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fgiga [73]
3 years ago
5

What is the relation between the gravitational force acting between any two bodies object with their masses and the distance bet

ween their centres?​
Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

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Which part of the water cycle would follow step D in the diagram shown?
Dmitry [639]

Condensation.


iGreen, more like iNOTGreen amirite?

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3 years ago
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Beams used in Heavy Timber construction are sometimes firecut. This is done to: a) allow air circulation at the beam’s end in an
ladessa [460]

Answer:

option C

Explanation:

The correct answer is option C

Fire cut of fireman cut is diagonal cut which is provided at the end of the beam to prevent the fall of masonry wall if a fire breaks out in the building.

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Without fire cut, the burnt beam will rotate downward affecting the connection of beam and wall and leading to damage it.

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3 years ago
What is electromagnet and list the types of electromagnet.​
sineoko [7]

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radio waves, micro wave, x-rays

Explanation:

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4 years ago
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A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi
Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

      $=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

    $= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

    = 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

                                           $ =\frac{1}{2} \times 1.225  \times 200^2$

                                          $=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

     $=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

     = 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

                  $=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

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3 years ago
What is cosmic microwave background radiation?
chubhunter [2.5K]
Cosmic, or background, radiation is the small amount of high energy radiation which is mostly left over from the big bang or from supernovas. It is mostly single protons, but also alpha particles and even sometimes heavier elements. It can also refer to the low levels of electromagnetic radiation present all over the universe.
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