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3 years ago

Three 5 Ohm resistors are connected in parallel to a 9 Volt power supply. What is the current through each resistor?

Physics

1 answer:

Sedaia [141]3 years ago

3 0

Answer:

I₁ = I₂ = I₃ = 12 A

Explanation:

Since the resistors are connected in parallel across a battery. Hence, the voltage across each of them will be the same, which will be equal to the voltage of the power supply. Since each resistor has the same resistance of 5 Ω. Therefore. the current across each resistor will also be the same. We can easily calculate these current values by the use of Ohm's Law, as follows:

$V = IR\\I_{1}=I_{2}=I_{3}=\frac{V}{R} \\\\I_{1}=I_{2}=I_{3}=\frac{9\ Volt}{5 \Omega}\\\\$

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if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago

When a car suddenly stops at a red light, a book lying on the car seat slides forward. Why does the book continue to move forwar

goblinko [34]

Answer:

Explanation:

The answer is c because Newton’s third law states “ for every action there is an equal and opposite reaction.” Since the car quit moving the book did not stay at rest because there was a reaction.

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3 years ago

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A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.

nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R$

where $m=4.10\,\rm kg$ , $v=2.85\frac{\rm m}{\rm s}$ , and $R$ is the radius of the circular path.

As shown in the diagram, we can see that

$\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)$

where $r=1.69\,\rm m$ , so that

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}$

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

$F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}$

Solve for $\theta$ :

$\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0$

Complete the square:

$\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}$

Plugging in the known quantities, we end up with

$\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27$

The second case has no real solution, since $-1\le\cos(\theta)\le1$ for all $\theta$ . This leaves us with

$\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}$

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2 years ago

Please help me i’ll give you brainlist

Diano4ka-milaya [45]

Answer:

True

Explanation:

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3 years ago

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Why are the alkali metals likely to react with group 17 elements?

Rudiy27

Answer:

Because alkali metals are so reactive, they are found in nature only in combination with other elements. They often combine with group 17 elements, which are very “eager” to gain an electron.

Explanation:

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3 years ago