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mixer [17]
3 years ago
11

A projectile is launched at ground level with an initial speed of 53.0 m/s at an angle of 35.0° above the horizontal. It strikes

a target above the ground 2.50 seconds later. What are the x and y distances from where the projectile was launched to where it lands?
Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

x = 108.5 m      y = 45.37 m,  

Explanation:

This is a projectile launching exercise, the horizontal distance it reaches is called the range and can be found by the expression

           x = v₀ₓ t

           

the initial hairiness is

          v₀ₓ = v₀ cos θ

           

           x = v₀ cos θ  t

  let's calculate

           x = 53 cos 35 2.50

           x = 108.5 m

the height of the projectile can be calculated with

            y =v₀ t - ½ g t²

             y = vo sin tea t - ½ g t²

            y = 53 sin 35 2.50 - ½ 9.8 2.5²

            y = 45.37 m

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Anni [7]

Answer:

I=38.181\ A is the current through the body of the man.

E=34.5\ J energy dissipated.

Explanation:

Given:

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<u>Now, from the Ohm's law we have:</u>

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I=\frac{21000}{550}

I=38.181\ A is the current through the body of the man.

<u>Energy dissipated in the body:</u>

E=I^2.R.t

E=38.181^2\times 550\times 43\times 10^{-6}

E=34.5\ J

5 0
4 years ago
which chemical formula shows how atoms of the elements calcium (ca) and sulfur (s) would join together ionically? a.cas b.ca2s c
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Calcium ions have a +2 charge and sulfur ions have a -2 charge, so the formula is CaS
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Three perfectly polarizing sheets are spaced 2?cm {\rm cm} apart and in parallel planes. The transmission axis of the second she
Agata [3.3K]

Answer:

I₃/Io % = 0.8.59

Explanation:

A polarizer is a complaint sheet for light in the polarization direction and blocks the perpendicular one. When we use two polarizers the transmission between them is described by Malus's law

      I = I₀ cos² θ

Let's apply the previous exposures in our case, the light is indicatively not polarized, so the first polarized lets half of the light pass

      I₁ = ½ I₀

The light transmitted by the second polarizer

     I₂ = I₁ cos²  θ

     I₂ = (½ I₀) cos2 28

The transmission by the polarizing  third is

     I₃ = I₂ cos²  θ₃    

The angle of the third polarizer with respect to the second is

     θ₃ = 90-28

      θ₃ = 62º

I₃ = (½ I₀ cos² 28 cos² 62)

Let's calculate

I₃ = Io ½ 0.7796 0.2204

I₃ = Io 0.0859

I₃/Io= 0.0859    100

I₃/Io % = 0.8.59

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4 years ago
a Ferrari with an initial velocity of 10 m/s, comes to a complete stop in 5 seconds, what will its acceleration be?
11111nata11111 [884]

10 m/s divided by 5 s

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6 0
4 years ago
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The pilot of an aircraft wishes to fly due west in a wind blowing at 45 km/h toward the south. If the speed of the aircraft in t
DiKsa [7]

Answer:

164.2°

Explanation:

speed of wind (w) = 45 km/h due south

speed of aircraft (a) = 165 km/h

in what direction (in degrees) should the aircraft head in so as to fly due west?

To get the direction the pilot should fly, we can form a triangle with the data available where

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  • the direction the pilot would have to fly so he can end up at he west serves as the hypothenuse
  • θ is the angle between the direction the pilot would have to fly and the direction the pilot wishes to fly.
  • the direction the pilot wishes to fly ( west) will serve as the adjacent side
  • all this can be seen from the attached diagram.

now sin θ = \frac{speed of the wind}{speed of the aircraft}

sin θ = \frac{45}{165}

θ = sin^{-1} 0.2728

θ = 15.8°

since we are to use the counter-clockwise from east convention our measurement would have to be taken anticlockwise from the east direction, therefore the direction of the aircraft (Ф) = 180-15.8 = 164.2°

3 0
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