a 2.0 kg block on an incline at a 60.0 degree angle is held in equilibrium by a horizontal force, what is the magnitude of this
2 answers:
Answer:
Horizontal force is 16.97 N.
Explanation:
It is given that,
Mass of the block, m = 2 kg
It is on an incline at a 60 degree angle is held in equilibrium by a horizontal force. We need to find the magnitude of horizontal force. The force acting on the block is its weight mg.
The horizontal component of force is,
The vertical component of force is,
So, the horizontal component of force is,
So, the horizontal component of force is 16.97 N.
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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( /
)
where intensity = power / area carried by wave
= 10⁻¹² W/m² { minimum threshold intensity }
Now,
intensity = power / area carried by wave = P/A = P/4πr² { spherical }
given that; β = 80.0 dB and P = 10 W
so
β(dB) = 10log₁₀( /
)
we substitute
80 = 10log₁₀( P / 4πr²× )
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