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2 years ago

What is an advantage of using wind for producing electricity instead of coal?

Physics

2 answers:

crimeas [40]2 years ago

8 0

Answer:

B. Obtaining Wind does not harm the environment, but Obtaining coal does

Explanation:

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

svetoff [14.1K]2 years ago

7 0

Answer:

Obtaining

Explanation:

<u>VERY VERY biased question and answers....</u>

Wind turb<u>i</u>nes do significant damage to the environment.....from all of the mining required to obtain rare earth metals for the generator magnets to the non-recyclable blades to the enormous amounts of concrete required in pedestal construction ( mega - co2 emissions to make the concrete) to the transportation costs to the construction site . The also kill a significant number of birds and bats AND contribute to noise pollution. They ALSO require significant investment/costs/construction of transmission lines across the land to get from where the wind is to where the electricity is used..... They are NOT great....but I think the answer they want you to spout is the second one. But do not be brainwashed into believing they can replace fossil fuels.

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A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di

Lyrx [107]

Answer:

Explanation:

From the information given:

The angular speed for the block $\omega = 50 \ rad/s$

Disk radius (r) = 0.2 m

The block Initial velocity is:

$v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s$

Change in the block's angular speed is:

$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

However, on the disk, moment of inertIa is:

$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

The time t = 10s

∴

Frictional torques by the wall on the disk is:

$T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m$

Finally, the frictional force is calculated as:

$F = \dfrac{T}r{}$

$F= \dfrac{0.6}{0.2} \\ \\ F = 3N$

8 0

3 years ago

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave

Marina86 [1]

Answer:

$a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree$

Explanation:

Puck A's initial speed is $v_a_1=14.6m/s$ and move in a direction $\theta_b=28.0\textdegree$ after the collision.

# $P_1=P_2$ since there's no external force on the system( $P=mv).$

#The collision equation can be written as;

$m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2$

The kinetic energies before and after the collision are expressed as:

$K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2$

$0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2$

Let +x along A's initial direction and +y along A's final direction makes the angle $62\textdegree$

$\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j$

#Substitute in $v_{b2}^2=213.16-v_{a2}^2$ :

$\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\ \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \ v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\$

$v_{a2}=12.891\ m/s$

Hence, the the speed of puck A after the collision is 12.891 m/s

#. The velocity of A after the collision is;

$\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j$

Substitute $\dot v_{a2}$ into $\dot v_{b2}=(14.6i)-\dot v{a2}$ :

$\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j$

This is the velocity of puck B after the collision, it's speed is:

$v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s$

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

$\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree$

Hence, the direction of B's velocity after the collision is 62°

4 0

4 years ago

What is different between an ionic bond and a covalent bond?

Natali [406]

Ionic Bonding: The formation of an Ionic bond is the result of the transfer of one or more electrons from a metal onto a non-metal. Covalent Bonding: Bonding between non-metals consists of two electrons shared between two atoms.

5 0

4 years ago

Read 2 more answers

Can someone help me with science

tamaranim1 [39]

The continuious physical force per unit area that one region of a gas, liquid, or solid exerts on another.

6 0

4 years ago

What is the potential energy of an object that has mass of 410 kg at a height of 12m

serious [3.7K]

PE = (weight) x (height)

   = (mass) x (gravity) x (height)

   = (410 kg) x (9.8 m/s²) x (12 m)

   = (410 x 9.8 x 12) kg-m²/s²

   = 48,210 joules (rounded)

8 0

3 years ago