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TEA [102]
2 years ago
11

Help please!

Physics
1 answer:
JulsSmile [24]2 years ago
7 0

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

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How far would a spacecraft moving in a circular orbit 500 km above Pluto's surface travel during that time?
Ilia_Sergeevich [38]

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7 0
3 years ago
A slit 0.240mm wide is illuminated by parallel light rays with a wavelength 600nm . The diffraction pattern is observed on a scr
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To solve this exercise, it is necessary to apply the concepts on the principle of superposition, specifically on constructive interference,

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dsin\theta=m\lambda

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m = The integer m is called the interference order and is the number of wavelengths by which the two paths differ.

\lambda = wavelenght

d = Distance

For smaller angles sin\theta = tan\theta,

d tan\theta = m \lambda

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d*\frac{y}{L} = m\lambda

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y  = \frac{(1)(600*10^{-9}(3.5m)}{0.240*10^{-3}m}

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PART B) To calculate the intensity it is necessary to find the angle between the previously calculated height and distance in order to calculate the phase angle, in other words,

tan\theta' = \frac{y}{L}

tan\theta' = \frac{1/2(8.75*10^{-3})}{3.5}

\tetha = 0.07161\°

Therefore phase angle is

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The intensity formula would then be given by,

I = I_0 \frac{sin\frac{\gamma}{2}^2}{\frac{\gamma}{2}}

I = 4.1*10^{-6}(\frac{4}{\pi})

I = 1.66*10^{-6}W/m^2

3 0
3 years ago
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