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wariber [46]
3 years ago
10

Which gas would have the slowest rate of diffusion, assuming that all gases are at the same temperature?

Physics
2 answers:
lianna [129]3 years ago
6 0

A. krypton (atomic mass 83.8 amu)

B. argon (atomic mass 39.95 amu)

C. xenon (atomic mass 131.3 amu)

D. radon (atomic mass 222 amu)

hope this helps!

sergij07 [2.7K]3 years ago
3 0

Answer:

Which of the following gases would have the slowest rate of diffusion, assuming all of the gases are at the same temperature?

The answer is- Kr (Krypton)

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Two plane mirrors are separated by 120°, as the drawing illustrates. If a ray strikes mirror M1 at a θ1 = 64° angle of incidence
tino4ka555 [31]

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

  = 90° - 64°

  = 26°

β + γ + 120° = 180°

γ = 180° - 120° - β

γ = 180° - 120° - 26°

γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

δ = 56°

According to the law of reflection,

angle of incidence = angle of reflection

θ2 = δ = 56°

Therefore, angle θ2 at which the light leaves mirror 2 is 56°

8 0
3 years ago
You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.
Vikki [24]

Answer:

A) Average speed = 18.75 m/s

B) More time is spent at 15 m/s than at 25 m/s.

Explanation:

Let the first distance be d1 and the second distance be d2.

We are given;

d1 = 10 km = 10000 m

d2 = 10 km = 10000 m

Speed; v1 = 15 m/s

Speed; v2 = 25 m/s

Now, the formula for distance is; Distance = speed x time

Thus:

d1 = v1 x t1

t1 = d1/v1 = 10000/15 = 666.67 seconds

Also,

d2 = v2 x t2

t2 = d2/v2 = 10000/25 = 400 seconds

Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s

From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;

More time at 15 m/s than at 25 m/s.

5 0
4 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
Help please I don't understand this
BARSIC [14]
It's either staying there or is going at the same pace
4 0
4 years ago
1) Write Newton's three laws of motion (20 points). 2) Define acceleration (20 points). 3) Write the Month and Day of the Assign
vaieri [72.5K]

Answer:

Hey

I have no idea when YOUR assignment is due.

Newtons 1rst law:

An object that has constant motion will remain at that speed unless acted on by an external force.

Newtons 2nd law:

F=ma (force=mass*acceleration)

Newtons 3rd law:

when a force is applied to an object, there will be an opposite but equal reaction.

Acceleration:

How much your speed increases/decreases per unit of time.

I wrote all that^

4 0
3 years ago
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