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klemol [59]
3 years ago
5

What will happen to a periodic wave acted upon an external damping force?

Physics
2 answers:
Svetllana [295]3 years ago
6 0

It'll decrease the frequency
OverLord2011 [107]3 years ago
4 0
The wave will decrease in its frequency due to a disturbing force acting upon it.
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Regardless of their frequency, wavelength, or energy, all electromagnetic waves: A. travel only through the vacuum of space. B.
agasfer [191]
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too. 
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3 years ago
What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c
Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

   = 413V / 0.0795 m

    = 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

6 0
3 years ago
. A person weighing 750 N gets on an elevator.
Kobotan [32]

 

F = 750 N  (Force)

d = 10 m  (displacement )

t = 25 s   (time)

L = ?   (Mechanical work )  =  (Energy)

P = ?   (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

5 0
2 years ago
The weight of an apple at 4r from earth is of that on earth
iris [78.8K]
W = m.g = weight
g = Gme/Re**2 where G= universal gravitational constant , Re= radius of the earth
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4 0
3 years ago
Read 2 more answers
An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro
kupik [55]

Answer:

a) Q_{in} = 13.742\,kW, b) \Delta S = 370.15\,\frac{kJ}{K}

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

Q_{in} +U_{sys,1} - U_{sys,2} = 0

Q_{in} = U_{sys,2} - U_{sys,1}

Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})

Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)

Q_{in} = 13.742\,kW

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

\Delta S = \frac{Q_{in}}{T_{in}}

\Delta S = \frac{13.742\,kJ}{370.15\,K}

\Delta S = 370.15\,\frac{kJ}{K}

3 0
3 years ago
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