According to classical physics what should happen as a filament in a light bulb gets hotter?

A student plucks a fixed-end string, creating a standing wave with 6.00 nodes (including any nodes at the ends). The string is t

Vesna [10]

1) 2.5 wavelengths

2) 0.208 m

3) 1731 Hz

Explanation:

1)

Standing waves are waves that do not propagate, but instead the particles of the medium just oscillate around a fixed position. Examples of standing waves are the waves produced on a string with fixed ends.

The points of a standing wave in which the amplitude of the oscillation is always zero are called nodes.

The two fixed ends of the string are two nodes. In this problem, we have a total of 6 nodes along the string: this means that there are 4 additional nodes apart from the two ends of the string.

Therefore, this also means that the string oscillate in 5 different segments.

One wavelength is equal to 2 segments of the oscillation: therefore, since here there are 5 segments, this means that the number of wavelengths that we have in this string is

$n=\frac{5}{2}=2.5$

2)

The wavelength of a wave is the distance between two consecutive crests (or throughs) of the wave.

The wavelength of a standing wave can be also measured as the distance between the nth-node and the (n+2)-th node: so, basically, the wavelength in a standing wave is twice the distance between two nodes:

$\lambda = 2 d$

where

$\lambda$ is the wavelength

d is the distance between two nodes

Here the length of the string is

L = 0.520 m

And since it oscillates in 5 segments, the distance between two nodes is

$d=\frac{L}{5}=\frac{0.520}{5}=0.104 m$

And therefore, the wavelength is

$\lambda=2d=2(0.104)=0.208 m$

3)

The frequency of a wave is the number of complete oscillations of the wave per second.

The frequency of a wave is related to its speed and wavelength by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

In this problem:

v = 360 m/s is the speed of the wave

$\lambda=0.208 m$ is the wavelength

Therefore, the frequency is

$f=\frac{v}{\lambda}=\frac{360}{0.208}=1731 Hz$

3 0

3 years ago

Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

olga_2 [115]

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

6 0

3 years ago

How can you tell high frequency on a wave

Svetradugi [14.3K]

A wave travelling with high frequency has short wavelength and high pitched !

5 0

4 years ago

North Dakota Electric Company estimates its demand trend line (in millions of kilowatt hours) to be: D = 75.0 + 0.45Q, whe

Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line is

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

We need to calculate the demand forecast for winter

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times101)\times0.80$

$D=96.36\ millions\ KWH$

We need to calculate the demand forecast for spring

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times102)\times1.20$

$D=145.08\ millions\ KWH$

We need to calculate the demand forecast for summer

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times103)\times1.40$

$D=169.89\ millions KWH$

We need to calculate the demand forecast for fall

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times104)\times0.60$

$D=73.08\ millions KWH$

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0

3 years ago

When you do work on an object some of the energy is ___ to that object

solniwko [45]

Answer:

The answer is transferred. So your answer will be: When you do work on an object some of the energy is transferred to that object.

5 0

3 years ago