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natta225 [31]
2 years ago
9

What is a stack of membranes that packages chemicals

Physics
1 answer:
jeyben [28]2 years ago
5 0
That would be the Golgi Apparatus
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What type of plant tissue is NOT part of the vascular bundle?
jek_recluse [69]

Answer:

Cortex

Explanation:

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2 years ago
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What is the mechanical advantage of the machine shown below? a 5 b 4 c 3 d 2
Alex

Answer:

Explanation: Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device preserves the input power and simply trades off forces against movement to obtain a desired amplification in the output force.

4 0
2 years ago
Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The
tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q  x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q  /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0
2 years ago
A wooden block has a mass of 562 g and a volume of 72 cm3. What is the density?
dangina [55]
Mass/volume is density so it’s 562g/72cm^3 so it’s roughly 7.805g per cubic centimeter
3 0
2 years ago
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A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t
fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

A \rightarrow (A-\frac{A}{9} )

A \rightarrow \frac{8A}{9}

As we know,

Energy will be:

⇒  E_{1}=\frac{1}{2}KA^2

and,

⇒  E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2

          =\frac{64KA^2}{162}

⇒  \Delta E=E_1-E_2

On putting the estimated values, we get

           =\frac{1}{2}KA^2-\frac{64KA^2}{162}

⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

          =\frac{40}{162}

          =0.246

3 0
2 years ago
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