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aleksklad [387]
3 years ago
8

Your friend turns the volume up on their speaker system, so that the sound intensity is 4 times greater than it was beforehand.

By what factor has the energy output of the speaker system increased?
Physics
1 answer:
erica [24]3 years ago
4 0

Answer:

4

Explanation:

We know that intensity I = P/A where P = power and A = area through which the power passes through.

Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.

So, I = P/A and I₀ = P₀/A

Now, if I = 4I₀,

P/A = 4P₀/A

P = 4P₀

Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t

Substituting P and P₀ into the equation, we have

P = 4P₀

E/t = 4E₀/t

E = 4E₀

Since the energy is four times the initial energy, the energy output increases by a factor of 4.

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Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

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Read 2 more answers
There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres
Greeley [361]

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.

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