All categories

3 years ago

Sequence the steps used to extract metal from the earths core

1 answer:

4 0

Methods of extraction include: extract by electrolysis, extract by reaction with carbon or carbon monoxide, and extracted by various chemical reactions. 《I hope tis help you》

You might be interested in

Complete the wording of the Law of Conservation of Matter and Energy (as defined in the text).

Aleks [24]

Answer:

Neither "matter" nor "energy" can be created or destroyed, but they can be changed from one form to the other.

Explanation:

6 0

3 years ago

We want to get a better, more concrete idea of the strength of the electric force. So imagine you could remove all the electrons

Advocard [28]

Answer:

Explanation:

solution answered below

5 0

3 years ago

Read 2 more answers

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

3 years ago

A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00

svetoff [14.1K]

Answer:

a) 200 m

b) 100 m/s

c) 709 m

d) -118.2 m/s

e) 26.24 s

Explanation:

The rocket flies upward with constant acceleretion.

The equation for position under constant acceleration is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

V0 = 0

Y(t) = 1/2 * 25 * t^2

Y(t) = 12.5 * t^2

And speed under constant acceleration:

Vy(t) = Vy0 + a * t

Vy(t) = 25 * t

It burns for 4 s and runs out of fuel

Y(4) = 12.5 * 4^2 = 200 m

V(4) = 25 * 4 = 100 m/s

Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity

It will be under constant acceleration. These new equations will have different starting constants.

Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2

Vy(t) = Vy4 + g * (t - 4)

When it reaches its highest point it will have a speed of zero.

0 = Vy4 + g * (t - 4)

0 = 100 - 9.81 * (t - 4)

100 = 9.81 * (t - 4)

t - 4 = 100 / 9.81

t = 10.2 + 4 = 14.2 s

At that moment it will have a height of:

Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m

The rocket will fall and hit the ground:

Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2

0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)

0 = -4.9 * t^2 + 139.2 * t -278.4

Solving this equation electronically:

t = 26.24 s

At that time the speed will be:

Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s

6 0

3 years ago

Complete the statement by filling in the Blanks:

Anika [276]

Answer:

induced current

Explanation:

intentionally manipulated.

5 0

3 years ago