The efficiency of the steam engine is 78.9%.
<h3>What is the efficiency of a machine?</h3>
The efficiency of a machine is a measure of the useful work done by the machine as against the work input into the machine.
- Efficiency of a machine = work output/work input × 100%
Work output of the steam engine = 225 K
Work input of the steam engine = 285 K
The efficiency of the machine = 225/275 × 100% = 78.9%
Therefore, the efficiency of the steam engine is 78.9%.
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Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
Answer:
I tend to slap thicc trees a#ss
Explanation:
Solar energy - A
nuclear energy - B
fossil fuel energy - C
wind energy - D
geothermal energy - E
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV