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Bond [772]
3 years ago
8

The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th

em are talking at the same time as loudly as she is, what sound level are you being exposed to?
Physics
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

=10^{-5.9}

intensity of sound of 5 persons

I=5\times 10^{-5.9}

dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

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Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge
Vanyuwa [196]

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k k \frac{ q_{1}  \  q_{2} }{ r^{2} }

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = F_{bc x}

Y axis  

       F_{y}Fy = F_{ab} - F_{bc y}

let's find the magnitude of each force

     F_{ab} = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

   

      F_{bc} = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + F_{y} j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = \sqrt{ F_{x}^2 + F_{y}^2 }

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= \frac{ F_{y}  }{  F_{x} }

       

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

3 0
3 years ago
What happens to the density of gas when it compressed
Stella [2.4K]
When a gas is compressed, its particles are pushed closer together so it is more dense.
3 0
3 years ago
Explain what is meant by “In Phase” and “Out of Phase?
LekaFEV [45]
In phase would mean both waves are at a positive peak, out of phase would mean one is at a positive whilst the other is at a negative. Out of phase would mean the waves cancel each other out
5 0
3 years ago
Read 2 more answers
Which statement describes a property of a magnet? A. It attracts ferrous materials. B. It could have only one pole (north or sou
hodyreva [135]

Answer:

It attracts ferrous materials

Explanation:

A magnet attracts ferrous materials A ferrous materials are metallic substances or conductors that can conduct heat and electricity. Example of this ferrous materials includes iron, metal etc. Since magnets only can attracts metallic substance to itself, then we can also conclude that they attract ferrous materials since ferrous materials. possesses properties of a metal.

Magnets possesses both north and south poles.

The same of the bar magnets are known to repel each other while unlike poles attract each other.

7 0
3 years ago
A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met
AleksandrR [38]

Answer:

h = 36.4 cm

Explanation:

given,

spring constant = 2.5 x 10⁴ N/m

compressed distance = 11.2 cm = 0.112 m

mass of the child = 44 kg

maximum height = ?

by energy of conservation

K.E_i +P.E_i= K.E_f + P.E_f

\dfrac{1}{2}kx^2 = mgh

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

156.8 = 44 \times 9.8 \times h

h = \dfrac{156.8}{44 \times 9.8}

h = 0.364 m

h = 36.4 cm

7 0
3 years ago
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