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aalyn [17]
3 years ago
8

A ball is dropped from the roof of a building. the mass of the ball is 3.0 kg. what is the potential energy of the ball in the i

nstant that it is exactly 18.0 meters above the ground?
Physics
1 answer:
vitfil [10]3 years ago
6 0
According to given condition there is no height(m) given from roof of building to the ground, there height given 18 m at a point above the ground.                         So, h=18m  ,                mass=3kg     ,         g=9.8m/s2                                                                      P.E=mgh                                                                                                                P.E=(3)(9.8)(18)                                                                                                      P.E=529J
                                      
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a bird flies 25.0 m in the direction 55° east of south to its nest. the bird then flies 75.0 m in the direction 55° west of nort
son4ous [18]

The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.

<h3>Displacement of the bird</h3>

The displacement of the bird is the change in the position of the bird.

<h3>Vertical component of the bird's displacement </h3>

Vy₁ = -25 m  x   sin(55)

Vy₁ = -20.48 m

Vy₂ = 75 m   x    sin(55)

Vy₂ = 61.44 m

Total vertical displacement = 61.44 m - 20.48 m = 40.96 m

<h3>Horizontal component of the bird's displacement </h3>

Vx₁ = -25 m  x   cos(55)

Vx₁ = -14.34 m

Vx₂ = 75 m   x    cos(55)

Vx₂ = 43.02 m

Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m

Learn more about displacement here: brainly.com/question/2109763

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8 0
1 year ago
Define 1 pascal pressure​
Rasek [7]

Answer:For example, standard atmospheric pressure (or 1 atm) is defined as 101.325 kPa. The millibar, a unit of air pressure often used in meteorology, is equal to 100 Pa. (For comparison, one pound per square inch equals 6.895 kPa.)

Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.

I hope this helps.... I'm sorry if it doesn't

5 0
2 years ago
What process is mostly responsible for the blue appearance of the sky
malfutka [58]

Answer:reflection by dust particles in air

7 0
3 years ago
A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

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4 0
1 year ago
Which type of wave is not a light wave
AlekseyPX

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4 0
3 years ago
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