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3 years ago

PART ONE

Physics

1 answer:

Lina20 [59]3 years ago

4 0

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass. Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

$\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right$

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

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3. The velocity of sound is 332 m/s. Answer the following questions:

Angelina_Jolie [31]

Answer:

20 Hz, 20000 Hz

0.0166 m, 16.6 m

Explanation:

The minimum frequency that a human ear can hear is 20 Hz

The maximum frequency that a human ear can hear is 20000 Hz.

v = Velocity of sound = 332 m/s

Wavelength is given by

$\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{332}{20}\\\Rightarrow \lambda=16.6\ \text{m}$

The longest wavelength that can be heard by the human ear is 16.6 m

$\lambda=\dfrac{332}{20000}\\\Rightarrow \lambda=0.0166\ \text{m}$

The shortest wavelength that can be heard by the human ear is 0.0166 m.

3 0

2 years ago

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

2 years ago

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2