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goldenfox [79]
3 years ago
13

I need help please in my homework

Physics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

Correct answer: vₓ = 30V , i₀ = 5A

Explanation:

You first calculate the equivalent resistance of the parallel connection of two resistances of 12 and 6 ohms.

1/ Re₁ = 1/12 + 1/6 = (12 + 6) / (12 · 6) => 1/Re₁ = 18/72 => 1/Re₁ = 1/4 =>

Re₁ = 4Ω

then you calculate the equivalent resistance of a series connection of two resistances of 8 and Re₁ = 4Ω

Re₂ = Re₁ + 8 = 4 + 8 = 12Ω

then you calculate the electric current according to Oh's law

i₀ = vc / Re₂ = 60 / 12 = 5A

at the end you calculate the voltage drop vₓ on the resistor of 6Ω

vₓ = i₀ · 6Ω = 5A · 6Ω = 30V

God is with you!!!

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A motor must lift a 1500-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.
natka813 [3]

Answer:

12900 W

24200 W

Explanation:

Given:

v₀ = 0 m/s

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t = 2.0 s

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Δx = ½ (v + v₀) t

Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)

Δx = 1.3 m

While accelerating:

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∑F = ma

F − mg = ma

F = m (g + a)

F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)

F = 19855 N

Power = work / time

P = W / t

P = Fd / t

P = (19855 N) (1.3 m) / (2.0 s)

P ≈ 12900 W

At constant speed:

Newton's second law:

∑F = ma

F − mg = 0

F = mg

F = (1500 kg + 400 kg) (9.8 m/s²)

F = 18620 N

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5 0
4 years ago
Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not
d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},

where  I_0(\text{at }1000\text{ Hz}) is the hearing threshold at 1000 Hz.

Therefore we have the following

0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}

I_1 is the threshold at 1000Hz and I_2 is the threshold at 100Hz.

By exponentiating we have

10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.

Now dividing these two equations we get

\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.

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gavmur [86]

Answer:

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A) Field lines cannot be crossed. Otherwise, at the point where the lines cross, there would be two different electric field vectors, which makes no physical sense.

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