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3 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]3 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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A sodium atom will absorb light with a wavelength near 589 nm if the light is within 10 MHz of the resonant frequency. The atomi

Troyanec [42]

Answer:

i)20369 photons

ii) 40 ps

Explanation:

Momentum of one Sodium atom:

$P=m*v =600m/s*23amu*\frac{1 kg}{6.02*10^{23}amu}\\P=2.29*10^{-23}kgm/s$

In other to stop it, it must absorb the same momentum in photons:

$P=2.29*10^{-23}kgm/s=n_{photons}*\frac{h_{planck}}{\lambda}\\=n*\frac{6.63*10^{-34}}{589*10^{-9}} \\==>n=20369 photons$

Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:

$t=n*T=n*\frac{\lambda}{c} =20369*\frac{589nm}{3*10^{8}m/s}=4*10^{-11} second=40 ps$

7 0

3 years ago

Can someone help me with these questions please. I will mark brainliest

VladimirAG [237]

Answer:

3: I can´t see the text/image, but it depend on the mass and the force applied to the ball, if both are too high, it will be harder to make a home run. (Second law)

4:It would be easier to make a home run because there is no interruption between the ball and the space the same travels. (Third law)

Explanation:

3 0

3 years ago

A 250 - ω resistor is connected in series with a 4.80 - μf capacitor. the voltage across the capacitor is vc=(7.60v)⋅sin[(120rad

galben [10]

<span>553 ohms The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is Xc = 1/(2*pi*f*C) where Xc = Reactance in ohms pi = 3.1415926535..... f = frequency in hertz. C = capacitance in farads. I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance. The 120 rad/s frequency is better known as 60 Hz. Substitute known values into the formula. Xc = 1/(2*pi* 60 * 0.00000480) Xc = 1/0.001809557 Xc = 552.6213302 Rounding to 3 significant figures gives 553 ohms.</span>

6 0

3 years ago

Read 2 more answers

An Astronaut lands on an Earthlike planet and drops a small lead ball with a mass from the top of her spaceship. The point of re

bearhunter [10]

First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,

s = ut +(gt^2)/2

Distance s = 25m

Time t = 5 s

Velocity u = 0

By putting these values,

25 = 1/2.g.(5)²

g = 2

So the gravity on that planet is 2. Lets find out the weight of the astronaut.

Mass of the astronaut on earth m = 80 kg

Weight of astronaut on earth W = mg = (80)(9.8) = 784 N

Weight of astronaut on earth like planet = (80)(2) = 160 N

x = 160N

5 0

2 years ago

Five moles of an ideal monatomic gas with an initial temperature of 121 ∘C expand and, in the process, absorb an amount of heat

liq [111]

Answer:

T₂ ≈ 107.85∘C

Explanation:

The question didn't state if the volume is constant or not as such, we can apply the first law of thermodynamic

From the first law of thermodynamic,

ΔU = Q - W

where ΔU = Internal Energy, Q = Quantity of heat absorbed, W = Amount of work done.

Q = 1200 J and W = 2020 J

∴ ΔU = 1200 -2020 = -820 J.

Using the ideal gas equation,

ΔU = 3/2nRΔT...................................equation 1

where n = number of moles, R = Molar gas constant, ΔT = Change in temperature = (T₂ - T₁).

Modifying equation 1,

ΔU = 3/2nR(T₂ -T₁)...............................equation 2.

making T₂ the subject of the relation in equation 2,

T₂ = {2/3(ΔU)/nR}+T₁........................ equation 3

where T₁=121∘C, R= 8.314 J / mol, n=5 moles, ΔU=-820 J

Substituting these values into equation 3,

∴ T₂ ={ 2/3(-820)/(5×8.314)}+121

T₂ = {2×(-820)/ (3×5×8.314)}+121

T₂={-1640/124.71}+ 121

T₂ = {-13.151} + 121

∴T₂ = 121 - 13.151 = 107. 849∘C

T₂ ≈ 107.85∘C

3 0

3 years ago