It is raining outside, so a student lifts their textbook vertically 1.2 meters above
1 answer:
Answer:
Work done is 50.4J.
Explanation:
The work done by the student in lifting the book 1.2 meters vertically is
When the student begins to walk horizontally, the net force on the book is zero <em>(gravity is pulling it down, the student is lifting it up)</em>. Thus, although the student is displacing the book horizontally, but because no force is exerted on it, the work done is zero.
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ANSWER:100dB
f<em>rom the sound intensity level,the sound intensity is calculated as:</em>
<em /><em>₁=</em>
<em>=(10dB)㏒₁₀(l₁/l₀)</em>
<em>inserting numbers:</em>
<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>
<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>
<em>can be used to solve for l₁ and get</em>
<em> l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>
<em>since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>
<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>
<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>
<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>
<em> </em><em>₂=</em>
<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>
<em> </em><em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>
<em /><em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>
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