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kramer
3 years ago
11

A rocket is launched from a height of 3 m with an initial velocity of 15 m/s What is the maximum height of the rocket? When will

this occur?
​
Physics
1 answer:
Fudgin [204]3 years ago
5 0

If no extra acceleration is added to the rocket, then its velocity at time <em>t</em> is

<em>v</em> = 15 m/s - <em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Also, recall that

<em>v</em>² - <em>u</em>² = 2 <em>a </em>∆<em>x</em>

where <em>u</em> is initial speed, <em>v</em> is final speed, <em>a</em> is acceleration, and ∆<em>x</em> is net displacement.

At the rocket's maximum height ∆<em>x</em>, the velocity is 0. So, the maximum height is

0² - (15 m/s)² = 2 (-<em>g</em>) ∆<em>x</em>

∆<em>x</em> = (15 m/s)² / (2 * (9.80 m/s²)) ≈ 11.48 m

But this assumes the rocket is launched from the ground. We're given that the rocket is launced from 3 m above the ground, so we need to add this to the height above. So the maximum height is closer to 14.48 m.

As mentioned before, this happens when vertical velocity is 0:

0 = 15 m/s - <em>g t</em>

<em>t</em> = (15 m/s) / (9.80 m/s²) ≈ 1.53 s

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Answer:

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the observation is that there is a change of visible color

Explanation:

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refraction n = 1.33

wall thickness (t) = 290 nm

2nt = (2m +1 ) ∝/2 -----equation 1

note when m = 0

therefore ∝ =  4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this

when m = 1

equation 1 becomes

∝ = 4nt/3 =( 4 * 1.33 * 290) /  3 = 1542.8 / 3 = 514.27 ( wavelength )

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B) the wavelength when the wall thickness is 340 nm

∝ = 4nt / 2m +1

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the observation is that there is a change of visible color

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3 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

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Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

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        t = √ (10 / 4.9)

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Now let's use the first equation and the last one

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         0  = v₀ sin θ - 9.8 t

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         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

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          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

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thisis the correct

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zaharov [31]

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A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC, is on the x axis at x=0.500 m.

No, there is no point at a finite distance where the electric potential is zero.

Hence, Option D) is correct.

What is electric potential?

Electric potential is the capacity for doing work. In the electrical case, a charge will exert a force on some other charge and the potential energy arises. For example, if a positive charge Q is fixed at some point in space, any other positive charge when brought close to it will experience a repulsive force and will therefore have potential energy.

It is also defined as the amount of work required to move a unit charge from a reference point to a specific point against an electric field.

To learn more about electric potential, refer to:

brainly.com/question/15764612

#SPJ4

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