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3 years ago

What are the characteristics of open pit mining

Physics

1 answer:

Ostrovityanka [42]3 years ago

4 0

Answer:

The dominant feature in an open pit or quarry operation is the pit, a broad, deepened, often funnel-shaped area where overburden has been removed to access the material to be mined. This is where material extraction takes place.

Explanation:

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Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.1

ludmilkaskok [199]

Answer:

$1.189eV$

Explanation:

The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:

$U=\frac{kq_{e}q_{p}}{r}$

where $q_{e}$ is the electron charge, $q_{p}$ is the proton charge, r is the separation distance between the charges and k is the coulomb constant.

Knowing this, we can calculate how much electric potential energy was lost:

$\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV$

5 0

4 years ago

An athlete performing a long jump leaves the ground at a 32.9 ∘ angle and lands 7.78 m away.part a )what was the takeoff speed ?

MatroZZZ [7]

Answer:

Explanation:

Given

Inclination $\theta =32.9^{\circ}$

Distance of landing point $R=7.78\ m$

Considering athlete to be an Projectile

range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

where u=launch velocity

$7.78=\frac{u^2\sin (2\times 32.9)}{9.8}$

$u^2=83.58$

$u=\sqrt{83.58}$

$u=9.142\ m/s$

(b)If u is increased by 8% then

new velocity is $u'=9.87\ m/s$

$R'=\frac{u'^2\sin 2\theta }{g}$

$R'=9.07\ m$

3 0

4 years ago

Read 2 more answers

A negative velocity, approaching zero, represents a negative acceleration. True or False

ruslelena [56]

Answer:

False.

Explanation:

Lets assume our positive direction to the right (this reasoning works for any direction). A negative velocity would then be then directed to the left. If it varies as such that it aproaches to zero, it means that the variation is directed to the right, and that is where the direction of the acceleration must be pointing. In other words, its losing its velocity, so the acceleration must point opposite to the velocity. Then it means the acceleration is positive.

8 0

4 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$