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Taya2010 [7]
3 years ago
7

Identify areas where winds are weak

Physics
1 answer:
tiny-mole [99]3 years ago
6 0
Winds are weak in High Pressure area's
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Suppose a ray of light traveling in a material with an index of refraction n a reaches an interface with a material having an in
KATRIN_1 [288]

Answer: C and D

Explanation: One of the first rule for total internal reflection to occur is that the ray must move from a dense to a less dense medium, hence refractive index of medium a must be greater than that of b.

When a ray moves from a dense to a less dense medium, the refracted ray moves away from the normal thus increasing the size of the angle of refraction (total internal refraction occurs when the angle of refraction is 90° and the angle of incidence at this point is known as the critical angle), hence the angle of incidence must be greater than the critical angle.

These points verifies option C and D

5 0
3 years ago
when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal
Juli2301 [7.4K]

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

  • The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

learn more:

Reactivity brainly.com/question/6496202

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4 0
3 years ago
A plane is going at a speed of 300 km/h at 63 W of N. The wind hits the plane at a direction of 65 km/h at 52 S of E. What is th
katovenus [111]
It’s around the g force so it’s gonna be around 54 km/h
3 0
2 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
Hii can someone please doo this! 50 pointss.
andrey2020 [161]

Answer:

The periodic table illustrate some of the elements from Hydrogen to Calcium

8 0
3 years ago
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