What kind of energy does a skier have standing still at the top of a hill?

Sonar is a device that uses reflected sound waves to measure underwater depths. If a sonar signal has a wavelength of 5.9 m and

dmitriy555 [2]

The frequency of the signal is = 2.5 Hz

<h3>Calculation of frequency</h3>

The wavelength of the sonar device = 5.9 m

The velocity of the sound = 15 m/s

The frequency = ?

But the formula for the velocity of sound wave = frequency × wavelength

Male frequency the subject of formula,

f = v/ wavelength

f = 15/5.9

f = 2.5Hz

Therefore, the frequency of the signal is = 2.5Hz

Learn more about sound waves here:

brainly.com/question/1199084

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6 0

2 years ago

The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined pl

Scorpion4ik [409]

Given :-

A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
$\mu_s = 0.3 \ \& \ \mu_k = 0.2$

To Find :-

Would the block move ?
If it moves what is its speed after it has descended a distance of 5m down the plane .

Solution :-

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,

$\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)$

Substitute ,

$\longrightarrow \theta_{repose}= tan^{-1}( 0.2)$

Solve ,

$\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}$

Hence ,

$\longrightarrow\theta_{plane}>\theta_{repose}$

Hence the block will slide down .

Now assuming that block is released from the reset , it's initial velocity will be 0m/s .

And the net force will be ,

$\longrightarrow F_n = mgsin53^o - \mu_k N$

Substitute, N = mgcos53⁰ ( see attachment)

$\longrightarrow ma_n = mgsin53^o - \mu_k mgcos53^o$

Take m as common,

$\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)$

Simplify ,

$\longrightarrow a_n = gsin53^o - \mu_k g cos53^o$

Substitute the values of sin , cos and g ,

$\longrightarrow a_n = 10( 0.79 - 0.2 (0.6))$

Simplify ,

$\longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}$

Now using the Third equation of motion namely,

$\longrightarrow2as = v^2-u^2$

Substituting the respective values,

$\longrightarrow2(6.7)(5) = v^2-(0)^2$

Simplify and solve for v ,

$\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}$

Hence the velocity after covering 5m is 8.18 m/s .

7 0

3 years ago

PLEASE ANSWER ASAP! 10 POINTS

nevsk [136]

Answer:

reactants

Explanation:

7 0

3 years ago

Calculate the force of earth's gravity on a 1kg mass at earth's surface. the mass of the earth is 6.0

jonny [76]

Force of earth gravity on 1 kg mass is 9.8 N.

Since the mass is at earth surface, therefore its distance from the centre of earth is equal to the radius of the earth i.e. 6400 km.

The force of earth gravity is calculated as

F=GMm/r^2 =(6.67*〖10〗^(-11)*6*〖10〗^24*1)/〖(64*〖10〗^6)〗^2 =9.8 N

4 0

3 years ago

A car drives to the east in a time of 6 hours. Then, immediately (not realistic, but just assume this is the case for this probl

garri49 [273]

Answer:

Average speed of car in the first trip is 10 km/hr

Explanation:

It is given that first the car drives 6 hours to the east

Then travels 12 km to west in 3 hours

Average speed for the entire trip = 8 km/hr

Total time = 3+6 = 9 hour

So distance traveled in 9 hour = 9×8 = 72 km

As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km

Time by which car traveled in east = 6 hour

So speed $=\frac{distance}{time}=\frac{60}{6}=10km/hr$

So average speed of car in the first trip is 10 km/hr

7 0

4 years ago