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koban [17]
3 years ago
14

A force can exist between two charged particles or objects even if they're as small as subatomic particles. Between which of the

se would an attractive force exist?
A. proton-proton
B. neutron-proton
C. proton-electron
D. electron-electron
Physics
2 answers:
ASHA 777 [7]3 years ago
3 0
If both particles have the SAME electrical charge, then they repel.
If they have DIFFERENT electrical charge, then they attract.

Protons have  +  charge .
Electrons have  -  charge .

So two protons (A) or two electrons (D) push apart.

One proton and one electron (C) pull together.
Zolol [24]3 years ago
3 0

Answer :

The attractive force exist between proton - electron.

(C) is correct option.

Explanation:

If both particles have like charges then the particles repel each other .

If both particles have unlike charges then the particles attract to each other.

Proton and electron have positive and negative charge respectively.

Neutron has no charge.

So, The attractive force exist between two charged particles proton and electron.

Hence, The attractive force exist between proton - electron.

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Answer:

motion

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3 years ago
A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an
ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

Z= R\sqrt{2} \\\\R  = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms

<h3>Resonant frequency</h3>

f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)

<h3>At driven frequency</h3>

X_l- X_c = R\\\\\omega L - \frac{1}{\omega C}  = 707.1\\\\2\pi f L -  \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\

<em>solve 1 and 2 together</em>

2\pi(8000) L - \frac{L}{2\pi (8000)(7.034 \times 10^{-10})} = 707.1\\\\50272L - 28279.48L = 707.1\\\\L = 0.032 \ H

Learn more about impedance of RLC circuit here: brainly.com/question/372577

7 0
2 years ago
What are the five plants you can see from earth without a telescope?
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The five planets that you can see from Earth without a telescope are Mercury, Venus, Mars, Jupiter and Saturn. 
3 0
3 years ago
he capacitor can withstand a peak voltage of 550 VV . If the voltage source operates at the resonance frequency, what maximum vo
anygoal [31]

Answer:

The maximum voltage is 39.08 V.

Explanation:

Given that,

Voltage = 550 V

Suppose, In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is 1.20\times10^{-2}\mu F

We need to calculate the resonant frequency

Using formula of resonant frequency

f=\dfrac{1}{2\pi\sqrt{LC}}

Put the value into the formula

f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}

f=2356.8\ Hz

We need to calculate the maximum current

Using formula of current

I=\dfrac{V_{c}}{X_{c}}

I=2\pi f C\times V_{c}

Put the value into the formula

I=2\pi\times2356.8\times1.20\times10^{-8}\times550

I=0.0977\ A

We need to calculate the impedance of the circuit

Using formula of impedance

Z=\sqrt{R^2+(X_{L}-X_{C})^2}

At resonant frequency , X_{L}=X_{C}

So, Z = R

We need to calculate the maximum voltage

Using formula of voltage

V=IR

Put the value into the formula

V=0.0977\times400

V=39.08\ V

Hence, The maximum voltage is 39.08 V.

4 0
3 years ago
Help no link and answer please
Tom [10]

Answer:

12. you measure mass with a balnece. such as a triple balence beam or an electronic balance.

i dont know 11 sorry

8 0
2 years ago
Read 2 more answers
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