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Stolb23 [73]
3 years ago
9

Kinetic energy can be calculated using the formula KE = 1/2mv2. What is the kinetic energy of a bicycle that has a mass of 20 kg

and is traveling at a velocity of 10 m/sec?​
Physics
1 answer:
Alenkasestr [34]3 years ago
8 0
<h3>Answer:</h3>

KE = 1000 J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Physics</u>

<u>Energy</u>

Kinetic Energy Formula: \displaystyle KE = \frac{1}{2}mv^2

  • Energy is in Joules
  • m is mass (in kg)
  • v is velocity (in m/s)
<h3>Explanation:</h3>

<u>Step 1: Define</u>

m = 20 kg

v = 10 m/s

<u>Step 2: Find KE</u>

  1. Substitute [KE]:                    \displaystyle KE = \frac{1}{2}(20)(10)^2
  2. Exponents:                          \displaystyle KE = \frac{1}{2}(20)(100)
  3. Multiply:                               \displaystyle KE = (10)(100)
  4. Multiply:                               \displaystyle KE = 1000
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A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

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So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

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In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\&#10;x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

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B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

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where we have

\omega=0.20 rad/s

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6 0
3 years ago
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

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3 years ago
A 330.-ohm resistor is connected to a 5.00-volt battery. What is the current through the resistor?
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I = 0.0152 A = 15.2 mA

Explanation:

Using Ohm's law,

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Explanation:

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  • Evaporation of a liquid will also cause the liquid to change its state from liquid to gas. This means molecules will go far away from each other leading to an increase in the entropy.
  • Sublimation is a process of conversion of a solid into gaseous phase without going through liquid phase. So, in this case also entropy will increase due to gain in energy by the molecules of a solid.
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