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11111nata11111 [884]
4 years ago
7

A ball is thrown vertically into the air with a initial velocity of 20 m/s. Find the maximum height of the ball and find the amo

unt of time needed to reach the maximum height.
Physics
1 answer:
REY [17]4 years ago
8 0

Answer:

The maximum height of the ball is 20 m. The ball needs 2 s to reach that height.

Explanation:

The equation that describes the height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration

v = velocity at time t

When the ball is at its maximum height, its velocity is 0, then, using the equation of the velocity, we can calculate the time at which the ball is at its max-height.

v = v0 + g · t

0 = 20 m/s - 9.8 m/s² · t

-20 m/s / -9.8 m/s² = t

t = 2.0 s

Then, the ball reaches its maximum height in 2 s.

Now,  we can calculate the max-height obtaining the position at time t = 2.0 s:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 20 m/s · 2 s - 1/2 · 9,8 m/s² · (2 s)²

y = 20 m

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Marina CMI [18]

Answer:

The answer is option B, horn.

Explanation:

The landform horn was formed because when glaciers erode down to form a sharp peak which is called a horn. All of the other options are false because none of those land forms are formed because of the erosion of glaciers

4 0
3 years ago
A 0.0223 m diameter coin rolls up a 12.0◦ inclined plane. The coin starts with an initial angular speed of 49.3 rad/s and rolls
zlopas [31]

Answer:

h = 0.0231 m

Explanation:

The movement of the coin is modelled after the Principle of Energy Conservation. The kinetic energy of the coin is the sum of the components associated with translation and rotation and there are no non-conservative forces. In addition, the coin starts moving at height of zero

K_{rot} + K_{tr} = U_{g}

\frac{1}{2} \cdot m_{coin} \cdot \omega^{2} \cdot R^{2} + \frac{1}{4} \cdot m_{coin} \cdot R^{2} \cdot \omega^{2} = m_{coin} \cdot g \cdot h\\

\frac{3}{4} \cdot R^{2} \cdot \omega^{2} = g \cdot h

The maximum vertical height is isolated in the previous equation:

h = \frac{3 \cdot R^{2}\cdot \omega^{2}}{4\cdot g}

h = \frac{3 \cdot (0.01115 m)^{2} \cdot (49.3 \frac{rad}{s} )^{2}}{4 \cdot (9.807 \frac{m}{s^{2}} )} \\h = 0.0231 m

5 0
4 years ago
How should you approach a dock when the wind or current is pushing you away from the dock?
erastova [34]

Answer:

If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.

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3 years ago
PLease Help!
scoundrel [369]

Answer:

Option A is the correct answer.

Explanation:

The instantaneous acceleration = Change in velocity in velocity/Time taken

The slope of the graph should give instantaneous acceleration.

 Slope of a graph = Change in value of Y -axis / Change in values of X -axis

 Comparing both the equations

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