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DedPeter [7]
3 years ago
5

What is the biggest current disadvantage to using gas hydrates as a form of energy?

Physics
2 answers:
ohaa [14]3 years ago
6 0
When gas hydrates are brought to the surface, they evaporate quickly. This would be the biggest, current disadvantage to using gas hydrates as a form of energy.
Rama09 [41]3 years ago
6 0
ANSWER)
they evaporate quickly when brought to the surface :-()
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Classify the type of plate boundary where the appalachian mountains formed . how have they changed since their formation
disa [49]
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6 0
3 years ago
S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to
Gre4nikov [31]

By calculation, the diameter of the wire  is 2.8 * 10^-3 m.

<h3>How do we obtain the length?</h3>

The following data are given in the question;

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

Area of the wire = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

6 0
1 year ago
A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?
il63 [147K]

Answer:

\huge\boxed{\sf P.E. = 39.2\ Joules}

Explanation:

<u>Given Data:</u>

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

<u>Required:</u>

Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = mgh

<u>Solution:</u>

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
5 0
3 years ago
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