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kotykmax [81]
3 years ago
10

State two factors that affect the rate of diffusion of a substance

Physics
1 answer:
Ronch [10]3 years ago
4 0

Explanation:

<em>Two</em><em> </em><em>factors</em><em> </em><em>that</em><em> </em><em>affect</em><em> </em><em>the</em><em> </em><em>rater</em><em> </em><em>of</em><em> </em><em>diffusion</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>substance</em><em> </em><em>are</em><em>:</em><em> </em>

  • <em>Diffusion</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>plays</em><em> </em><em>an</em><em> </em><em>important</em><em> </em><em>role</em><em> </em><em>on</em><em> </em><em>cellular</em><em> </em><em>transport</em><em> </em><em>in</em><em> </em><em>plants</em><em>.</em><em> </em>
  • <em>Diffusion</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>passive</em><em> </em><em>movement</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>from</em><em> </em><em>a</em><em> </em><em>region</em><em> </em><em>of</em><em> </em><em>higher</em><em> </em><em>concentration</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>region</em><em> </em><em>of</em><em> </em><em>lower</em><em> </em><em>concentration</em><em>. </em>
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11)A 1100 kg car travels on a straight highway with a speed of 30 m/s. The driver sees a red light ahead and applies her
cestrela7 [59]

Answer:

Time taken = 8.25 second

Explanation:

Given:

Force = 4000 N

Force = ma

4,000 = (1100)(a)

Acceleration = 3.6363 m/s²

v = u + at

0 = 30 + (3.6363)t

Time taken = 8.25 second

3 0
3 years ago
An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c
Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

                      \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0
3 years ago
What mountains range can be seen from vancouver wa?
kirill115 [55]
The North Shore mountains
8 0
3 years ago
Read 2 more answers
A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s
DanielleElmas [232]

a) 1.48 m/s

The tuning fork is moving by simple harmonic motion: so, the maximum speed of the tip of the prong is related to the frequency and the amplitude by

v_{max}=\omega A

where

v_{max} is the maximum speed

\omega is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s, where f is the frequency

A=0.600 mm=6\cdot 10^{-4} m is the amplitude

Therefore, the maximum speed is

v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s

b)  3.0\cdot 10^{-5} J

The fly's maximum kinetic energy is given by

K=\frac{1}{2}mv_{max}^2

where

m=0.0270 g=2.7\cdot 10^{-5} kg is the mass of the fly

v_{max}=1.48 m/s is the maximum speed

Substituting into the equation, we find

K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J

7 0
3 years ago
A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic
valina [46]

The spring has been stretched 0.701 m

Explanation:

The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring with respect to its equilibrium position

For the spring in this problem, we have:

E = 84.08 J (potential energy)

k = 342.25 N/m (spring constant)

Therefore, its elongation is:

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(84.08)}{342.25}}=0.701 m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0
3 years ago
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