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zloy xaker [14]
3 years ago
13

Question I. SUHU ULT

Physics
1 answer:
Lilit [14]3 years ago
3 0

Answer:

Feathers are great thermal insulators.

Explanation:

Feathers are great thermal insulators. The loose structure of down feathers traps air.

As a result, energy cannot be transmitted easily through down feathers. This means birds are insulated from cold air outside, plus their body heat doesn't escape easily either.

Human beings discovered that down feathers are good for insulation long ago. For example, documents from the 1600s show that Russian merchants sold “bird down" to the Dutch hundreds of years ago.

Today, down is used in all sorts of products, including coats, bedding, and sleeping bags, to help better insulate the user from cold weather. Down can be collected from many different types of birds, but most of today's supply comes from domestic geese.

If you have a down coat or comforter, is it all down? In the United States, laws require that products labeled “100 percent down" contain only down feathers.

If your product is labeled “down," it can contain a mixture of both down feathers and synthetic fibers. Not all down feathers are created equal, though.

Down insulation is rated on a measure called “fill power." The higher the fill power, the more the down insulates.

The highest fill-power rating — 1200 — goes to eiderdown, which comes from the Common Eider duck. Eiderdown tends to be expensive.

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- vocational aim
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2 years ago
) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in
Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}  

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0
3 years ago
How far will a ball travel that goes 35 meters per second for 18 seconds?
katen-ka-za [31]
Theoretically, 35 x 18 = 630
7 0
3 years ago
a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative densi
ki77a [65]

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

6 0
3 years ago
Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
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