Ask question

Ask question

All categories

3 years ago

12

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.

You might be interested in

Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t

aivan3 [116]

Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

$K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}$

Where $K, U, W$ are linear kinetic energy, gravitational potential energy and work, respectively.

8 0

3 years ago

What is weight?

Over [174]

Answer:

1

Explanation:

plzzzzzzz Mark my answer in brainlist and follow me

4 0

3 years ago

Read 2 more answers

A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show

Lady bird [3.3K]

Answer:

change in height is 1.664 mm

Explanation:

Given data

drops = 3.00 mm

diameter = 5.00 cm = 0.05 mm

decrease = 350 cm^3

temperature = 95°C to 44.0°C

to find out

the decrease in millimeters in level

solution

we will calculate here change in volume so we can find how much level is decrease

change in volume = β v change in temp ...............1

here change in volume = area× height

so = $\pi$ /4 × d² h

so we can say change in volume = $\pi$ /4 × d² × change in height .......2

so from equation 1 and 2 we calculate change in height

( β(w) -β(g) )× v× change in temp = $\pi$ /4 × d² × change in height

change in height = 4 × ( β(w) -β(g) ) v× change in temp / $\pi$ /4 × d²

put all value here

change in height = 4 × ( 210 - 27 )(350 ) $10^{-6}$ × (95-44) / $\pi$ /4 × 0.05²

change in height is 1.664 mm

5 0

3 years ago

PLZ ANSWER FAST WILL GIVE BRAINILEST TO FIRST PERSON WHO ANSWERS:P

ankoles [38]

Answer:

4

Explanation:

6 0

3 years ago

Read 2 more answers

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago