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andre [41]
2 years ago
14

Suppose a ceiling fan has a mass of 7.5 kg and is 9.7 m above the ground. What is the gravitational potential energy of the ceil

ing fan?
Physics
1 answer:
EleoNora [17]2 years ago
8 0

The gravitational potential energy (G.P.E) of the ceiling fan is 712.95 Joules.

<u>Given the following data:</u>

  • Mass of ceiling fan = 7.5 kg
  • Height = 0.7 m

<u>Scientific data:</u>

  • Acceleration due to gravity = 9.8 m/s^2

To calculate the gravitational potential energy (G.P.E) of the ceiling fan:

<h3>What is gravitational potential energy?</h3>

Gravitational potential energy (G.P.E) can be defined as the energy that is possessed by an object or body due to its position (height) above planet Earth.

Mathematically, gravitational potential energy (G.P.E) is given by this formula;

GPE = mgh

<u>Where:</u>

  • G.P.E is the gravitational potential energy.
  • m is the mass of an object.
  • g is the acceleration due to gravity.
  • h is the height of an object.

Substituting the given parameters into the formula, we have;

GPE = 7.5 \times 9.8 \times 9.7

GPE = 712.95 Joules.

Read more on potential energy here: brainly.com/question/8664733

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How to solve arctan[tan(7pi /4)] ...?
Svet_ta [14]
Well, 
arctan is a bijection from R into (-pi/2 , pi/2)*and 
pi is a period of tangent function: 
so 
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4) 
we finally get : 

<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4 
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
7 0
3 years ago
The period of the wave is
Karo-lina-s [1.5K]

Answer:

2.5 s

Explanation:

3 0
3 years ago
before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.
Virty [35]

The momentum of block B after the collision is -50 kg m/s.

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

p_A + p_B = p'_A + p'_B

where:

p_A = -100 kg m/s is the momentum of block A before the collision

p_B = -150 kg m/s is the momentum of block B before the collision

p'_A = -200 kg m/s is the momentum of block A after the collision

p'_B is the momentum of block B after the collision

Solving for p'_B, we find:

p'_B = p_A + p_B - p'_A = -100 +(-150) -  (-200)=-50 kg m/s

So, the momentum of block B after the collision is -50 kg m/s.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

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2 years ago
Is a measure of how rapidly or slowly molecules move?
horsena [70]
Temperature is a measure of how rapidly or slowly molecules move.<span />
8 0
3 years ago
Read 2 more answers
Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol
Tju [1.3M]

Answer:

R \approx 2.418\times 10^{9}\,km

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

T = \frac{2\pi}{\omega}

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

\omega = \sqrt{\frac{a_{r}}{R} }

Ahora se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }

La aceleración experimentada por el planeta es:

a_{r} = G\cdot \frac{M_{sun}}{R^{2}}

Se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }

La distancia del planeta con respecto al sol es finalmente despejada:

R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}

R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}

Finalmente, se sustituyen las variables y se determina la distancia:

R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}

R \approx 2.418\times 10^{12}\,m

R \approx 2.418\times 10^{9}\,km

4 0
3 years ago
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