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2 years ago

#27 question

Physics

1 answer:

Furkat [3]2 years ago

4 0

So, the initial altitude of the parachuter is approximately <u>(C). 123 m</u>.

<h2>Introduction</h2>

Hi ! In this question, I will help you. In this question, you will learn about the fall time of the free fall motion. Free fall is a downward vertical motion without being preceded by an initial velocity. When moving in free fall, the following equations apply:

<h3>The equation for calculating the height (h)</h3>

$\boxed{\sf{\bold{h = \frac{1}{2} \cdot g \cdot t^2}}}$

<h3>The equation for calculating the time (s)</h3>

$\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}$

<h3>The equation for calculating the velocity (v)</h3>

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$

With the following condition :

t = interval of the time (s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)
v = velocity (m/s)

<h2>Problem Solving</h2>

We know that :

t = interval of the time = 5 s
g = acceleration of the gravity = 9.81 m/s²

What was asked :

h = height or displacement at vertical line = ... m

Step by Step :

$\sf{h = \frac{1}{2} \cdot g \cdot t^2}$

$\sf{h = \frac{1}{2} \cdot 9.81 \cdot 5^2}$

$\sf{h = \frac{245.25}{2}}$

$\boxed{\sf{h = 122.625 \: m \approx 123 \: m}}$

<h3>Conclusion</h3>

So, the initial altitude of the parachuter is approximately 123 m (C.)

Time that needed for hearing the splash of fallen rock in the well brainly.com/question/26485521
The speed of the object at a certain height (free fall motion) brainly.com/question/26377041
The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625

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Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

$\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s$

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The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

$\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km$

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

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m = mass of an electron = $9.1\times 10^{-31}\ kg$
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p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

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