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2 years ago

10

When the volcano Krakatoa erupted in 1883, it was heard 5000 km away. Which statement about the sound from the volcano is not co

rrect?
a. If such a loud sound were to be made today, an astronaut orbiting in space (a vacuum) at a height of 400 km could hear it.
b. People further from the volcano heard the sound later than people nearer to the volcano.
c. The amplitude of the sound waves would have been smaller further from the volcano.
d. The sound was very loud because a lot of energy was transferred to vibrations of the air.

1 answer:

lina2011 [118]2 years ago

7 0

Answer a) is incorrect as sound does not travel in a vacuum.

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The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = $\frac{excess initial cost}{\Delta M}$

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2 years ago

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) <u>On doubling the mass:</u>

New mass, $m'=2m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

7 0

1 year ago

In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being

german

Answer: Got It!

<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.

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2 years ago

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the

hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as $x(t)=Asin(\omega t+\phi)$

'A' is the amplitude = 6 inches (given)

$\omega =\sqrt{\frac{k}{m}}$ is the natural frequency of the system

At equilibrium we have

$mg=kx\\\\k=\frac{mg}{x}$

Applying values we get

$k=40 lb/ft$

thus natural frequency equals

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Thus the equation of motion becomes

$x(t)=6sin(8t+\phi)$

At time t=0 since mass is at it's maximum position thus we have

$A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})$

Thus the position of mass at the given times is as follows

1) at $\frac{\pi}{12}$ $x(t)=5.99inches$

2) at $\frac{\pi}{8}$ $x(t)=5.9909inches$

3) at $\frac{\pi}{6}$ $x(t)=5.98397inches$

4) at $\frac{\pi}{4}$ $x(t)=5.9639inches$

5) at $\frac{9\pi}{32}$ $x(t)=5.954inches$

4 0

1 year ago

A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di

Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

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Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

$\int (80 - (40/3)t) dt = 80(v_f - 12)$

$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$

6 0

1 year ago