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3 years ago

Find the current if R is 470 ohms and the volts is 3.5V

Physics

1 answer:

boyakko [2]3 years ago

8 0

Answer:

I = 0.00744 A (Amps)

Explanation:

Hi there !

Ohm's law formula

I = V/R

I = 3.5V/470Ω

I = 0.00744 A

Good luck !

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MA_775_DIABLO [31]

The answer is (A) hope it helps

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3 years ago

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51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

3 years ago

The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it

horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

Where P = Power, I = rms current, V = rms voltage.

make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

V = 1500(4.525)

V = 6787.5 V

Hence the rms voltage = 6787.5 V

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3 years ago

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Aleks [24]

I don't understand the question but impulse and momentum is the same. So maybe is the force same too

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3 years ago

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