All categories

3 years ago

Which is most likely the length of a student’s textbook?

Physics

2 answers:

UkoKoshka [18]3 years ago

3 0

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

$30 mm = (30) * (\frac{1}{1000}) = 0.030 m$

$30 cm = (30) * (\frac{1}{100}) = 0.30 m$

$30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m$

Therefore, physically the correct measure is:

$0.30 m = 30 cm$

Answer:

the length of a student's textbook most likely is:

30 centimeters

LenKa [72]3 years ago

3 0

Answer:

30 centimeters

Explanation:

I just took the quiz and I got a 100 on it

You might be interested in

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$

Where,

$I_ {0}$ indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

$\theta$ indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

$I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)$

Replacing the values,

$I=100*cos^2(20)*cos^2(40-20)$

$I=100*cos^4(20)$

$I=77.91W/m^2$

Therefore the intesity of the light after it has passes through both polarizers is $77.91W/m^2$

7 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

Two guitarists attempt to play the same note of wavelength 6.48 cmcm at the same time, but one of the instruments is slightly ou

k0ka [10]

The answer is λ₂ = 6.48 cm or 6.52 cm.

The out-of-tune guitar may have a wavelength between "6.48 cm" and "6.52 cm."

fb = |f2 − f1|

f₁ = 343/0.064

= 5276Hz

f₂ = 5276.9 Hz ± 17 Hz

f₂ = 5293.9 Hz or 5259.9 Hz

Now, calculating the possible wavelengths:

λ = 343/ 5259.9 or 343/ 5293.9

λ₂ = 6.48 cm or 6.52 cm

<h3>Why is beat frequency important?</h3>

When two waves with almost identical frequencies traveling in the same direction collide at a certain location, beats are produced. The opposing beneficial and harmful disruption causes the sound to alternatively be loud and weak whenever two sound waves with different frequencies reach your ear. This is referred to as beating.

The entire value of the frequency difference between the two waves is the beat frequency.

The following formula yields the beat frequency:

fb = |f2 − f1|

Learn more about beat frequency here:

brainly.com/question/14705053

#SPJ4

5 0

2 years ago

Read 2 more answers

Gravity is defined as: Select one: a. an apple falling from a tree toward the ground. b. ability to accomplish a job with the le

Scorpion4ik [409]

Answer: