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Feliz [49]
2 years ago
7

A table tennis ball has a mass of about 2.45 g. Suppose the ball is hit across the table with a speed of 4.0 m/s. What is it’s k

inetic energy?
Physics
1 answer:
cluponka [151]2 years ago
3 0

Answer:

0.0196 j

Explanation:

i) The formula for kinetic energy is as follows: 0.5*m*v^2

ii) Since we have all the values all that's left is to plug them into the equation

iii) First, WE MUST, Convert grams into kgs as this is the SI unit of mass so 2.45/1000

iv) All that's left now is to plug it into the equation so:

0.5* (s.45/1000)*(4^2)

v) Lastly we add the unit joules at the end as we're talking about energy

Hope this was useful! :)

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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

P_i=10 atm

P_f=2 atm

Work done in iso-thermal process=P_iV_iln\frac{P_i}{P_f}

P_i=initial pressure

P_f=Final Pressure

W=10\times 2.463\times ln\frac{10}{2}=39.64 J

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is=P\Delta V

V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L

V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L

\Delta W=1\times (12.315-2.463)=9.852 J

\Delta q=\Delta W=9.852 J

\Delta U=0

8 0
4 years ago
After you enlarge a map, which one of the following scale remains correct?
9966 [12]

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

  • Verbal scale use of words to represent scale information on the map.  The distance or linear units are used for depicting this scale on the map.  For example: 1 inch = 1 Kilo meter.
  • Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio.  Example: 1: 10,000 or 1/10,000
  • In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given.  Example: 1:500
  • Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map.  Example: 1:25,000
  • A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
What is the primary ingredient of power yoga?
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4 0
4 years ago
A thin flake of mica (n=1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the wiew
Llana [10]

To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))

Where

L = Thickness

n = Index of refraction of each material

\lambda = Wavelength

Our values are given as

\Phi = 7(2\pi)

L=t

n_1 = 1.58

n_2 = 1

\lambda = 550nm

Replacing our values at the previous equation we have

\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))

7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))

t = \frac{7*550}{1.58-1}

t = 6637.931nm \approx 6.64\mu m

Therefore the thickness of the mica is 6.64μm

3 0
3 years ago
How much heat in joules is required to heat a 50 g sample of aluminum from 71 ∘f to 142 ∘f?
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A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
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