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4 years ago

What is the connection between temperature and kinetic energy

Physics

1 answer:

joja [24]4 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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nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago

Assume that both the supply of and the demand for a good are relatively price elastic. The imposition of a per-unit excise tax o

sergey [27]

Answer:P:Increase Q:Decrease

Explanation: The price of the product will increase as the imposed per-unit excise tax on the sale of the good. This is true for both Perfectly-elastic and relatively elastic products.

The quantity demanded of the product will decrease as the price of the product increases, this his is true for both Perfectly-elastic and relatively elastic products. It will follow the law of Demand and Supply.

5 0

4 years ago

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Please help!

Kisachek [45]

B. Inelastic collision.

In elastic collision , both momentum and kinetic energy are conserved while in inelastic collision only momentum is conserved. there is some loss of energy in inelastic collision during collision.

During the collision of bat with baseball, some energy gets lost to heat and sound. hence the kinetic energy is not conserved although the momentum is conserved.

3 0

3 years ago

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A baseball has an approximate mass of 0.15 kg. If a bat strikes the baseball with a force of 6 N, what is the acceleration of th

Oksanka [162]

Answer:

a = 40 [m/s²]

Explanation:

These kinds of problems can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force = 6 [N]

m = mass = 0.15 [kg]

a = acceleration [m/s²]

$a=F/m\\a=6/0.15\\a=40[m/s^{2} ]$

3 0

3 years ago

Two billiard balls of identical mass move toward each other. Assume that the collision between them is perfectly elastic. If the

anzhelika [568]

Answer:

-20.0 m/s and 30.0 m/s

Explanation:

Momentum is conserved:

m (30.0) + m (-20.0) = m v₁ + m v₂

30.0 − 20.0 = v₁ + v₂

10.0 = v₁ + v₂

Since the collision is perfectly elastic, energy is also conserved. Since there's no rotational energy or work done by friction, the initial kinetic energy equals the final kinetic energy.

½ m (30.0)² + ½ m (-20.0)² = ½ mv₁² + ½ mv₂²

(30.0)² + (-20.0)² = v₁² + v₂²

1300 = v₁² + v₂²

We now have two equations and two variables. Solve the system of equations using substitution:

1300 = v₁² + (10 − v₁)²

1300 = v₁² + 100 − 20v₁ + v₁²

0 = 2v₁² − 20v₁ − 1200

0 = v₁² − 10v₁ − 600

0 = (v₁ + 20) (v₁ − 30)

v₁ = -20, 30

If v₁ = -20, v₂ = 30.

If v₁ = 30, v₂ = -20.

So either way, the final velocities are -20.0 m/s and 30.0 m/s.

3 0

3 years ago

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