Please i need help what happens to water once it reaches the earths surface!!!??????

Explain the conditions through which friction can be increased.

Rufina [12.5K]

Answer:

Friction can be increased the longer two objects are rubbed together, since the longer they are rubbed is the more heat produced.

Explanation:

8 0

3 years ago

Read 2 more answers

A powerful motorcycle can accelerate from rest to 24.8 m/s (55 mi/h) in only 5.90 s. (a) What is its average acceleration in m/s

mel-nik [20]

Answer

given,

initial speed,u = 0 m/s

final speed,v = 24.8 m/s

time, t= 5.9 s

a) acceleration,

$a=\dfrac{v-u}{t}$

$a=\dfrac{24.8 - 0}{5.9}$

a = 4.20 m/s²

b) distance to travel in that time

using equation of motion

v² = u² + 2 as

24.8² = 0² + 2 x 4.20 x s

8.4 s = 615.04

s = 73.22 m

the motorcycle will travel 73.22 m

6 0

4 years ago

An object moving with constant acceleration near the surface of the earth

ziro4ka [17]

The answer is C because if it’s moving with a constant speed it’s gonna be balanced and there will be no force

6 0

3 years ago

Suppose Person A runs off the edge of the cliff at 2 m/s and Person B runs off the edge at 1 m/s. Which will hit the ground fart

djverab [1.8K]

Answer:

<h2>Person A will hit a distance father</h2>

Explanation:

Based on the fact that the velocity of person A is more than that of person B, that is from the question, person A has a velocity of 2m/s and person B has a velocity of 1m/s, this clearly shows that person A has the tendency to hit a distance farther from the cliff than person B.

3 0

4 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

5 0

3 years ago