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lesantik [10]
3 years ago
9

6.1.115 To approximate the speed of the current of a​ river, a circular paddle wheel with radius 4 ft. is lowered into the water

. If the current causes the wheel to rotate at a speed of 18 revolutions per​ minute, what is the speed of the​ current?
Physics
1 answer:
shtirl [24]3 years ago
5 0

Answer:

1.2 ft/s or 72 ft/min

Explanation:

To solve this problem, we can use the relationship between the angular and linear speed for this particular movement (rotation). As the speed of the current is tangent to the trajectory of the wheel we can define as follows:

v=  \omega r=18  \frac{rev}{min} \times \frac{1min}{60s}\times 4 ft\\v= 1.2 \frac{ft}{s}= 72 \frac{ft}{min}

The speed of the current is 72 ft/min

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Answer:

0.43 m

Explanation:

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tan Θh = L / (x (eye) + xw)

xw = Distance of the mirror from the window

L'/ x (eye) = L / ( x (eye) + xw)

L' = L*x (eye) / ( x (eye) + xw)

L' = (2*0.5) / (0.5 + 1.8)

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4 years ago
Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both
Nuetrik [128]

Answer:

Part a)

Width of the slit is

a = 580 nm

Part b)

Ratio of intensity is given as

\frac{I}{I_o} = 0.81

Explanation:

Part a)

As we know by the formula of diffraction we will have

a sin\theta = \lambda

so we have

\theta = 90

\lambda = 580 nm

so we will have

a sin90 = 580 nm

a = 580 nm

Part b)

As we know that the intensity in diffraction pattern is given as

I = I_o (\frac{sin\theta}{\theta})^2

\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2

so for angle 45 degree

\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

\frac{I}{I_o} = 0.81

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The answer is (a).

Golf balls have a higher density than water hence sink to the bottom.

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jeka57 [31]

F = force applied to lift the box = weight of the box = 88 N

P = power produced while lifting the box upward = 72 Watt

v = speed of the box = ?

Speed of the box is given as

v = \frac{P}{F}

inserting the values

v =  \frac{72}{88}

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6 0
3 years ago
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