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3 years ago

13

Name two effects of using coal as energy

2 answers:

german3 years ago

7 0

Answer:

Here is some useful background facts...

Explanation:

They include mercury, lead, sulfur dioxide, nitrogen oxides, particulates, and various other heavy metals. Health impacts can range from asthma and breathing difficulties, to brain damage, heart problems, cancer, neurological disorders, and premature death

Xelga [282]3 years ago

6 0

Answer:

Coal can hurt the environment severely. The mining of coal is bad for the ground and you pollute the air when you burn the coal.

Coal is also unhealthy for your lungs. If you burn coal and breathe in the smoke you can damage your lungs, especially if you are constantly around that smoke.

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Considering only the earth's rotation, determine how much later the asteroid would have had to arrive to put the explosion above

tangare [24]

Answer:

5.1 hours

Explanation:

The only fact we need to know about such a question is that when gazing down at the north pole, the earth spins longitudinally at 360 degrees / day in the clockwise direction.

The planet would have to spin an additional 77 ° to strike the asteroid at 25° E. If the earth rotates in 24 hours 360 degrees, then it must it rotates in 5.1 h at 77 degrees.

8 0

3 years ago

What is your theory on frame of reference?

____ [38]

What is your theory on frame of reference?

Answer:

It is the idea of seeing something from a different perspective

Explanation:

Hope this helps!

8 0

3 years ago

A nurse pushes a cart by exerting a force on the handle at a downward angle of 36 degrees below the horizontal. the loaded cart

Troyanec [42]

Since it was stated that it must move at constant velocity, so the only force it must overpower is the frictional force.

So the equation is:

F cos θ = Ff

F cos 36 = 65 N

F = 80.34 N

<span>So the nurse must exert 80.34 N of force</span>

4 0

2 years ago

The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m

Angelina_Jolie [31]

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

$U(x)=2x^3-15x^2+36x-23$

and we know force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

$F=-(2\times 3x^2-15\times 2x+36)$

For Force to be zero F=0

$\Rightarrow 6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow x^2-2x-3x+6=0$

$\Rightarrow (x-2)(x-3)=0$

Therefore at x=2 and x=3 Force on particle is zero.

8 0

2 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago