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satela [25.4K]
3 years ago
9

If the potential energy of a particle at some point in space is zero, must the force on it also be zero at that point?

Physics
1 answer:
sweet-ann [11.9K]3 years ago
3 0
Answer is no. Force has nothing to do with gravitational potential energy.
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A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height.
aleksandrvk [35]

Answer:

  • 430 J
  • 6.26 m/s

Explanation:

A. The kinetic energy is the same as the initial potential energy:

  PE = mgh = (215 N)(2.0 M) = 430 J

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B. The velocity achieved by falling from a height h is given by ...

  v = √(2gh)

  v =  √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)

  v ≈ 6.26 m/s

8 0
3 years ago
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Ivan

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6 0
3 years ago
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A race car makes three and half laps around a circular track in 4.5 min. What is the car’s angular speed?
slavikrds [6]

Answer:

the angular speed is around 45

Explanation:

7 0
3 years ago
Can density be used to identify what a substance is made of?
Korolek [52]
Possibly, if you have list of densities and you have to match it. I can't think of any other scenarios in which it would be able to.

Hope I helped! :)
7 0
3 years ago
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
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