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tangare [24]
3 years ago
9

While Robert was helping his father in the garden, he pushed a shovel into the ground to dig a hole in the dirt. He picked up a

shrub, carried it to the
hole, and held it while he lowered it into the hole. During which parts of this process does Robert NOT do work? Choose the two statements that apply.
A. He pushed a shovel into the ground to dig a hole.
B. He picked up a shrub.
C. He carried it to the hole
D. He held it while he lowered it into the hole
Physics
1 answer:
makvit [3.9K]3 years ago
7 0

Work done is given by the change in kinetic energy of an object

  • The kinetic energy of the shovel, the shrub, and in Robert's movement were changed, therefore, work is done in the given processes,

Reason:

Work is done when the total energy of object is affected by the application of force on the object over a distance

Therefore;

  • In option <em>A</em>, pushing the shovel into ground (to dig out the dirt) the requires the application of a force (push) over a distance, (into and out of the ground) therefore work is done
  • In option <em>B</em>, picking the shrub up gives it gravitational potential energy, therefore, work is done
  • In option <em>C</em>, carrying the shrub to the hole does visible work
  • In option <em>D</em>, holding the shrub while lowering it into the hole does work by preventing the shrub from falling randomly

Therefore, <u>work is done in the given processes</u>

Learn more about work-energy theorem here:

brainly.com/question/10063455

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A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so
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ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<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>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

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Explanation:

Similar to another Brainly answer :O

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