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3 years ago

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A student holds an egg outside of an open window and let's go. The window is 40 meters above the ground and the egg is falling u

nder the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:
How far will the egg travel after 2 seconds? Be sure to use the 5 steps to solve the problem. Show which equation you used, what each variable equals, and answer the question in a complete sentence.

1 answer:

melisa1 [442]3 years ago

5 0

The distance traveled by the egg after 2 seconds is 19.6 m.

The given parameters:

<em>Height of the window, h = 40 m</em>

The distance traveled by the egg after 2 seconds is calculated as follows;

$h = v_0_y t + \frac{1}{2} gt^2\\\\$

where;

<em /> $v_0_y$ <em> is the initial vertical velocity of the egg = 0</em>
<em>g </em><em>is acceleration due to gravity = 9.8 m/s²</em>
<em>t </em><em>is the time of motion, = 2 s</em>

$h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m$

Thus, the distance traveled by the egg after 2 seconds is 19.6 m.

Learn more about second equation of motion here: brainly.com/question/7977693

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How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?

matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be $f_{1}$

So, paper will attain terminal velocity when mg = $f_{1}$

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = $f_{2}$

Rock will attain terminal velocity when Mg = $f_{2}$

Mg > mg

so, $f_{2}$ > $f_{1}$

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

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3 years ago

a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$

Solving for $\ddot{x}$ :

$\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}$

where:

$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}$

Integrating to get x with initial conditions x(0) = 0:

$x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}$

When t=T:

$x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}$

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})$

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3 years ago

a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin

Daniel [21]

This question involves the concepts of Wein's displacement law and characteristic wavelength.

The blackbody temperature will be "3.22 x 10⁵ k".

<h3>WEIN'S DISPLACEMENT LAW</h3>

According to Wein's displacement law,

$\lambda_{max} T = c\\\\T=\frac{c}{\lambda_{max}}$

where,

$\lambda_{max}$ = characteristic wavelength = 9 μm = 9 x 10⁻⁹ m
T = temperature = ?
c = Wein's displacment constant = 2.897 x 10⁻³ m.k

Therefore,

$T=\frac{2.897\ x\ 10^{-3}\ m.k}{9\ x\ 10^{-9}\ m}$

T = 3.22 x 10⁵ k

Learn more about characteristic wavelength here:

brainly.com/question/14650107

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3 years ago

An example of a folkway violation would be driving 70 mph in a 50-mph zone. Group of answer choices True False

shutvik [7]

Answer:

Yes.

Explanation:

That is also a law violation.

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3 years ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

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3 years ago

Read 2 more answers