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madam [21]
3 years ago
14

A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan

ce?
Physics
1 answer:
ASHA 777 [7]3 years ago
5 0

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of a \;\;m/s^2 from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, a=\frac{v-u}{t}=\frac{5-4.2}{t}

\Rightarrow at=0.8\cdots(i)

Now, from the equation of motion

s=ut+\frac 12 at^2

\Rightarrow s=ut+\frac 12 at(t)

\Rightarrow 115=4.2t+\frac 12 \times 0.8 t [ from equation (i)]

\Rightarrow 115=(4.2+0.4)t

\Rightarrow t= 115/4.6 = 25 seconds.

Hence, she takes 25 seconds to walk the distance.

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A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos
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Answer:

The possible thickness of the soap bubble = 1.034\times 10^{-7}\ m.

Explanation:

<u>Given:</u>

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Let the thickness of the soap bubble be t.

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

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Since the soap bubble is appearing very bright, the order should be 0, as 0^{th} order interference has maximum intensity.

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2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.

It is the possible thickness of the soap bubble.

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