclude that higher frequency means higher energy.

In our case ultraviolet has higher frequency and microwave has lower frequency.

So ultraviolet has higher energy and microwave has lower energy.

microwave < infrared < visible < ultraviolet

Therefore, the order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

7 0

2 years ago

Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s

Black_prince [1.1K]

The initial momentum of the system can be expressed as,

$p_i=m_1u_1+m_{2_{}}u_2$

The final momentum of the system can be given as,

$p_f=m_1v_1+m_{2_{}}v_2$

According to conservation of momentum,

$p_i=p_f$

Plug in the known expressions,

$\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}$

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

$\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}$

Thus, the final velocity of second mass is 2.99 m/s.

3 0

11 months ago

An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

2 years ago

Fan blades would be an analogy for which atomic model? Thomson's Bohr's electron cloud Dalton's

nata0808 [166]

The correct answer would be Electron Cloud

5 0

2 years ago

A circuit consists of a conducting movable bar and a lightbulb connected to two conducting rails as shown in the figure below. A

IrinaVladis [17]

Moving the bar to the left or the right will cause emf to be generated, making the bulb light up. The first two options are correct.
Moving parallel to the field or not moving at all and increasing or decreasing the intensity of the field will not make the bulb light up because no current will be induced.

6 0

2 years ago