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3 years ago

Which of these is a possible disadvantage of using wind as a source of

Physics

2 answers:

vitfil [10]3 years ago

4 0

Answer:

D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D DD

Explanation:

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

D. Wind turbines take up a lot of space.

Explanation:

In wind turbines the kinetic energy received by the air molecules is converted into electrical energy by the use of turbines

So here in order to get more kinetic energy from air we need more crossectional area of the wind mill to interact with the air

So here we need the large size of turbines

so this is the main disadvantage of the wind turbines because it needs large area to install the whole setup also the efficiency of this turbine is small so it needs large number of wind mills to setup good output power

so correct answer will be

D. Wind turbines take up a lot of space.

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Goldie Doldfish, a speed swimmer, loves to rece around the parks pond, which is 0.5 miles around. If she can swim around the tra

rosijanka [135]

Answer:

Her average speed is 0.25 miles / hour.

Explanation:

I did 0.5 divided by 2 = 0.25 miles / hour.

<em>Hope this helps! :)</em>

8 0

2 years ago

A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t

viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

$mv+MV=(m+M)v'$

Here , $v'$ is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

$0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s$

Hence , this is the required solution .

3 0

3 years ago

Which statement best describes the difference between acceleration and deceleration?

natima [27]

Acceleration is the rate at which an object picks up speed. deceleration is the rate at which an object loses speed.

7 0

3 years ago

The picture shows an object resting on a balance.

Maslowich

Answer:

4.90kgm^-2

Explanation:

4 0

2 years ago

Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga

Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0

3 years ago