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2 years ago

How does the role of blood protect the body

Physics

1 answer:

maksim [4K]2 years ago

4 0

Well it transports white blood cells to infected areas.

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What is the magnetic flux linkage, in units of Weber, for a coil of 360 turns and cross sectional area of 0.133 m^2 when the mag

zalisa [80]

Answer:

83.3 Wb

Explanation:

The magnetic flux linkage through the coil is given by:

$N\phi = BAN sin \theta$

where

B is the magnetic field strength

A is the cross sectional area

N is the number of turns in the coil

$\theta$ is the angle between the direction of the field and the normal to the coil

In this problem:

B = 1.74 T

A = 0.133 m^2

N = 360

$\theta=90^{\circ}$

Therefore, the magnetic flux linkage is

$N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb$

7 0

3 years ago

22 points !

nordsb [41]

Evaporation. The glass of water was less full after 10 minuets because the water got evaporated

7 0

3 years ago

A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th

Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0

2 years ago

An 81.5-kg man stands on a horizontal surface.

Elza [17]

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10² N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

$P=\dfrac{F}{A}$

$P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2$

P=181.11 x 10² N/m²

4 0

3 years ago

With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t

VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

$v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N$

Therefore, the tension of the rope is 34.95 N

4 0

2 years ago