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wariber [46]
2 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]2 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]2 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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A bike rider was moving to the right at a constant speed. Suddenly the wind then starts blowing against her with 3 N of force to
stiks02 [169]

Answer:

The correct option is D

Explanation:

This question can be better understood when discussed using the Newton's first law of motion which states that an object would continue to move with a uniform speed (in a straight line) unless acted upon by an external force. What happens here (in the question) is that the bike rider would have continued moving at a constant speed (to the right) if not for the opposing force of the wind that acted against her (to the left). <u>This wind/force would cause her speed to reduce or slow down (as posited by the law)</u>.

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2 years ago
A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must
mezya [45]
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction. 
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end) 
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
4 0
3 years ago
Read 2 more answers
X=Y=MC scourer that would be the answer for you mam or sir
Alecsey [184]

Me das mas información?

5 0
2 years ago
Read 2 more answers
3. An object of mass 90 kg travels down a slide.
Gwar [14]

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

E_{k}=\frac{1}{2} *m*v^{2}  \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]

4 0
2 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
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