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Lady_Fox [76]
3 years ago
10

You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear

a whisper of about 20 dB intensity. You move closer to bring the conversation up to a sound level of about 50 dB. How close do you need to be?
A. 100 cm
B. 333 cm
C. 90 cm
D. 47 cm
Physics
1 answer:
Marat540 [252]3 years ago
7 0

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

r_2 =15 \times 10^{-1.5}

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

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Lina20 [59]
S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

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7 0
3 years ago
If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53° to the positive direction of the x axis, then what are (a) the x comp
Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

(c) F(x, y) = 1.464 i + 1.952 j

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a)   Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

(c) Net force in vector notation

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

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For details visit www.brainly.com

6 0
2 years ago
What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s^2? (a) 104N (b) 338 N
Brilliant_brown [7]

Answer:

b) 338 N

Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.

the force required to accelerate the gymnast is given by:

F = m×a

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  = 337.5 N

Therefore, the force a trampoline has to apply is 138 N.

6 0
3 years ago
A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.
mr_godi [17]

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

brainly.com/question/1714663

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4 0
1 year ago
Am I correct?? Will give brainliest
Anna35 [415]
Yes that looks correct to me. good luck!!
5 0
3 years ago
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