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GrogVix [38]
3 years ago
13

Gas hydrates are used ___

Physics
2 answers:
cupoosta [38]3 years ago
7 0
For a source of energy 

murzikaleks [220]3 years ago
4 0
For a source of energy
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A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capac
Pavlova-9 [17]

Answer:

k =  6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance  due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

7 0
2 years ago
a water-balloon launcher with mass 4 kg fires a 0.5 kg balloon with a velocity of 3 m/s to the east. what is the recoil velocity
kotykmax [81]
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 =  4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s

Therefore; 4 V1 = 0.5 × 3
                   4V1= 1.5
                     V1= 1.5/4
                          = 0.375 m/s










5 0
3 years ago
Read 2 more answers
What is the work done by the friction when the body slides against a rough horizontal surfaces?
katovenus [111]

Negative work

Hope this helps :)

4 0
2 years ago
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

8 0
2 years ago
oe finds that the temperature of a substance is 12 degrees Celsius. What does this tell Zoe about the substance? Its internal en
Roman55 [17]

Its internal energy is less than 12 degrees

5 0
3 years ago
Read 2 more answers
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