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3 years ago

Which sub-atomic particle is positively charged and found in the nucleus?

Physics

2 answers:

arsen [322]3 years ago

8 0

Answer:

The proton. There are three main types of subatomic particles. Protons, Neutrons, and Electrons. Protons are positively charged and found in the nucleus. Neutrons are not charged at all, and found in the nucleus, and electrons are negatively charged and found outside the nucleus in regions known as orbitals.

IrinaK [193]3 years ago

4 0

The proton. There are three main types of subatomic particles. Protons, Neutrons, and Electrons. Protons are positively charged and found in the nucleus. Neutrons are not charged at all, and found in the nucleus, and electrons are negatively charged and found outside the nucleus in regions known as orbitals.

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3 years ago

A 1.5 m wire carries a 9A current when a potential difference of 84 v is applied. what is the resistance of the wire

WARRIOR [948]

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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4 years ago

Read 2 more answers

What is the centripetal force

Lelechka [254]

Answer:90N

Explanation:

Mass=30kg

Centripetal acceleration=3m/s^2

centripetal force=mass x centripetal acceleration

Centripetal force=30 x 3

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3 years ago

4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s

melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

$V_o=Vrms*\frac{1}{12}\\V_o=10Vrms$

the peak voltage before the bridge rectifier is given by:

$V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V$

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

$V_l=V_{op}-2(0.7)\\V_l=12.74V$

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

$V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V$

Using Ohm's law:

$I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA$

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3 years ago