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xz_007 [3.2K]
2 years ago
7

You are sitting on the beach and wondering about the properties of mechanical waves. Describe them in terms of ocean waves.

Physics
1 answer:
Rudiy272 years ago
7 0

Answer:

The ocean waves is a mechanical wave that transmits mechanical energy in the wave by the synchronized and repeated oscillation of the waters about an equilibrium level such that as the wave approaches the shoreline, and the water depth decreases, the height of the wave also increases reflecting the effective transmission of energy while the medium which is the water through which the wave propagates, move back and forth within a small region

Explanation:

A mechanical wave like other waves is the oscillation of a field about an equilibrium level. In mechanical waves, the field consists of the oscillating matter such that the wave transmits energy through a medium. The displacement of the medium through which the wave energy is limited such that the wave energy is conserved to travel far.

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
2 years ago
Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time. Express
Inga [223]

Answer:

a=-\dfrac{k}{m}x(t)

Explanation:

From Newton's second law,

F = ma

where F is the force, m is the mass and a is the acceleration.

From Hooke's law,

F = -kx(t)

where k is the spring constant and x(t) is the displacement function measured from the origin. The negative sign indicates the force acts in opposite direction to the displacement. In fact, it is a restoring force; it acts to return the spring to its original undisturbed position.

Since both forces are the same,

F = ma= - kx(t)

a=-\dfrac{k}{m}x(t)

The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.

The ratio \dfrac{k}{m} is a constant except in situations where the mass is varying (say, the mass on the spring is a decaying material).

4 0
2 years ago
"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f
ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c  = 25 ms = 25 × 10⁻³ s

The initial temperature T_1 = 273 K

The final temperature T_2 = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume =  0.4 mV/K × 473 K =  189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2  + (189.2 - 109.2)( 1 - \mathbf{e^{-t/c}})mV

y(t) = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram  attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 ×  189.2 mV

=  170.28 mV

170.28 mV = (109.2 +  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

170.28 mV - 109.2 mV = 80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

61.08 mV =  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

0.7635  mV = ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

t = 1.44 × 25  × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = \dfrac{189.2-170.28  }{189.2}

The error fraction = 0.1

The error fraction = 10%

8 0
2 years ago
You drop a 6.0x10^-2 kg ball from height of 1.0m above hard flat surface. Ball strikes surface and energy decreases by 0.14J, th
harina [27]
If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45. 
3 0
2 years ago
Read 2 more answers
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7 

The formula that is applicable from physics is 
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes 
7 0
2 years ago
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