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Flura [38]
1 year ago
13

What type of motion would produce positive acceleration in an object? What type of motion would produce negative acceleration?

Physics
1 answer:
Bumek [7]1 year ago
4 0

Answer:

positive acceleration: releasing an object from a height - acceleration due to the force of gravity, the speed of the object will increase as it falls

negative acceleration: applying breaks on the tires to slow down the car - deceleration due to frictional forces

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What is the Morgan-Keenan Spectral Classification system?
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Answer: A method of classifying stars by their temperatures and compositions.

Explanation:

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3 years ago
What is the definition of Cubit
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Answer:

See below.

Explanation:

Cubit is a unit of length based on the length of the forearm from the elbow to the tip of the middle finger and usually equal to about 18 inches (46 centimeters).

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Other similar units of measurements are displayed in the image below. Thanks!

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2 years ago
What type of lens is used to make objects appear smaller and clearer convex or concave?
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2 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}

7 0
3 years ago
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