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3 years ago

Please please help i have to turn in

Physics

1 answer:

larisa [96]3 years ago

6 0

Hi,

I've found a link that should assist you or answer your question.
http://click.dji.com/ANbvbbP7bwUWtSACp6U_?pm=link&as=0004

Have a nice day!

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A ball is being thrown straight upward and rises to a maximum height of 12.0m above its launch point. At what height above it’s

AlexFokin [52]

Answer:

9.0 m

Explanation:

Let the initial velocity be 'u'.

Given:

Final velocity is half of initial velocity.

Maximum height reached by ball (H) = 12.0 m

Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)

Now, first, we will find initial velocity of the ball using equation of motion given as:

$v^2=u^2+2a(\Delta y)$

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, $v=0\ m/s$

Also, $\Delta y=H=12\ m$

Now, plug in all the values and solve for 'u'.

$0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s$

Now, consider the motion of the ball till the velocity reaches half of initial velocity.

So, final velocity (v) = $\frac{u}{2}=\frac{15.34}{2}=7.67\ m/s$

Now, again using the same equation and finding the new height now. Let the new height be 'h'.

So, equation of motion is given as:

$v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m$

Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.

8 0

3 years ago

How many complete wave cycles are shown?

skelet666 [1.2K]

Answer:

2.0

Explanation:

8 0

4 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is: