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Arlecino [84]
3 years ago
9

A(n) 15.9 kg rock is on the edge of a(n

Physics
1 answer:
Mariulka [41]3 years ago
5 0

The potential energy (P.E) of the rock relative to the base of the cliff is 31,054.926 Joules..

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

  • Mass of rock = 7.5 kg
  • Height = 0.7 m
  • Acceleration due to gravity = 9.8 m/s^2

To calculate the potential energy (P.E) of the rock relative to the base of the cliff:

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

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

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

P.E = mgh

<u>Where:</u>

  • P.E is the 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;

P.E = 15.9 \times 9.8 \times 199.3

P.E = 31,054.926 Joules.

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

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You observe a light ray move from one piece of glass to another (a different type of glass) and the light ray bends towards the
kodGreya [7K]

Answer: C

Explanation: When light rays moves from one medium to another with a change in its direction (bending towards media interface), it is called refraction.

The angle the ray in the second medium (refracted ray) makes with the medium interface (normal) explains the bending of light and is dependent on the following the refractive index (n), wave speed in the medium (v) and other properties such as wavelength, and angle of incidence.

This question is focused on the relationship between refractive index and wave speed.

Refractive index (n) is inversely proportional to wave speed (v). This implies that a ray of light moving from a dense medium (say air) to a more dense medium (say glass) has it wave speed decreased and if reversed ( from glass to air) the wave speed increases.

A change in refractive index also affects the bending of the refracted ray.

A move from a dense to a more dense medium makes the refracted ray move towards the normal thus decreasing the angle of refraction (angle the refracted ray makes with the normal)

So for our question, since light ray (refracted ray) moves towards the glass to glass interface (normal) it means that light ray had moved from a dense to a more dense medium (that is glass 2 has higher refractive index than glass one) and the wave speed will decrease since there is an increase in refractive index (that is light travels faster in glass 1 than glass 2)

3 0
3 years ago
Write this number in standard notation. 4.702 x 10–4
sergeinik [125]

Answer:

0.0004702

Explanation:

5 0
3 years ago
determine the quantity of work done when a crane lifts a 100-n block from 2m above the ground to 6m above the ground
Mekhanik [1.2K]
100n divided by 8m because 2m and 6m are like terms. Hope that helps:)
6 0
3 years ago
Read 2 more answers
point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is
Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, \rho_{w} = 1000\ kg//m^{3}

Acceleration due to gravity, g = 9.8\ m/s^{2}

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = \frac{H + D}{2}

h = \frac{3 + 2.5}{2} = 2.75\ m

Volume of water in the pool, V = \pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}

Mass of water, m_{w} = \frac{\rho_{w}}{V}

m_{w} = \frac{1000}{311.02} = 3.215\ kg

Work done is given by the potential energy of the water as:

W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done,W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J

7 0
2 years ago
The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm
lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: n_0 = 1.52

Wave length of light = λ = 570 nm = 570\times10^{-9}\ m

\text{ Thickness}=\dfrac{\lambda}{4n}

=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0
3 years ago
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