solve the following system by any method 2x - 6y = 24 ... -5x + 6y = -6 ... A. (0,-6) B. (4,-1) C. (-6,-6) D. (6,-1)

What are four drawbacks to using fossil fuels in Alberta thermal power stations?

puteri [66]

Answer:

Fossil fuels have been used for centuries to generate power, but there are many disadvantages associated with their use:

Fossil fuels pollute the environment.

Fossil fuels are non-renewable and unsustainable.

Drilling for fossil fuels is a dangerous process.

Explanation:

can you help me please? brainly.com/question/21745542

6 0

3 years ago

Physicists and engineers from around the world came together to build the largest accelerator in the world, the Large Hadron Col

Vsevolod [243]

Answer:

Explanation:

From the question,

The kinetic energy of protons

K = 7eV.

Mass of proton and the speed of light are the given parameters.

Please kindly go through the files I attached for a step by step explanation of the answer required.

5 0

3 years ago

A wire is used as a heating element that has a resistance that is fairly independent of its temperature within its operating ran

dedylja [7]

Answer:

Double the current

Explanation:

The energy delivered by the heater is related to the current by the following relation:

E= $I^{2}R t$

let R * t = k ( ∴ R and t both are constant)

so E= k $I^{2}$

Now let:

E2= k I₂^2

E2= 4E

⇒ k I₂^2= 4* k $I^{2}$

Cancel same terms on both sides.

I₂^2= 4* $I^{2}$

taking square-root on both sides.

√I₂^2 = √4* I^2

⇒I₂= 2I

If we double the current the energy delivered each minute be 4E.

3 0

2 years ago

1. Two forces act on a box as follows: F= 100 N at 02 = 170. and F2= 75 N at

Setler [38]

Answer:

yfv.

Explanation:

7 0

2 years ago

A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Ede4ka [16]

When acceleration is constant, the average velocity is given by

$\bar v=\dfrac{v+v_0}2$

where $v$ and $v_0$ are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}$

where $x,x_0$ are the final/initial displacements, and $t,t_0$ are the final/initial times, respectively.

Take the car's starting position to be at $t_0=0\,\mathrm s$ . Then

$\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t$

So we have

$x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m$

You also could have first found the acceleration using the equation

$v=v_0+at$

then solve for $x$ via

$x=x_0+v_0t+\dfrac12at^2$

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of $a$ anyway.

7 0

2 years ago