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3 years ago

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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Match each situation

EastWind [94]

A: is potential
C: is losing kinetic energy and gaining potential energy
B: kinetic energy is at its highest
D: is loosing potential energy and gaining kinetic energy

3 0

3 years ago

Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto

Irina-Kira [14]

Answer:

W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

V = A L

V = π (R_ext ² - R_int ²) L

let's calculate

V = π (4² - 2²) 10⁻⁴ 3

V = 1.13 10⁻² m³

m = ρ V

m = 8966 1.13 10⁻²

m = 1.01 10² kg

the weight of the tube

W = mg

W = 1.01 10² 9.8

W = 9.93 10² N

4 0

2 years ago

Explore the role of friction when creating safety ramps and escalators. "Why is it important for people and object to move slowl

faust18 [17]

The role of friction is of great importance when creating safety ramps and escalators because with the help of friction things move.

<h3>Why is it important to move objects slowly on ramps and escalator?</h3>

It is important to move objects slowly on ramps and escalator because the ramps and escalator moves object in the opposite direction of gravity. If we did not move objects slowly, then the objects or a person get damaged.

So we can conclude that the role of friction is of great importance when creating safety ramps and escalators because with the help of friction things move.

Learn more about friction here: brainly.com/question/24338873#SPJ1

3 0

2 years ago

our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure

Dafna1 [17]

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass $m = 59$ Kg

Frictional force $F_{s} = 318.3$ N

From the formula of frictional force,

$F_{s} = \mu_{s} mg$

Where $\mu _{s} =$ coefficient of static friction, $g = 9.8 \frac{m}{s^{2} }$

Put the above values and find the coefficient of static friction.

$318.3 = \mu_{s} \times 59 \times 9.8$

$\mu_{s} = 0.55$

Therefore, the coefficient of static friction between your partner and the floor is 0.55

4 0

3 years ago

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

7 0

3 years ago