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Vadim26 [7]
3 years ago
12

A ball of mass 12 kg is released from a height of 2.5 m

Physics
1 answer:
VMariaS [17]3 years ago
4 0

Answer:

300 J

Explanation:

From conservation of energy;

Gravitational potential energy at top = Kinetic Energy at bottom

mgh = ½mv²

We are given;

Mass; m = 12 kg.

Height; h = 2.5 m

g = 10 m/s²

Thus;

mgh = 12 × 2.5 × 10 = 300 J

Since mgh = Kinetic Energy at bottom.

Thus; kinetic energy of the ball once it reach the ground = 300 J

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What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?
ivann1987 [24]

Answer:

2.46\cdot 10^5 J

Explanation:

The energy of a single photon is given by:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

For the photon in this problem,

\lambda=486 nm=4.86\cdot 10^{-7}m

So, its energy is

E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J

One mole of photons contains a number of photons equal to Avogadro number:

N_A = 6.022\cdot 10^{23}

So, the total energy of one mole of photons is

E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J

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3 years ago
Which statement best reflects a change in climate? It's very windy today and will rain later. Droughts in California have become
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The first one it's very windy today and will rain later.
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Activity that uses 150 calories of energy per day, or 1,000 calories per week, describes ___________.
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4 years ago
A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
Llana [10]

Answer:

a) C = 40.138\,pF, b) q = 16.056\,nC, c) U = 3.212\,\mu J

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

C = K\cdot \epsilon_{o}\cdot \frac{A}{d}

Where:

K - Dielectric constant.

\epsilon_{o} - Vaccum permitivity.

A - Plate area.

d - Distance between plates.

Hence, the capacitance of the system is:

C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)

C = 4.014\cdot 10^{-11}\,F

C = 40.138\,pF

b) The charge can be found by using the definition of capacitance:

q = C\cdot V_{batt}

q = (4.014\times 10^{-11}\,F)\cdot (400\,V)

q = 1.606\times 10^{-8}\,C

q = 16.056\,nC

c) The energy stored in the charged capacitor is:

U=\frac{1}{2}\cdot Q\cdot V_{batt}

U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)

U = 3.212\times 10^{-6}\,J

U = 3.212\,\mu J

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Answer:

Explanation:

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