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2 years ago

How is the temperature of an object related to the average kinetic energy of its particle??

Physics

2 answers:

dusya [7]2 years ago

6 0

By It is directly proportional to the average kinetic energy.

topjm [15]2 years ago

4 0

It follows the same ratio.

For one Same physical state, the agitation of the particles is related to temperature. Thus, the Temperature is closely linked to the average kinetic energy of the particles that make up the body. A higher temperature indicates greater agitation of the particles and, therefore, higher kinetic energy average. (The higher the thermal agitation, the higher the temperature).

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Strongest Element on Earth

Alex777 [14]

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Titanium has a tensile strength of 63,000 PSI. ...

Chromium, on the Mohs scale for hardness, is the hardest metal around.

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2 years ago

What is the difference between a physical quantity and a unit

Shalnov [3]

Answer: What is the difference of physical quanity and a unit?

Explanation: If you are working science or math problems, the answer to this question is that quantity is the amount or numerical value, while the unit is the measurement. For example, if a sample contains 453 grams, the quantity is 453 while the unit is grams.

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2 years ago

What is the dimensional formula of young modulas

LekaFEV [45]

Answer:

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2 years ago

Read 2 more answers

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

4 0

2 years ago

A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he

zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water, u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0

2 years ago