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Aliun [14]
4 years ago
14

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

ogs sprint away in different directions while Daniel stands still. From Daniel's point of view, Newton runs due North at 3.40 m/s, but from Pauli's point of view, Newton appears to be moving at 1.10 m/s due East. What must Pauli's velocity relative to Daniel be for this to be true? Express your answer in terms of the ???? ‑ and ???? ‑components if North is the +???? ‑direction and East is the +???? ‑direction.
Physics
1 answer:
DedPeter [7]4 years ago
7 0

Answer:

Explanation:

3.4 m/s due North, -1.1 m/s due East

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PLEASE HELP ME on this question.
VladimirAG [237]
The second one is the answer
4 0
4 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
What must be the distance in meters between point charge q1 = 23.5 µc and point charge q2 = -64.2 µc for the electrostatic force
svlad2 [7]
The answer for your problem is shown on the picture.

7 0
3 years ago
What is the correct Lewis structure for group 5A element, arsenic?
Lena [83]
The correct answer is A
7 0
3 years ago
Read 2 more answers
Find the gravitational force between Earth (5.97 x 1024 kg) and the Sun (1.99 x 1030 kg) knowing they are 1.48 x 1011 m apart.
guapka [62]

Using the gravitational force of F= G(m1*m2/r^2)

m1= 5.97 x 10^24 kg (Earth)

m2= 1.99 x 10^30 kg (Sun)

r= 1.48 x 10^11 m

G is a known value, it is 6.672 x 10^-11

All units are proper. Therefore plug in the values and you get 3.16 x 10^22 N.

Let me know if I calculated this wrong and it is something else so I can delete this. Thank you. I don't want to make other students put down the wrong answer.

7 0
3 years ago
Read 2 more answers
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