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Andrei [34K]
3 years ago
11

Walking along a 6-meter beam without falling helps to develop __________.

Physics
2 answers:
Gala2k [10]3 years ago
7 0

Walking along a 6-meter beam without falling helps to develop C. BALANCE.


IrinaVladis [17]3 years ago
4 0
C is correct answer.


Walking along a 6 meter beam without falling helps to develop balance.


Hope it helped.

-Charlie
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Evgesh-ka [11]

Answer:

False

Explanation:

Solid water (ice) has a lesser density than liquid water. Hence, the statement is false.

3 0
2 years ago
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A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency
Anvisha [2.4K]

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

d=2A

Where, d = distance

A = amplitude

Put the value into the formula

d=2\times0.08190

d=0.1638\ m

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

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3 years ago
A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h
alukav5142 [94]

Answer:

Total height (s) = 176.4 m

Explanation:

Given:

Initial velocity (u) = 0 m/s

Time taken (t) = 6 sec

Acceleration due to gravity = 9.8 m/s²

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Total height (s)

Computation:

s = ut + [1/2]gt²

s = (0)(6) + [1/2][9.8][6²]

s = 176.4 m

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Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi
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To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

I = I_0 cos^2\theta

Where,

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I = Resulting intensity

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From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

I_1= \frac{I_0}{2}

In the case of the second polarizer the angle is directly 60 degrees therefore

I_2 = I_1 cos^2\theta

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I_2 = 0.125I_0

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Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

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