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Vladimir [108]
3 years ago
15

What type of energy transfer occurs through light or electromagnetic waves? A) conduction B) radiation C) induction

Physics
2 answers:
puteri [66]3 years ago
6 0
A) conduction should be the right answer
Lunna [17]3 years ago
4 0

A) Conduction should be the answer Your welcome :)

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According to the Theory of Continental Drift, what did the Earth’s continents look like 255 million years ago? What is the name
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Theres an image of how this supercontinet looked

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_______ occurs when an object travels in a cruved path
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Circular motion occurs when an object travels in a curved path.
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A net external nonconservative force does positive work on
Sergio039 [100]

Answer with Explanation:

From work energy theorem we have

\Delta Energy=\int F.ds

Now since it is given that the work done on the object by the force is positive hence we conclude that the term on the right hand of the above relation is positive hence \Delta Energy>0

Part a)

Hence we conclude that the mechanical energy of the particle will increase.

Part b) Since the mechanical energy of the particle is the sum of it's kinetic and potential energies, we can write

\Delta K.E+\Delta P.E=\Delta E\\\\\therefore \Delta K.E+\Delta P.E>0

Now since the sum of the 2 energies ( Kinetic and potential ) is positive we cannot be conclusive of the individual values since 2 cases may arise:

1) Kinetic energy increases while as potential energy decreases.

2)  Potential energy increases while as kinetic energy decreases.

Hence these two cases are possible and we cannot find using only the given information which case will hold

Hence no conclusion can be formed regarding the individual energies.

8 0
3 years ago
A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 40
11111nata11111 [884]

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ = \frac{\sqrt{4km - y^2} }{2m}

= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}

(c)

Quasi period:

T = 2π / μ

T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6}  }{12}

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

8 0
3 years ago
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