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Tju [1.3M]
2 years ago
12

The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is

right next to the speakers at a loud concert and you are four times as far away from the speakers, how does the loudness of the music at your position compare to the loudness at your friend's position?
Physics
2 answers:
kirill [66]2 years ago
5 0
Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
                    L = k/(d^2)
For Case 1,
                       L1 = k/(d1^2)
For Case 2,
                       L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2. 
Therefore, the loudness at your friend's position is 16 times that of yours. 
defon2 years ago
3 0

Answer:

Loudness at our position is 1/16 times smaller than the loudness at your friend position

Explanation:

As we know that the loudness decreases as we move away from the speaker

and the loudness is inversely depending on the square of the distance from the speaker

So here we know

L = \frac{k}{r^2}

so here we can compare two loudness at two different positions

\frac{L_1}{L_2} = (\frac{r_2}{r_1})^2

here we know that

r_2 = 4 r_1

now we know that

\frac{L_1}{L_2} = 4^2

L_2 = \frac{L_1}{16}

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Fields of Point Charges Two point charges are fixed in the x-y plane. At the origin is q1 = -6.00 nC . and at a point on the x-a
My name is Ann [436]

Answer:

Part A) Electric fields at the point due to q₁ and q₂:

E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Net electric field at P (Ep)

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ = -6.00 nC = -6 *10⁻⁹C

q₂ = +3.00 nC = +3*10⁻⁹C

d₁ = 4cm = 4 *10⁻²m

d_{2} =\sqrt{(4*10^{-2})^{2}+((3*10^{-2})^{2} }

d₂ = 5 *10⁻²m

Part A) Calculation of the electric fields at the point due to q₁ and q₂

Look at the attached graphic:

E₁: Electric Field at point  P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge

E₂: Electric Field at point  P(0,4) cm  due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² =  10.8*10³ N/C

E₁ = 33.75*10³ N/C (-j)

E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C

E₂y=E₂sinβ = 10.8*(4/5) =  8.64*10³ N/C

E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Calculation of the net electric field at P (Ep)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep=Epx (i) + Epy (j)

Epx= E₂x= 6.48*10³ N/C (-i)

Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

3 0
2 years ago
A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.
shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

F = \mu W

we know that F = ma

ma =  \mu (2/3 mg)

a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2

the new power is p  = 70\% P_[max} = 0.7 P_{max}

P =f_{max} v

0.7P_{max} = ma_{max} v

solving for speed v

v =0.7 \times \frac{P_{max}}{ma_{max}}

v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}

v = 11.5689 m/s

7 0
2 years ago
A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the en
Nina [5.8K]

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

         Em₀ = U = m g h₁

Final point. Lower (slide bottom)

           Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

            mgh₁ = ½ m v² + mgh₂

            v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

          x = v₀ₓ t

         y =  t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

        t = √ 2h₂ / g

We substitute in the other equation

        d = √ (2g (h₁-h₂))  √ 2h₂ / g

        d = √ (4 h₂ (h₁-h₂))

        H = h₁ + h₂

        h₁ = H -h₂

        d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

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Mazyrski [523]
a = 2.7t 
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