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3 years ago

13

Which natural hazards are most likely to affect Florida, due to its geography? Check all that apply. PLS HELP! I WILL GIVE BEST

ANSWER BRAINLIEST! 15 POINT EACH! PLS HELP!

1 answer:

zhenek [66]3 years ago

6 0

Answer:

drought, floods, rip currents, tropical cyclones, wildfires

Explanation:

right on edge

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What are the two factors that affect the period of a pendulum?

klemol [59]

The two factors that affect the period of a pendulum are the length of the string and the distance in which the pendulum falls.

Hope this helps! I would greatly appreciate a brainliest! :)

3 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

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3 years ago

Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

Pachacha [2.7K]

The gravitational force between two objects is given by
$F=G \frac{m_1 m_2 }{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, $m_1 = 175 kg$ , $m_2 =15 kg$ and $r=2 m$ , therefore the gravitational force between the two objects is
$F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N$

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3 years ago

Why is the perfect mechine is not possible in the real life?

Vinvika [58]

Answer:

Because the mechanical advantage of the machine is affected by friction and weight but velocity ratio is not. So, mechanical advantage is less than velocity rate. Thus, the machine's efficiency is less than 100% and can't be a perfect machine

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3 years ago

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drif

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uhhhh idk cheif...

thats a big oof right there.

ged is always an option

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4 years ago