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Ostrovityanka [42]
3 years ago
7

Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam

are primed (x', t') and those made by Ned are not primed (x, t). v is the velocity of Pam (the other frame of reference) as measured by Ned. What is the interpretation of v'?
Physics
1 answer:
kenny6666 [7]3 years ago
8 0

This is a problem based on the logic and interpretation of the variables. From the measured data taken

what is collected by the two individuals is expressed as,

- NED reference system: (x, t)

- PAM reference system: (x ', t')

From the reference system we know that ν is the speed of PAM (the other reference system) as a measurement by NED.

Then ν' is the speed of NED (from the other system of the reference) as a measurement by PAM.

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

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3 0
4 years ago
A 6.20 g bullet moving at 929 m/s strikes a 850 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
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Answer:

(a) Final speed of block = 3.2896 m/s

(b) 6.7350 m/s is the speed of the bullet-block center of mass?

Explanation:

Given that:

Mass of bullet (m₁) = 6.20 g

Initial Speed of bullet (u₁) = 929 m/s

Final speed of bullet (v₁) = 478 m/s

Mass of wooden block (m₂) = 850g

Initial speed of block initial (u₂) = 0  m/s

Final speed of block (v₂) = ?

<u>By the law of conservation of momentum  as:</u>

<u>m₁×u₁ + m₂×u₂ = m₁×v₁ + m₂×v₂</u>

6.20×929 + 850×0 = 6.20×478 + 850×v₂

Solving for v₂, we get:

<u>v₂ = 3.2896 m/s</u>

Let the V be the speed of the bullet-block center of mass. So,

V = [m₁* u₁]/[m₁ + m₂]  (p before collision = p after collision)

   = [6.2 *929]/[5.2+850]

<u>V = 6.7350 m/s </u>

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