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2 years ago

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If cells are placed in a hypertonic solution containing a solute to which the membrane is impermeable what could happena)cells w

ould shrink then later reach equilibriumb)cell swell and ultimately burstc)cell show no changed)loose water and shrink

1 answer:

elena55 [62]2 years ago

6 0

Answer:

cell swell and burst(b)

Explanation:

This process is called hemolysis. it also occurs in a red blood cell.

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The maximum electric field strength in air is 3.0 Mv/m . Stronger electric fields ionize the air and create a spark. What is the

adoni [48]

Answer:

The value is $P_{max} = 2.11 * 10^{6} \ W$

Explanation:

From the question we are told that

The maximized electric field strength is $E = 3.0 \ MV/m = 3.0 *10^{6} \ V/m$

The diameter is $d = 1.5\ cm = 0.015 \ m$

Generally the cross -sectional area is mathematically represented as

=> $A = \frac{\pi * d^2 }{4}$

=> $A = \frac{3.142 * 0.015^2 }{4}$

=> $A = 0.0001767 \ m^2$

Generally the maximum power is mathematically represented as

$P_{max} = \frac{E_{max}^2}{ 2 * \mu_o * c } * A$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

$\mu_o$ is the permeability of free space with value $\mu_o = 4 \pi *10^{-7} \ H/m$

$P_{max} = \frac{ (3.0 *10^{6})^2}{ 2 * (4 \pi *10^{-7}) * 3.0 *10^{8} } * 0.0001767$

=> $P_{max} = 2.11 * 10^{6} \ W$

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2 years ago

Which correctly ranks, from fastest to slowest, the time it will take one wavelength to pass a certain point?

dsp73

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How long will it take them to glide to the edge of the rink?

Ganezh [65]

At the center of a 50 m diameter circular ice rink, if a 77 kg skater traveling at 2.3 m/s and then collides with a 63 kg skates traveling at 3.7 m/s. This is how long it will take them to glide to the edge of the rink:

Speed after the collision= √{[77(2.3)77^2] + [63(3.7)^2]} / (77+63)=2.09 m/s

For them to be able to get to the edge which is 50 m away it will take them 23.9 seconds.

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2 years ago

A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele

Korvikt [17]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=0.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}$

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natulia [17]

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