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Aleonysh [2.5K]

4 years ago

8

lipids are stored until they are needed by cells. there are molecules in the cells that break the lipids down and convert them i

nto

1 answer:

juin [17]4 years ago

4 0

The basic monomer of a lipid is a triglyceride. It breaks down into one glycerol molecule and 3 fatty acid tails.

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What happens to the Connecticut Energy of a snowball as it rolls across the lawn and gains mass?

statuscvo [17]

I think it will go down like decrease minus or whatever you call it

6 0

4 years ago

What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is 5.90 x 10^5 K?

OLEGan [10]

<u>Answer:</u> The average kinetic energy of helium atoms is $1.222\times 10^{-17}J$

<u>Explanation:</u>

To calculate the average kinetic energy of the atom, we use the equation:

$K=\frac{3}{2}kT$

where,

K = average kinetic energy = ?

k = Boltzmann constant = $1.3807\times 10^{-23}J/K$

T = temperature = $5.9\times 10^5K$

Putting values in above equation, we get:

$K=\frac{3}{2}\times 1.3807\times 10^{-23}J/K\times 5.9\times 10^5K\\\\K=1.222\times 10^{-17}J$

Hence, the average kinetic energy of helium atoms is $1.222\times 10^{-17}J$

4 0

3 years ago

A train moves moves from rest to a speed of 25 m/s in 30.0 sec . what is its acceleration

spayn [35]

A = Delta v/Delta t
Delta v = 25 - 0 = 25
Delta t = 30
25/30 = 5/6 = 1.66 repeating

7 0

3 years ago

Read 2 more answers

Consider the cylindrical weir of diameter 3 m and length 6m. If the fluid on the left has a specific gravity of 0.8, find the ma

sladkih [1.3K]

This question is incomplete, the complete question is;

Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.

Answer:

- the magnitude of the resultant force is 557.32 kN

- the direction of resultant force is 48.29°

Explanation:

Given the data in the question and the diagram below,

First we work on the force on the left hand side.

Left Horizontal

$F_{LH$ = βgAr

here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²

we substitute

$F_{LH$ = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N

Left Vertical

$F_{LV$ = ( βgπh² / 2 ) × W

we substitute

$F_{LV$ = [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N

Now we go to the right hand side

Right Horizontal

$F_{RH$ = βgAh

here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²

we substitute

$F_{RH$ = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N

Right Vertical

$F_{RV$ = ( βgπh² / 4 ) × W

we substitute

$F_{RV$ = [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N

Hence

Fx = $F_{LH$ - $F_{RH$ = 52974 N - 423792 N = 370818 N

Fy = $F_{LV$ + $F_{RV$ = 332845.458 N + 83211.36 N = 416056.818 N

R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N

R = 557.32 kN

Therefore, the magnitude of the resultant force is 557.32 kN

Direction of resultant force;

tanθ = Fy / Fx

we substitute

tanθ = 416056.818 N / 370818 N

tanθ = 1.121997

θ = tan⁻¹( 1.121997 )

θ = 48.29°

Therefore, the direction of resultant force is 48.29°

4 0

3 years ago

In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre

sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

Explanation:

Given that,

Wave length = 500 nm

Radius $d= 40\ \mu m$

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

$\sin\theta=\dfrac{y}{D}$

$\sin\theta=\dfrac{y}{6}$

Using formula for dark fringe

$(n-\dfrac{1}{2})\lambda=2d\sin\theta$

Put the value into the formula

$(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}$

$y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}$

$y=0.05625\ m$

$y=5.625\times10^{-2}\ m$

Hence, The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

6 0

3 years ago