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3 years ago

What is Hooke's law? Also, what does it say about how elastic materials respond to force? Explain.

1 answer:

3 0

It states that the force F needed to extend or compress a spring by some distance X is proportional to that distance.

For elastic materials, they extend more in same amount of force, (as they are directly proportional), due to it's elastic nature (presence of large deforming force)

Hope this helps!

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In the four-stroke internal combustion engine, fuel is ignited during the ____ stroke.

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The four strokes in order are the intake stroke, the compression stroke, the power stroke, and the exhaust stroke. Fuel is ignited during the power stroke.

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A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then

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3 years ago

You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le

BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7 × 10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5 × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) × 5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0

3 years ago

A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective dura

Genrish500 [490]

Power of the lamp = (voltage) x (current)

0.62 watt = (3.0 v) x (current)

Current = (0.62 watt) / (3.0 volt)

Current = 0.207 Ampere

Charge = (current) x (time)

Charge = (0.207 Amp) x (0.32 sec)

a). Charge = 0.066 Coulomb

Energy discharge through the lamp = (charge) x (voltage)

Energy = (0.66 C) x (3.0 volt)

Energy = 0.1984 Joule

Energy stored in a capacitor = (1/2) (Capacitance) (voltage²)

0.1984 Joule = (1/2) (Cap) (3.0 v)²

Cap = 2 x 0.1984 / 9

b). Cap = 0.044 Farad (yikes !)

Resistance of the lamo = (voltage) / (current)

Resistance = (3.0 volt) / (0.207 Amp)

c). Resistance = 14.49 Ω

All of this feels OK except for that ridiculous 44,000 μF, 3v capacitor. Sadly, I have checked my work without finding an error, and I am too lazy to check it again. So I'll call this my final answer, and leave it to some better Brainly person to find where I went wrong.

6 0

3 years ago