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Anni [7]
3 years ago
10

How many neutrons are in the nucleus of an atom with an atomic mass of 80 A.M.U. and an atomic number of 35?

Physics
1 answer:
belka [17]3 years ago
4 0

Answer:

45

Explanation:

The mass number is 80

Proton number is 35

A-P=n

80-35=45

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A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the fo
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The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

<h3>Force required to pull one end at a constant speed</h3>

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

  • m is mass
  • a is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Learn more about constant speed here: brainly.com/question/2681210

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2 years ago
An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has
kolezko [41]

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is 1.50\times 10^{-3}\text{ C}.

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by

I(t) = I_0e^{-t/\tau}

Here, I_0 is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).

The maximum current is given by

I = \dfrac{V}{R_\text{eff}}

<em>V</em> is the emf of the battery and R_\text{eff} is the effective resistance.

In this question, R_\text{eff} = 10.0 Ω + 25.0 Ω = 35.0 Ω

I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}

(b) The maximum charge is given

<em>Q</em> = <em>CV</em>

where <em>C</em> is the capacitance of the capacitor

Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}

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