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elixir [45]
3 years ago
8

What ocean resource may be harvested by vacuuming or trawling?

Physics
2 answers:
Ksju [112]3 years ago
5 0

Answer:

B. fish

Explanation:

Vacuuming: to suck up something from the ocean. So in this case it would be sucking up whole fish with huge vacuums.

juin [17]3 years ago
3 0

Answer:

B, fish.

Fish may be harvested by vacuuming or trawling.

Add on:

hope this helped at all.

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A translucent object allows light to travel through its material.
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How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w
sergij07 [2.7K]

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω 

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

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Make I^2 the subject of formula

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I = 0.957 A

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4 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
8 0
3 years ago
The current through a 10-ohm resistor connected to a 120-V power supply is The current through a 10-ohm resistor connected to a
Mazyrski [523]

Answer:

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7 0
1 year ago
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swat32

Answer:

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3 0
3 years ago
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