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Stels [109]
3 years ago
15

A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictio

nless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion.
In one full cycle of its motion, the total distance traveled by the object is:

1) A
2) A/2
3) 2A
4) 4A
5) A/4
Physics
1 answer:
eduard3 years ago
7 0

Answer:4A

Explanation:

Given

Mass is displace x= A units from its mean position x=0'

When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A

One cycle is completed when block returns to its original position

so first block will go equilibrium position x=0 and then to x=-A

from x=-A it again moves back to x=0 and finally back to its starting position x=A

so it travels a distance of A+A+A+A=4A    

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A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of
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The low-frequency speaker of a stereo set produces 10.0 W of acoustical power. If the speaker projects sound uniformly in all di
Paha777 [63]

Answer:

the required distance is 89.125 m

Explanation:

Given the data in the question;

we know that, sound intensity B in decibels of sound is;

β(dB) = 10log₁₀( I / I_0 )

where intensity I = power / area carried by wave

I_0 = 10⁻¹² W/m² { minimum threshold intensity }

Now,

intensity I = power / area carried by wave = P/A = P/4πr² { spherical  }

given that; β = 80.0 dB and P = 10 W

so

β(dB) = 10log₁₀( I / I_0 )

we substitute

80 = 10log₁₀( P / 4πr²× I_0)

80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )

8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )  

8 = 1 - log₁₀( 4πr²× 10⁻¹² )

8 - 1 = -log₁₀( 4πr²× 10⁻¹² )

7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )

7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]

7 = -[ -10.9 + log₁₀( r² ) ]

7 = 10.9 - log₁₀( r² )

-log₁₀( r² ) = 7 - 10.9

-log₁₀( r² )  = - 3.9

log₁₀( r² ) = 3.9

2log₁₀r = 3.9

log₁₀r  = 3.9 /2

log₁₀r = 1.95

r = 89.125 m

Therefore, the required distance is 89.125 m

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