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Brut [27]
3 years ago
13

Lindsey wants to know if mineral water will make sunflowers grow taller than plain water. In Group A, she pours one cup of miner

al water on 30 sunflower plants every day. In Group B, she pours one cup of plain water on another 30 sunflower plants every day. She keeps all other conditions the same for both groups. After 90 days she measures the height of the sunflowers in both groups. Which term best describes Group A?
A.
control group

B.
independent group

C.
experimental group

D.
dependent group
Physics
1 answer:
Finger [1]3 years ago
8 0
I would go with C but that's just my personal opinion, I chose C because she only did it to group A
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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
joja [24]

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 />

3 0
4 years ago
A block is balanced on top of a frictionless sphere of radius R. When the block is given a slight nudge it starts to slide down
Nataly_w [17]

Answer:\frac{R}{3}

Explanation:

Given

Sphere of Radius R

Suppose mass of block is m

At any instant \theta Normal reaction(N) and weight(mg) is acting such that

mg\sin \theta -N=\frac{mv^2}{R}  , where v is velocity of block at any angle \theta

When block is just about to leave then N=0

therefore

mg\sin \theta =\frac{mv^2}{R}

v^2=gR\sin \theta-------------------1

Also by conserving Energy we get

Potential Energy=kinetic Energy of block

mgh=\frac{mv^2}{2}

here h=vertical distance traveled by block

From diagram

h=R-R\sin \theta

h=R(1-\sin \theta )

mgR(1-\sin \theta )=\frac{mv^2}{2}

2gR(1-\sin \theta )=v^2-----------------2

From 1 and  2

2(1-\sin \theta )=\sin \theta

3\sin \theta =2

\sin \theta =\frac{2}{3}

Thus from this value of h is

h=R(1-\sin \theta )

h=R(1-\frac{2}{3})

h=\frac{R}{3}

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