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3 years ago

Which type of tissue contains macrophages and mast cells?

Physics

2 answers:

ankoles [38]3 years ago

4 0

Answer:

Ordinary connective tissue

Vadim26 [7]3 years ago

4 0

Answer: Ordinary connective tissue

Explanation: Ordinary connective tissue includes two resident cell types with immunological function, mast cells and macrophages.

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Explanation:

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3 years ago

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When driving straight down the highway at a constant velocity you have to give the engine a little gas (which means an added ext

Alona [7]

Answer:

Explanation:

This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.

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4 years ago

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t

marin [14]

Answer:

$T=66262.4s$

Explanation:

From the question we are told that:

Altitude $A=2.90 *10^7$

Mass $m=5.97 * 10^{24} kg$

Radius $r=6.38 *10^6 m.$

Generally the equation for Satellite Speed is mathematically given by

$V=(\frac{GM}{d} )^{0.5}$

$V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}$

$V=3354.83m/s$

Therefore

Period T is Given as

$T=\frac{2 \pi *a}{V}$

$T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}$

$T=66262.4s$

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3 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$