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3 years ago

8

A gas-powered lawnmower is rated at 98 db. how many times more intense is the sound of this mower than that of an electric-power

ed mower rated at 70 db?

1 answer:

givi [52]3 years ago

6 0

The gas- powered mower is

(98dB - 70dB) = 28 dB louder.

28 dB = 10^2.8

= 631 times as much sound power.

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A space vehicle is traveling at 5000 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward

Art [367]

Answer:

the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr

Explanation:

Given:

speed of space vehicle =5000 km/hr

rocket motor speed = 71 km/hr relative to the command module

mass of module = m

mass of motor = 4m

By conservation of linear momentum

Pi = Pf

Pi= initial momentum

Pf= final momentum

Since, the motion is only in single direction

$MV_i=4mV_{mE}+mV_{cE}$

Where M is the mass of the space vehicle which equals the sum of motor's mass and the command's mass, Vi its initial velocity, V_mE is velocity of motor relative to Earth, and V_cE is its velocity of the command relative to Earth.

The velocity of motor relative to Earth equals the velocity of motor relative to command plus the velocity of command relative to Earth.

V_mE = V_mc+V_cE

Where V_mc is the velocity of motor relative to command this yields

$5mV_i = 4m(V_{mc}+V_{cE})+mV_{cE}$

$5V_i = 4V_{mc}+5V_{cE}$

substituting the values we get

$V_{cE} = \frac{5V_i-4V_{mc}}{5}$

$V_{cE} = \frac{5(5000)-4(71)}{5}$

= 4943.2 Km/hr

the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr

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C. you need to double your speed.

since there is gravity and the more mass the more gravity pulls it to the core of the earth; cars are heavy.

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The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

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Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

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A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net ra

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Answer:

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