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emmainna [20.7K]
3 years ago
8

A lamp hangs from the ceiling at a height of 2.6 m. The lamp has a mass of 3.8 kg. The screws holding the lamp break, and it fal

ls to the ground. a. How much gravitational potential energy does the lamp have before it falls? b. How much kinetic energy does the lamp have when it reaches the ground? c. How fast is the lamp moving when it hits the ground?
Whoever answers will get the Brainliest.
Physics
1 answer:
iVinArrow [24]3 years ago
4 0

Answer:

Explanation:

Given height of lamp from the ceiling = 2.6m

mass of the lamp = 3.8kg

acceleration due to gravity = 9.81m/s²

As the body falls to the ground, it falls under the influence of gravity.

Gravitational potential energy = mass*acc due to gravity * height

Gravitational potential energy  = 3.8*2.6*9.81

Gravitational potential energy  = 96.923 Joules

b) Kinetic energy = 1/2 mv²

m = mass of the body (in kg)

v = velocity of the body (in m/s²)

To get the velocity v, we will use the equation of motion v^{2} = u^{2}+2gh

v^{2} = 0^{2}+2(9.81)(2.6) \\v^{2} = 51.012\\v =\sqrt{51.012}\\ v = 7.14m/s

Since mass = 3.8kg

K.E =  1/2 * 3.8 *7.14^{2}\\ K.E = 96.86Joules

c) To know how fast the lamp is moving when it hits the ground, we will use the formula. When the body hits the ground, the height covered will be 0m. this means that the body is not moving once it hits the ground. It stays in one position. The energy possessed by the body at this point is potential energy. The correct answer is therefore 0 m/s

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Answer:

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Explanation:

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Given,

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Ep= 2 × 9.81 × 7

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
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Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

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How did planck find the correct curve for the specturm of light emitted by a hot obkect?
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Planck find the correct curve for the specturm of light emitted by a hot object by vibrational energies of the atomic resonators were quantized.

<h3>Briefing :</h3>
  • The energy density of a black body between λ and λ + dλ is the energy E=hc/λ of a mode times the density of states for photons, times the probability that the mode is occupied. 
  • This is Planck's renowned equation for a black body's energy density.
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Learn more about the Planck's constant with the help of the given link:

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Answer:

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