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2 years ago

What is the connection between temperature and kinetic energy

Physics

1 answer:

joja [24]2 years ago

8 0

Answer:

The relationship between temperature and kinetic energy is directly proportional.

Explanation:

The temperature is directly proportional to the kinetic energy of the gas molecules. We know that kinetic energy is the energy due to the motion and when gas molecules absorbs heat, they move with more speed and the kinetic energy of particles increases as a result.

According to the kinetic theory of gases

K = 3/2RT/Na

Where K is the average kinetic energy of gas molecules, T is the temperature of the gas, Na is Avogadro's number and R is the universal gas constant.

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An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

tamaranim1 [39]

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

5 0

2 years ago

A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and

natima [27]

The upward force exerted on the board by the support is 530.8 N.

<h3>Upward force exerted on the board by the support</h3>

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

Learn more about upward force here: brainly.com/question/6080367

#SPJ1

8 0

1 year ago

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0

2 years ago

With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.

viktelen [127]

Answer:

0.976 c

Explanation:

$v_{1e}$ = velocity of object 1 relative to earth = 0.80 c

$v_{21}$ = velocity of object 2 relative to object 1 = 0.80 c

$v_{2e}$ = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

$v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}$

$v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}$

$v_{2e}$ = 0.976 c

6 0

2 years ago

What is required to specify the position of an object?(๑•ᴗ•๑)♡

Mkey [24]

Answer:

Explanation:

To describe the motion of any object, we must be able to describe its position x. It reflects it at any particular time. In other words, we need to specify its position relative to the conventional frame of reference.

I hope it helps! ≧◉◡◉≦

8 0

1 year ago