An ideal gas is in a sealed rigid container. the average kinetic energy of the gas molecules depends most on

What is 18 m/s north an example of?

Mkey [24]

Answer:

It is an example of velocity

Explanation:

It is an example of velocity Don't ask how I know because I do know it I just don't know how to explain it.

8 0

3 years ago

A ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a dis

GaryK [48]

Answer: $3920\ J$

Explanation:

Given

mass of ball m=10 kg

It is placed at a height of 150 m

It is dropped from the height and allowed to free fall for 40 m

Velocity acquired by the ball during this fall is given by $v^2-u^2=2as$

Insert u=0, a=g

$\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s$

Kinetic energy at this instant

$K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J$

3 0

2 years ago

Can someone help me with this I don’t understand how to do this

olga nikolaevna [1]

Answer:

You are doing this wrong. Make sure that if you have 2 rows and on the top it has pounds then on the second bottom row you also have pounds so that they cancel each other out.

Explanation:

4 0

3 years ago

Charges of +2 µC and +3 µC are 4 mm from each other. Raina’s group draws four diagrams trying to represent the electrical force

ahrayia [7]

The correct diagram is shown below:

The charges of +2 µC and +3 µC are 4 mm from each other. The diagram below represents the electrical force between the charges. i.e. repulsive force. However the force of repulsion exerted by charge +3 µC on +2<span> µC will be more. The same charges repel each other and opposite charges attract each other.</span>

4 0

3 years ago

Read 2 more answers

Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa

Katarina [22]

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%

8 0

3 years ago