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forsale [732]
2 years ago
13

50 POINTS!! BRAINLEST

Physics
2 answers:
likoan [24]2 years ago
8 0

Answer: pls mark me as a brainliest friend pls, it would be a great help for me friend

Explanation:

1. B

2. The graph that shows students data is graph b. It is constant so it’s velocity is constant.

Hope it’s useful friend…

Good luck and have a great day buddy…

Stay safe and happy…

Keep rocking !!

sergey [27]2 years ago
4 0

Answer:2 one

Explanation:

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Answer please help me ​
aksik [14]

Answer:

It is showing the wavelength.

Explanation: Hope it helps you:)))

have a good day

7 0
2 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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=157^{\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-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

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

here, \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}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

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

8 0
3 years ago
A speed-time graph is shown below:
VladimirAG [237]

Answer: 0.5 m/s^{2}

Explanation:

Average acceleration a_{ave} is the variation of velocity \Delta V over a specified period of time \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

\Delta V=V_{f}-V_{o} being V_{o}=0 cm/s the initial velocity and V_{f}=4 cm/s the final velocity  (according to the information given from the described graph)

\Delta t=8 s

Then:

a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}

a_{ave}=0.5 m/s^{2}

5 0
3 years ago
The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit
Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V=  30 m/s  ( + x direction )

d)  λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

As we know that standard form of wave equation given as

y=A sin(\phi -\omega t)

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s                     (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s  ( + x direction )

V = f λ

30 = 95.49 x  λ

 λ = 0.31 m

We know that speed is the rate of displacement

U=\dfrac{dy}{dt}

U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0
3 years ago
Describe at least TWO hazardous conditions that exist in space AND the technological features a spacecraft must have in order to
nexus9112 [7]

gamma radiation and heat flares from the sun, they use refelective gold sheets

3 0
3 years ago
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