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3 years ago

Describe the unique role of water in chemical and biological systems.

1 answer:

7 0

Of the important biological molecules only the non-polar lipids (fats and oils) and large polymers (e.g. polysaccharides, large proteins and DNA) do not dissolve. The water acts as a solvent for chemical reactions and also helps transport dissolved

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find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo

Ksivusya [100]

Answer:

x = 0.237

y = 0.0789

Explanation:

Vector with direction 18.4° and magnitude 0.250 has x and y components of:

x = 0.250 cos 18.4°

x = 0.237

y = 0.250 sin 18.4°

y = 0.0789

7 0

3 years ago

Read 2 more answers

Atoms contain empty space true or false

olga nikolaevna [1]

The answer would be false

5 0

3 years ago

A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn

Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

$AB$ = 155 ft

$Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft$

$Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft$

Using Pythagorean theorem in triangle ADC

$AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft$

$d$ = distance between the anchor points

distance between the anchor points is given as

$d = BD + CD = 70.4 + 93.4\\d = 163.8 ft$

5 0

3 years ago

Read 2 more answers

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po

coldgirl [10]

Answer:

the intensity of the light after passing through the two polarizing filters is 4.11 units

Explanation:

Given the data in the question;

the intensity of an unpolarized light; I₀ = 25.0 units

when the unpolarized light passes through the first polarizer, its intensity reduces to half of its initial value;

⇒ I₁ = I₀/2 = 25/2 = 12.5 units

the angle between the transmission axes of two polarizers is;

∅ = 55° - 0° = 55°

The intensity of the light after passing through two polarizing filters will be;

I₂ = I₁cos²∅

we substitute

I₂ = 12.5 × cos²(55)

I₂ = 12.5 × 0.3289899

I₂ = 4.11 units

Therefore, the intensity of the light after passing through the two polarizing filters is 4.11 units

6 0

3 years ago