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rewona [7]
3 years ago
15

PLZZZZZ I NEED HELP Which is an adaptation that helps birds maintain a stable body temperature? A. air sacs connected to lungs B

.large chest muscles C.down feathers D. nearly hollow bones
Physics
2 answers:
bagirrra123 [75]3 years ago
7 0

Answer: I thinks it‘s the feathers

Explanation:

Bc feathers are warm and the hollow bones are to make the bird lighter so it can fly, the chest muscles don’t make heat, and the air sacs are probably for storing air

aniked [119]3 years ago
5 0

Answer:

c

Explanation:

trust

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nika2105 [10]

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8 0
3 years ago
Two application of heat energy​
densk [106]

Answer:

Heat is very important in our daily life in warming the house, cooking, heating the water, and drying the washed clothes. The heat has many usages in the industry as making and processing the food and manufacture of the glass, the paper, the textile, and etc.

Explanation:

4 0
3 years ago
Technician A says that a tire will increase or decrease approximately 1 psi for each 10°F change of temperature. Technician B sa
bearhunter [10]

Answer:

Technician A

Explanation:

It is seen that a tire pressure will increase or decrease 1 psi for each 10^{\circ} F  change in temperature.

For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as  the temperature falls to normal .          

8 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
How will the motion of a falling whirligig compare to that of a falling paper ball?
kicyunya [14]

The motion of a falling whirligig is different to that of a falling paper ball due to spinning.

<h3>Type of motion performed by whirligig and falling paper ball </h3>

The motion of a falling whirligig is different from the motion of a falling paper ball because the paper ball falls on the ground without spinning while on the other hand, the whirligig falls on the ground along with spinning.

The falling whirligig performs two motion i.e. one is falling on the ground and the other is spinning during motion whereas paper ball performs one motion i.e. motion in the air towards the ground so we can conclude that the motion of a falling whirligig is different than of a falling paper ball.

Learn more about motion here: brainly.com/question/453639

4 0
3 years ago
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