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irina [24]
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
Physics
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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Scientists discovered how to use x-rays over 100 years ago. They found that x-rays
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A - to find injuries
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Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.
irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

4 0
3 years ago
An example of a single displacement reaction is
Elden [556K]
Single replacement would be represented by a single element being replaced.

This is shown in answer choice B
Where the positions of A and B are swapped
4 0
3 years ago
According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt
scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength(\lambda _1)=450 nm

Second wavelength(\lambda _2)=700 nm

According wien's diplacement law

\lambda T=constant

where \lambda =wavelength

T=Temperature

Let T_1 and T_2 be the temperatures corresponding to \lambda _1 & \lambda _2 respectively.

\lambda _1\times T_1=\lambda _2\times T_2

\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}

\frac{T_1}{T_2}=\frac{700}{450}=1.55

Thus object with \lambda 450 nm is 1.55 times hotter than object with wavelength \lambda =700 nm

8 0
3 years ago
S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to
denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

5 0
1 year ago
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