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nalin [4]
3 years ago
14

If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later

Physics
1 answer:
garik1379 [7]3 years ago
5 0
X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

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the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe
Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

          β = 10 log \frac{I}{I_{o}}

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

            I / I₀ = 10 (β / 10)

            I = I₀ 10 (β / 10)

trumpet

            I1 = 1 10⁻¹² 10 (94/10)

            I1 = 2.51 10⁻³ / cm²

Thrombus

           I2 = 1 10⁻¹² 10 (107/10)

           I2 = 5.01 10-2 W / cm²

low

           I3 =1 1-12    (113/10) W/cm²

            I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

           I_total = I₁ + I₂ + I₃

           I_ttoal = 2.51 10-3 + 5.01 10-2 + ​​1.995 10-1

           I_total = 0.00251 + 0.0501 + 0.1995

           I_total = 0.25211 W / cm²

let's bring this amount to the SI system

         β = 10 log (0.25211 / 1 10⁻¹²)

           β = 114 db

7 0
3 years ago
An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia
sergejj [24]

Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = \frac{1}{2}CV^{2}

Here V is the potential difference between the plates.

According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = \frac{1}{2}C_{1} V^{2} = \frac{1}{2}\frac{C}{2} V^{2}

U₁ = U/2

\frac{U_{1} }{U} = \frac{1}{2}

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