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NeX [460]
3 years ago
6

The energy a glass has as you are holding it still above a table is

Physics
2 answers:
My name is Ann [436]3 years ago
7 0
 it would be potential energy.
Novosadov [1.4K]3 years ago
5 0
Energy <span>due to position is potential energy</span>
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Explica de que tipo es cada oración según la actitud del hablante
Volgvan

Answer:

Según la actitud del hablante las oraciones se clasifican en enunciativas, interrogativas, etc. ... adverbios o expresiones que complementan a toda la oración (COr): ojalá, quizá.

Explanation:

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2 years ago
A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t
nexus9112 [7]

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m

So, the hayride will cover a distance of 68.18 m.

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3 years ago
What type of energy does a pv cell create ?
Alex_Xolod [135]

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6 0
1 year ago
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Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th
azamat
So the problem are asking to find the value of G base on the formula of the said equation of the magnitude of gravitational attraction on either body. Base on that, the possible answer or the derived formula of the said function is G = Fr^2/m1m2. I hope you are satisfied with my answer and feel free to ask for more 
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3 years ago
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A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
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