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3 years ago

Why was the microscope and its improvement an important part of the development of the cell theory?

Physics

2 answers:

netineya [11]3 years ago

8 0

Answer:it’s is part of the cell theory because they where studying cells and to see it you need a microscope

Explanation:basically in the answer area

garri49 [273]3 years ago

8 0

It helps scientists get a closer look at cells. The improvements help thing become clearer which gives a better explanation of the cells to the scientists to help expand the knowledge of the cell theory.

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Can someone please explain what inertia is?

Bad White [126]

If an object is in motion it will stay in motion

8 0

3 years ago

Read 2 more answers

Determine the nuclear radius (in fm) for each of the following nuclei.

tekilochka [14]

Answer:

(a) 2.75 fm

(b) 2.89 fm

(d) 7.12 fm

Explanation:

For a given element, the radius r of its nuclei is given by;

r = r₀ $A^{(1/3)}$

Where;

A = Atomic mass of the element

r₀ = 1.2 x 10⁻¹⁵m = 1.2fm

Now let's solve for the given elements

(a) ¹²₆C

Carbon element => This has an atomic mass number of 12

Therefore its radius is given by;

r = 1.2 x $12^{1/3}$

r = 1.2 x 2.29

r = 2.75 fm

(b) ¹⁴₇N

Nitrogen element => This has an atomic mass number of 14

Therefore its radius is given by;

r = 1.2 x $14^{1/3}$

r = 1.2 x 2.41

r = 2.89 fm

Cobalt element => This has an atomic mass number of 60

Therefore its radius is given by;

r = 1.2 x $60^{1/3}$

r = 1.2 x 3.92

r = 4.70 fm

(d) ²⁰⁸₈₂Pb

Lead element => This has an atomic mass number of 208

Therefore its radius is given by;

r = 1.2 x $208^{1/3}$

r = 1.2 x 5.93

r = 7.12 fm

4 0

3 years ago

Which description matches up correctly with each type of particle? Alpha - no mass and no charge; beta - mass of 4 and charge of

Veseljchak [2.6K]

The closest answer is
Alpha - mass of 4 and charge of +2; beta - no mass and charge of -1; gamma - no mass and no charge (consists of energy)

It’s not exactly correct because a beta particle has the (small) mass of an electron (also the positron). All other choices are way off, I’d go with this one.

8 0

3 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$