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stira [4]
4 years ago
6

What environmental factors affect kinetic energy and diffusion?

Physics
2 answers:
Alinara [238K]4 years ago
7 0
1) Temperature (High temperature <span> increase the kinetic energy of a substance, speeding </span><span>diffusion rates)
</span>
2) Particle size (<span>Larger particles require more kinetic energy to move, causing lower rates of diffusion)
</span>
3) Concentration 

Hope this helps!
omeli [17]4 years ago
3 0

The environmental factors that affect kinetic energy and diffusion are the surrounding temperature, the size of the solute inside a solution, the concentration of the solution.

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if the efficiency of an electric furnace is 96%, then 96% of the input energy is transformed into thermal energy. what is the ot
Nonamiya [84]
It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.
5 0
4 years ago
1. During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the tre
Grace [21]
Initial velocity u = 40 
Angle at launch = 55 degrees
 At maximum height v = 0, velocity equation v^2 = u^2 - 2gh,
 0 = (40 x sin55)^2 - 2 x 9.81 x h => (40 x 0.819)^2 = 19.62h => h = 32.76^2 /
19.62
 Maximum height = 54.7 m 
 We have v = u - gt => 0 = (40 x sin55) - 9.81 x t => t = 32.76 / 9.81 => t =
3.34 s
 Time taken to hit the ground is 2t = 2 x 3.34 = 6.68 s 
Distance from castle to trebuchet = utcos55 = 40 x 6.68 x 0.573 = 153.1 m
8 0
4 years ago
3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t
sattari [20]

Answer:

Work done is 12.3 J

Explanation:

We have,

Mass of puck, m = 0.35 kg

Force of friction acting on the puck when it slides is 0.15 N

Distance travelled by the puck is 82 m.

It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J

The work done on the puck is 12.3 J.

5 0
4 years ago
When we do dimensional analysis, we do something analogous to stoichiometry, but with multiplying instead of adding. Consider th
Ksivusya [100]

Answer:

The  dimension is  D =  L ^{2} T^{-1}

Explanation:

From the question we are told that

     J  =  -D \frac{dn}{dx}

Here  [J] = \frac{1}{L^2 T}

       [n] =\frac{1}{L^3}

        [x] = L

So

    \frac{1}{L^2 T} =  -D \frac{d(\frac{1}{L^3})}{d[L]}

Given that the dimension represent the unites of  n and  x then the differential  will not effect on them

So

\frac{1}{L^2 T} =  -D \frac{(\frac{1}{L^3})}{[L]}

=>   D =  \frac{L^{-2} T^{-1} * L }{L^{-3}}

=>   D =  L ^{2} T^{-1}

   

5 0
3 years ago
A constant current source supplies a electric current of 200 mA to a load of 2kΩ. When the load changed to 50Ω, the load current
serious [3.7K]

Answer:

700ma

Explanation:

all u do is have to add them up.

4 0
3 years ago
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