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2 years ago

What is the maximum current value that can be measured with an analog VOM? A. 5 A B. 20 A C. 10 A D. 15 A

Physics

2 answers:

Korvikt [17]2 years ago

8 0

Answer is b to ur question

zvonat [6]2 years ago

4 0

Answer:

B) 20 A

Explanation:

VOM full form is Volt Ohm milliammeter. It indicates that voltage, resistance and the current can be measured using this device. When the variations are very rapid , analog VOM is used instead of a digital VOM.

Using this current ( both A C and DC) can be measured in amperes.

Though the accuracy given by digital devices are more accurate, to identify changes in the circuit being tested and for trouble shooting, analog meters are more preferred.

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3 years ago

Balanced or Unbalanced?

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3 years ago

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suter [353]

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1 year ago

Read 2 more answers

Use the following photoelectric graph to answer the following question:

klio [65]

The photoelectric effect is obtained when you shine a light on a material, resulting in the emission of electrons.

The kinetic energy of the electrons depends on the frequency of the light:
K = h(f - f₀)

where:
K = kinetic energy
h = Planck constant
f = light frequency
f₀ = threshold frequency

Rearranging the formula in the form y = m·x + b, we get:
K = hf - hf₀
where:
K = dependent variable
f = <span>indipendent variable
h = slope
hf</span>₀ = y-intercept

Every material has its own threshold frequency, therefore, what stays constant for all the materials is h = Planck constant (see picture attached).

Hence, the correct answer is C) the slope.

5 0

3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago