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Bess [88]
3 years ago
6

A beam of light strikes a mirror reflects off at the angle in which it hit the mirror. which term does the diagram below illustr

ate A. Rotation B. Reflection C.refraction D.revolution
Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer:

B.) reflection

Explanation:

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Suppose you are in total equilibrium in water (in other words you are floating and not moving in any direction)What could you do
Eva8 [605]

Explanation:

According to newtons first law of motion:

'' a body will continue in its state of rest or uniform motion along a path unless it is acted upon by an external force".

A body in equilibrium that is floating will be stable and not move in any direction. Even if it moves, the motion will be constant wouldn't change.

  • To move the body in any direction, one has to swim.
  • Swimming is the application of an external force to counter the balanced forces at equilibrium on a body.
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learn more:

Newton brainly.com/question/11411375

#learnwithBrainly

5 0
3 years ago
An aircraft flying in a straight
Lelechka [254]

Answer:

t = 10.1 s

d = 2020 m

Explanation:

Time to drop from vertical rest

h = ½gt²

t = √(2h/g) = √(2(500)/9.8) = 10.1 s

d = vt = 200(10.1) = 2020 m  

3 0
2 years ago
2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
A certain metal wire has a cross sectional area of 1 mm2 and is 1 m long. when it is hung from the ceiling and a 10 kg mass is h
kvasek [131]
From the Hooke's law , the extension force of an elastic material is directly proportional to the extension. 
That is, F = k e, where F is the force , k is the constant and e is the extension
 F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
                = 100 N/ 0.001
                = 100000 N/m or 100 N/mm
5 0
3 years ago
What do all elements have in common?
Sindrei [870]
The common<span> feature is that the atoms of </span>all elements<span> consist of electrons, protons, and neutrons. Hope this helps!</span>
7 0
3 years ago
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