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3 years ago

What is the most important source of water vapor in the atmosphere?

1 answer:

8 0

The answer is Oceans. It is the most important source of water vapor in the atmosphere. <span>Given the huge amount of water they have and their huge surface areas, naturally the bulk of water which evaporates and enters the atmosphere is from oceans.</span>

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The unit of work is a derived unit prove that<br>

schepotkina [342]

Answer:

Explanation:

because I know the answer.

7 0

3 years ago

A solenoid coil 10cm long has 40 turns cm-1. The current rises from zero to its maximum value of 5.0000 A and 0000.01s when the

Crank

Answer:

Explanation:

Energy stored in a solenoid per unit volume = B₀² / (2μ₀)

B₀= μ₀ n I₀

n is no of turns per unit length .

Energy stored in a solenoid per unit volume = ( μ₀ n I₀)² / (2μ₀)

=μ₀n²I₀²/2

n = 40 turns / cm = 40 x 100 turns per meter.

I₀ = 5000A .

Putting the values

Energy stored in a solenoid per unit volume

= 4π x 10⁻⁷ x 40 x 10² x 5000² /2

= 1256 J/ m³

Volume of solenoid = cross sectional area x length

= 28 x 10⁻⁴ x .10 = 2.8 x 10⁻⁴ m³

Total energy stored = 1256 J / m³ x 2.8 x 10⁻⁴ m³

= .35 J .

3 0

3 years ago

How does laser works ?<br>

erastova [34]

Explanation:

Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.

6 0

3 years ago

Assume that charge −q−q-q is placed on the top plate, and +q+q+q is placed on the bottom plate. What is the magnitude of the ele

STatiana [176]

Answer:

Magnitude of electric field = E = q/Aε0

Explanation:

Consider plates are placed at a distance of d. As given in the question the charge stored on the plates have magnitude q and given by:

q = CV

And

V = q/C ……. (i)

The capacitance is given by the following equation:

C = Aε0/d ……. (ii)

Put equation (ii) in (i) ,

V = qd/ Aε0 …..(iii)

The electric field is defined as:

E = V/d …… (iv)

Put equation (iii) in (iv),

E = qd/ Aε0d

E = q/Aε0

Hence, the magnitude of electric field will be q/Aε0 .

8 0

3 years ago

A 0.0220 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block. (a) What is their v

nevsk [136]

Answer:

a) 16.86 m/s.

b) 15.40 m/s

c) 3.175 m/s

Explanation:

let pi be the initial momentum of the system, pf be the final momentum of the system, m1 be the mass of the bullet and V1 be the intial velocity of the bullet, m2 be the mass of the block and V2 be the intial velocity of the block

a) from the conservation of linear momentum:

pi = pf

m1×(V1)i + m2×(V2)i = Vf×(m1 + m2)

(0.0220)×(400) + (0.500)×(0) = Vf(0.0220 + 0.500)

8.8 = 0.522×Vf

Vf = 16.86 m/s

Therefore, the bullet-embedded block system will be moving with a speed of 16.86 m/s.

b) let Vf be the final velocity of the bullet-embedded block system, Wf be the work done by friction on the system, Vi be the initial velocity that the bullet-embedded block system initially moves with.

the total work done on the system is given by:

Wtot = Δk = kf - ki

Wf = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

f×ΔxΔcos(Ф) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

m×Δx×g×μ×(-1) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

m×g×Δx×μ = 1/2(Vi)^2 - 1/2(Vf)^2

1/2(0.0220 + 0.500)(Vf)^2 = 1/2(0.0220 + 0.500)(16.86)^2 - (0.0220 + 0.500)×(9.8)×(8)×(0.30)

(0.261)(Vf)^2 = 86.469

(Vf)^2 = 237.2196

Vf = 15.40 m/s

Therefore, bullet-embedded block system will have a speed of 15.40 m/s after sliding of the friction surface.

c) let Vf be the speed of the bullet-embedded before hitting the block, Vi be the initial velocity of the block and V be the velocity of the bullet-embedded block and block system.

M×Vf + m×Vi = (m + M)×V

(0.0200 + 0.50)×(15.40) + (2)×(0) = (0.022 + 0.50 + 2)×V

8.008 = 2.522×V

V = 3.175 m/s

Therefore, the bullet-embedded block and block system will be moving at a speed of 3.175 m/s.

5 0

3 years ago