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UkoKoshka [18]
3 years ago
12

You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 80-kg cylinder of radius r that ha

s a maximum angular speed of 430 rev/s. the flywheel must deliver an average of 2.00 kj of energy for each kilometer of distance. find the smallest value of r for which the car can travel 300 km without the flywheel needing to be recharged.
Physics
1 answer:
Oliga [24]3 years ago
8 0
E = 300km * 2000J/km = 600000J

The rotational energy is given by:
E = 0.5 * I * ω² 

I for a uniform cylinder is given by:
I  = 0.5 * m * r²

Resulting equation:
E = 0.25 * m * r² * ω²

Given values:
ω = 430 rev/s = 430 * 2π / s
E = 600000

Solve for r.
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Answer:

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Explanation:

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3 0
3 years ago
The law of universal gravitation states that a gravitational force exists between all objects in the universe. This force is dir
Ber [7]

Answer: D. it has been demonstrated to be without exception under certain stated conditions.

Explanation:

A <u>Law</u> is an affirmation (something established) based on repeated long-term observation of a phenomenon that has been studied and verified.  

That is: A law is present in all known theories and therefore is considered universal. In addition, a law can not be refuted, nor changed, because its precepts have been proven through various studies.  

<u>Unlike theory</u>, which is the set of rules and principles that describe and explain a particular phenomenon and <u>is subject to changes as new evidence emerges that gives meaning to it.  </u>

Then, based on what is explained above, the law of universal gravitation is a statement that exists because it was rigorously tested and verified, therefore it can not be refuted.

3 0
3 years ago
What is the potential energy of a 3kg ball that is on the ground?
ELEN [110]

This is where we have to admit that gravitational potential energy is
one of those things that depends on the "frame of reference", or
'relative to what?'.

         Potential energy = (mass) x (gravity) x (<em>height</em>).

So you have to specify <em><u>height above what</u></em> .

-- With respect to the ground, the ball has zero potential energy.
(If you let go of it, it will gain zero kinetic energy as it falls to
the ground.)

-- With respect to the floor in your basement, the potential energy is

                 (3) x (9.8) x (3 meters) = 88.2 joules.

(If you let go of it, it will gain 88.2 joules of kinetic energy as it falls
to the floor of your basement.)

-- With respect to the top of that 10-meter hill over there, the potential
energy is
                    (3) x (9.8) x (-10) = -294 joules

(Its potential energy is negative. After you let go of it, you have to give it
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5 0
3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.
Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

P_1 = Initial pressure = 55.1 mmHg

P_2 = Final pressure = 1 atm = 760 mmHg

T_2 = Boiling point

T_1 = Initial temperature = 35°C

\Delta H_{vap} = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K

The normal boiling point of the substance is 389.78681 K

3 0
3 years ago
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