All categories

4 years ago

Why is copper so widely sought world market and new york mercantile exchange?

2 answers:

6 0

It is present in the wires, microelectronics, computer chips, pipes, and other construction supplies. The increase and decline of copper values give a picture of the well-being of the whole world marketplace. It conducts electricity thoroughly than any metal besides silver, malleable, steady melting temperature, panics away bacteria.

tresset_1 [31]4 years ago

5 0

Copper is widely sought world market and New York mercantile exchange because of its very high thermal and electrical conductivity. It is mostly used as a conductor of heat and electricity, a component of various metals of alloy and as a building material.

You might be interested in

HELP PLEASE ASAP !!!!Question 5 of 15

lesantik [10]

I think the answer is C

6 0

3 years ago

Read 2 more answers

What process modifies light waves to vibrate in a single plane

AnnyKZ [126]

This process is called polarization.

8 0

3 years ago

Metals are used in many products because of the characteristics properties that most metals have. Which product requires the hig

nadezda [96]

Alright sounds good buddy

5 0

2 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

SCIENCE HELP! ASAP :)

Vinvika [58]

I believe it's D. The centripetal force moves away from the center, but the marble stays there due to the string.

6 0

3 years ago