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Savatey [412]
3 years ago
5

Why do atoms have an overall neutral charge?

Physics
1 answer:
Levart [38]3 years ago
4 0
Https://www.google.com/search?q=Why+do+atoms+have+an+overall+neutral+charge%3F&oq=Why+do+atoms+have+...

Neutrons are no exception. So, if an atom<span> has equal numbers of electrons and protons, the </span>charges<span> cancel each other out and the </span>atom<span> has a </span>neutral charge<span>.
</span>
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As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit ____________ (increases,
Anestetic [448]

Answer:

Decreases, Increases

Explanation:

Resistance is parallel can be calculated using

1/Req = 1/R1 + 1/R2 + 1/R3 +....

Then, as more resistor is added in parallel the equivalent resistance is reduced.  

Let use a simple sample

Let all the resistor have equal resistances

Let say R = R1 = R2 = R3 =...Rn

Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....

1/Req = 1/R + 1/R + 1/R +.... 1/Rn

Req = R/n

Check attachment on how I got that.

This implies that, the equivalent resistance will always be less than the original resistance, since n>1

So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.

B. Now, to know if the current reduces or increases

Using Ohms law

V = iR

Then, I = V/R

So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.

So current increase as we add more resistor in parallel to a circuit

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3 years ago
What are non-clastic?
igor_vitrenko [27]
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Name one benefit and one limitation of Comparative investigations
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Read 2 more answers
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

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