Polarized light passes through a polarizer. If the electric vector of the polarized light is horizontal what, in terms of the in

You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.

serg [3582]

Answer:

d = 2021.6 km

Explanation:

This distance problem can be solved using vector analysis; it's best to find each plane's position components before applying the Pythagorean theorem to calculate the separation between them.

For Airplane 1:

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

For Plane 2:

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 sin 25 = 8.452 103 m = 8452 m

To determine the distance between the planes using the Pythagorean theorem:

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Now, we perform the calculations:

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 + 9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km

7 0

2 months ago

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

ValentinkaMS [3465]

Answer:

a) Blood mass is $m= 5.7876kg$

b) The count of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the problem statement, we learn that

The blood volume is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of blood is $\rho_b = 1060 kg/m^3$

% of blood which consists of cells is = 45.0%

the % of blood that is plasma is = 55.0%

density of blood cells is $\rho_d = 1125kg/m^3$

% of cells that are white is = 1%

% of cells that are red is = 99%

The red blood cell diameter is = $7.5 \mu m = 7.5*10^{-6}m$

The red blood cell radius is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

The mass is generally represented mathematically as

$m = \rho_b * V_b$

Substituting values

$m = 1060 * 0.00546$

$m= 5.7876kg$

Cell mass is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The white blood cells volume is $V_w$ = 1% of the cells volume

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The red blood cells volume is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The total red blood cell count is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The total white blood cell count is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The overall number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

2 months ago

Read 2 more answers

We fully submerge an irregular lump of material in a certain fluid. The fluid that would have been in the space now occupied by

Ostrovityanka [3204]

Answer:

the lump descends

Explanation:

The full articulation of the query is

Upon fully submerging a 3 kg lump of material in a certain fluid, the fluid that would have occupied the space now filled by the lump weighs 2 kg. (a) When released, does the lump float up, sink, or remain steady

(a)

$F_{b}$ = Buoyant force acting upward on the lump