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kozerog [31]
3 years ago
11

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.​

Physics
1 answer:
Elza [17]3 years ago
6 0

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E =  ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E =  \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E =  \frac{1}{2}mv^2

E =  ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

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At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}

C=1,25\cdot 10^{5} kJ/^{\circ}C

I hope it helps!

6 0
3 years ago
Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What
Sauron [17]

Answer:

0.426 L

Explanation:

Boyles law is expressed as p1v1=p2v2 where

P1 is first pressure, v1 is first volume

P2 is second pressure, v2 is second volume.

Given information

P1=96 kPa, v1=0.45 l

P2=101.3 kpa

Unknown is v2

Making v2 the subject from Boyle's law

v2=\frac {p1v1}{p2}

Substituting the given values then

v2=\frac {96*0.45}{101.3}=0.4264560710760l\approx 0.426 l

Therefore, the volume is approximately 0.426 L

6 0
2 years ago
What happens to the particles of the liquid inside a thermometer when the thermometer is heated?
aev [14]
The particles of the liquid inside a thermometer speed up and spread apart when the thermometer is heated. In short, the particles expand from one another when they're heated, and become condensed and compact when chilled.

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Radda [10]

Answer:

\frac{p_{c}}{p_{b}}\approx 0.507

Explanation:

Since the cardinal and ball have the same kinetic energy, it is possible to determine the ratio between speeds. (c for cardinal, b for baseball)

K_{c} = K_{b}

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\frac{v_{c}}{v_{b}}=\sqrt{\frac{m_{b}}{m_{c}} }

The ratio is obtained by multiplying each side by \frac{m_{c}}{m_{b}}:

\frac{p_{c}}{p_{b}}=\frac{m_{c}}{m_{b}}\cdot \sqrt{\frac{m_{b}}{m_{c}} }

\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }

The value of this ratio is:

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3 years ago
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