What is scientific evidence and when do we have enough?

Discuss the difference between waveform graphs and vibration graphs

Kitty [74]

Difference exists mainly in the label for x axis.

Explanation:

Shapes of waveform and vibration graphs are same.
Vibration graphs shows the particle at a single location in the path of the wave when time passes.
Waveform graphs shows the particle at multiple locations at a single moment of time.

6 0

3 years ago

One mole of an ideal gas expands isothermally from 0.01 m3 to 0.05 m3. The final pressure of gas is 1.2 atm. [15 Marks] determin

erik [133]

Answer:

Initial pressure = 6 atm. Work = 0.144 J

Explanation:

You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J

4 0

3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

The body is gaining the speed, which means there is a change in kinetic energy.
The change in kinetic energy is equal to the work done.
The friction force is the product of coefficient of the friction and normal force.
The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

7 0

2 years ago

Read 2 more answers

When a gas is rapidly compressed (say, by pushing down a piston) its temperature increases. When a gas expands against a piston,

shusha [124]

Answer:

Explained in explanation

Explanation:

The first law of thermodynamics states that the change in internal energy of a system(ΔU) is equal to the sum of the net heat transfer into the system(Q) and the net work done on the system(W). In equation, this law is;

ΔU = Q + W

Now, when there's gas inside a container with a movable piston that's tightly fitting, we will assume that the piston can move up and down thereby compressing the gas or allowing the gas to expand against it.

Now these gas molecules inside the container possess kinetic energy. Thus, the internal energy(U) of the system is simply the sum of all the kinetic energies of the individual gas molecules present in the container.

Therefore, if the temperature(T) of the gas increases, then the speed and internal energy(U) of the gas molecules will also increase. In the same way, if the temperature of the gas decreases, the speed and internal energy of the gas molecules would also decrease.

Now, back to the question, when the piston is pushed down, it does work on the gas and the gas does negative work on the piston. Thus, the gas will be get compressed to a smaller space, and thereby making the gas molecules to hit the piston at a faster rate. Thus, there is a decrease in speed and as we saw earlier that when there is a decrease in speed, it means temperature has decreased.

Whereas, when the piston is moved up, the gas does positive work on the piston and the speed of the gas molecules will increase. Like I said earlier that increase in speed means increase in temperature.

4 0

3 years ago