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2 years ago

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Two containers hold an ideal gas at the same temperature and pressure. Both containers hold the same type of gas, but container

B has twice the number of particles and twice the volume of container A. The average kinetic energy per molecule in container B is :
A. Twice that for container A

B. The same as that for container A

C. Half that for container A

D. Impossible to determine

1 answer:

sergeinik [125]2 years ago

7 0

Answer:

B. The same as that for container A

Explanation:

To calculate the average Kinetic Energy per molecule,

$K_{avg} = 3/2 K_{B} T$

Where $K_{avg}$ is the average kinetic energy per molecule and $K_{B}$ is the Boltzmann constant

From the above equation it is seen that $K_{avg}$ only depends on the temperature of the gas

And since the temperatures of both gases are equal, their average kinetic energy per molecule is the same

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You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

2 years ago

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

3 0

2 years ago

A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a

Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

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2 years ago

In a certain process, the energy of the system decreases by 250 kJ. The process involves 480 kJ of work done on the system. Find

gregori [183]

Answer:

Q = - 730 KJ

730 KJ is transferred out of the system

Explanation:

According to the first law of thermodynamics, energy can neither be created nor destroyed, but can be transformed from one form to another.

For a particular process/system, the first law is interpreted as

ΔU = Q + W (depending on convention, some textbooks give this relation as ΔU = Q - W)

ΔU is the change in internal energy of the system, in my convention, it is positive if the internal energy increases and negative otherwise.

Q = Heat transferred into or out of the system. Q is positive for heat transferred into the system and negative for heat transferred out of the system.

W = workdone by the system or work done on the system. W is positive for workdone on the system and W is negative for when work is done by the system.

ΔU decreases by 250 KJ, that is, ΔU = - 250 KJ

Q = ?

W = 480 KJ (Work is done on the system)

- 250 = Q + 480

Q = - 250 - 480 = - 730 KJ

The heat is transferred out of the system.

6 0

2 years ago

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The graph shows projected changes in the populations of the world.

malfutka [58]

Answer: C. Africa

Explanation:

The data given on the graph shows that:

Asia will grow from around 1,300 million to 5,000 million in 2050 which is an increase of:

= 5,000 /1,300 = 3.84 times

Europe will decrease over that period.

Africa will go from around 300 million to 2,000 million which is an increase of:

= 2,000 / 300

= 6.67 times

Mexico

, Central America, Caribbean Islands:

= 900 / 120

= 7.5 times

United States, Canada, and Greenland:

= 400/120

= 3.33

Oceania:

= 50 / 13

= 3.85

From the options give, Africa will see the greatest increase at 6.67 times its population in 1950.

3 0

2 years ago

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