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3 years ago

7. Calculate the gravitational potential energy of a 10 kg box on the top shelf of a closet, 2 m above the floor.

Physics

1 answer:

snow_lady [41]3 years ago

5 0

Answer:

P.E = 196 J

Explanation:

Given,

The mass of the box, m = 10 Kg

The height of the box from the ground, h = 2 m

The acceleration due to gravity, a = 9.8 m/s²

The gravitational potential energy of the box is given by the formula

P.E = mgh joules

Substituting the given values in the above equation,

P.E = 10 x 2 x 9.8

= 196 J

Hence, the gravitational potential energy of box, P.E = 196 J

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You have a piece of cork with a volume of 2 cm^3 and a density of 210kg/m^3. You hold it under water and release it.

insens350 [35]

I am sorry if it didn't helped

answers;

Calculate the buoyant force of a piece of cork of 8cm3 that floats in water. Density of cork is 207kg/m3. ?

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Explanation:

From Archimedes Principle, any object partially or totally submerged in a fluid is buoyed upwards with a force equal to the weight of the displaced fluid.

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1000

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3 years ago

Describe seven physical properties that help distin-<br> guish one mineral from another.

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Answer:

Color, Streak, luster, cleavage and fracture, hardness, crystal shape, and density.

Explanation:

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3 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

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Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

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3 years ago

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Answer:

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Explanation:

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