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2 years ago

5

In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background sign

al level is 13 mV, the signal is 91 mV with no attenuator and is 25 mV with an attenuator in place. Calculate pt/pi including the background correction.

1 answer:

FrozenT [24]2 years ago

4 0

Answer:

The answer is " $15.38\%$ "

Explanation:

Background $= 13 \ mv\\\\$

corrected signal $= 91 \ mv-13\ mv= 78\ mv\\\\$

with attenuator $=25\ mv-13\ mv= 12\ mv\\\\$

$\to \frac{p_t}{p_i}=\frac{12}{78}\times 100= 15.38\%$

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“All dogs bark. Fido barks. Thus, Fido is a dog,” is an example of which of the following?

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3 years ago

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

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a. ρ $_\beta=1.996J/m^3$

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Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

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b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

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3 years ago

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

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3 years ago