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3 years ago

Anyone to assist me on it ...especial page7 and 8

Physics

1 answer:

julia-pushkina [17]3 years ago

5 0

Answer:

i needed points it was an emergency sorry

Explanation:

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If a display has a dynamic range of 20 dB and the smallest voltage it can handle is 200 mV, then the largest voltage it can hand

DedPeter [7]

Answer:

The largest voltage is 0.02 V.

(d) is correct option.

Explanation:

Given that,

Range = 20 dB

Smallest voltage = 200 mV

We need to calculate the largest voltage

Using formula of voltage gain

$G_{dB}=10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$20 =10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$2=log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$10^2=\dfrac{V_{out}^2}{V_{in}^{2}}$

$\dfrac{V_{out}}{V_{in}^}=10$

$V_{in}=\dfrac{V_{out}}{10}$

$V_{in}=\dfrac{200}{10}$

$V_{in}=20\ mV$

$V_{in}=0.02\ V$

Hence, The largest voltage is 0.02 V.

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3 years ago

2. Vehicle-braking distance is the distance your vehicle travels after you see a problem

babunello [35]

The answer is B. False

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3 years ago

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The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc

algol [13]

Answer:

Explanation:

Given

current required $I=110 mA$

Internal Resistance $R=300 \Omega$

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

$V\propto I$

$V=IR$

$V=110\times 10^{-3}\times 300$

$V=33 V$

5 0

3 years ago

Which of the following choices is not an example of an occupation that is projected to lose jobs?

Orlov [11]

Answer:

Explanation:

D is the most likely answer because as we become more modernized as a whole we will progressively need to fill more positions in support specialty fields.

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3 years ago

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What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a

Fiesta28 [93]

Answer:

a) $E_photon =0.306 eV$

b) $E_photon =0.166 eV$

Explanation:

The energy of the photon (E) for $n^th$ orbit of the hydrogen atom is given as:

$E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$

where,

$E_o$ = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

$E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})$

$E_photon =0.306 eV$

b) Now for the transition from n = 5 to n = 6

$E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})$

$E_photon =0.166 eV$

7 0

3 years ago