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3 years ago

Which of the following types of models could be used to represent a cell?

Physics

2 answers:

m_a_m_a [10]3 years ago

6 0

Snansnsnjsjjsncncnfjf

lesya [120]3 years ago

4 0

Answer:

d. all of the above

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If you separate vector B into its components. How many components will it have? Those components will be called?

Inessa [10]

The vector B will have two components and those components will be called resultant vectors.

<h3>What is a component vector?</h3>

A component vector is a unit vector that represents a given vector in a particular direction.

A vector can be represented in x - direction and y - direction.

x - component of the vector = Bcosθ
y - component of the vector = Bsinθ

Thus, the vector B will have two components and those components will be called resultant vectors.

Learn more about component vectors here: brainly.com/question/13416288

#SPJ12

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2 years ago

Seasonal changes bring about scenes like this one. Plant cells respond to changes in _________________ and as a result, photosyn

jekas [21]

Light intensity & temp change

4 0

3 years ago

Read 2 more answers

All of the orbitals in a given subshell have the same value of the __________ quantum number.

Gelneren [198K]

All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number

Hope this helps!</span>

3 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago

What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes

mr Goodwill [35]

The current is defined as the amount of charge Q that passes through a given point of a wire in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Since I=500 A and the time interval is
$\Delta t=4.0 min=240 s$
the charge is
$Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C$

One electron has a charge of $q=1.6 \cdot 10^{-19}C$ , therefore the number of electrons that pass a point in the wire during 4 minutes is
$N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23}$ electrons

3 0

3 years ago