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Anastasy [175]
2 years ago
9

The wavelength is 1. 87 x 10-7 m. What is the frequency?.

Physics
1 answer:
jeka57 [31]2 years ago
3 0

Answer:

11.7 im pretty sure. i did the work so it should be correct

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What would be the resulting condition if Earth's axis was perpendicular to the Sun?
blagie [28]

Answer:

I belive it would be B or D, but B seems more likely

Explanation:

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Freight trains can produce only relatively small acceleration and decelerations. (a) what is the final velocity (in m/s) of a fr
m_a_m_a [10]
(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s 
Acceleration a = 0.0600 m/s² 
time = 8 min = 8 x 60 = 480 s 
Final velocity 
= u + at 
= 5.00 + 0.0600(480) 
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The final velocity is 33.8 m/s
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Use a(n) ________ near water sources in the kitchen or bathroom for electrical safety.
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I think it would be GFI outlet

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3 years ago
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

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