The si base unit for length is the

A) Khi nào một vật được coi là chuyển động ? Cho ví dụ ?

Lady bird [3.3K]

Answer: ) Khi nào một vật được coi là chuyển động ? Cho ví dụ ?

b) Khi nào một vật coi là đứng yên? Cho ví dụ ?

c) Theo em câu phát biểu “ Khi khoảng cách từ vật đến vật mốc không thay đổi theo thời gian thì vật đứng yên so với vật mốc”. Có luôn đúng không, vì sao?

Explanation:

3 0

2 years ago

A person standing a certain distance from an airplane with four equally noisy jet engines is experiencing a sound level of 145 d

Katyanochek1 [597]

Answer:

The sound level is 138.97 dB.

Explanation:

Given that,

Sound level L= 145 dB

We need to calculate the intensity

Using formula of sound intensity

$L=10\log(\dfrac{I}{I_{0}})$

Put the value into the formula

$145=10\log(\dfrac{I}{I_{0}})$

$\log(\dfrac{I}{I_{0}})=\dfrac{145}{10}$

$\log(\dfrac{I}{I_{0}})=14.5$

$\dfrac{I}{I_{0}}=10^{14.5}$

$I=10^{14.5}\times I_{0}$

$I=10^{14.5}\times10^{-12}$

$I=316.2$

We need to calculate the noise of one engine

Using formula of intensity

$I'=\dfrac{I}{4}$

Put the value into the formula

$I'=\dfrac{316.2}{4}$

$I'=79.05$

We need to calculate the intensity level due to one engine

Using formula of sound intensity

$L'=10\log(\dfrac{I'}{I_{0}})$

$L'=10\log(\dfrac{79.05}{10^{-12}})$

$L'=138.97\ dB$

Hence, The sound level is 138.97 dB

5 0

2 years ago

What would make oppositely charged objects attract each other more?

Elina [12.6K]

Answer:

a: increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object

Explanation:

5 0

3 years ago

The waste product of photosynthesis is:

zalisa [80]

Answer:

The waste product of photosynthesis is oxygen

6 0

2 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