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2 years ago

Which type of mine involves digging tunnels and shafts deep underground?

Physics

1 answer:

mel-nik [20]2 years ago

7 0

The mine involves digging tunnels and shafts deep underground is the underground mining.

<h3>What is mining?</h3>

Mining is the digging of the Earth deep down in search of some precious elements or resources found on Earth.

Underground mining is the digging down into the earth in order to create tunnels. The shafts need to be inserted into those tunnels so that it reaches the deposits of resources.

Thus, underground mining involves digging tunnels and shafts deep underground.

Learn more about mining.

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A ball starts at rest and rolls down an inclined plane. The ball reaches 7.5 m/s in 3 seconds. What is the acceleration?

just olya [345]

Answer:

$a=2.5\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly over time.

The equation that describes the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

Solving the equation for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The ball starts at rest (vo=0) and rolls down an inclined plane that makes it reach a speed of vf=7.5 m/s in t=3 seconds.

The acceleration is:

$\displaystyle a=\frac{7.5-0}{3}$

$\boxed{a=2.5\ m/s^2}$

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3 years ago

A mechanic uses a jack to lift up a car. He exerts a force of 11,000 N at a distance of 3m from the axis of rotation. How much t

pshichka [43]

Answer:

<h2>The amount of torque put on the car is 33,000Nm</h2>

Explanation:

Formula for calculating torque is expressed as T = rFsin $\theta\\$ where;

r is the radius of the of the arm of the jack = 3m

F is the force exerted = 11000

$\theta\\$ is the angle of rotation = 90°

On substituting;

$T = 3*11000sin90^{o} \\T = 3*11000 (sin90^{o} =1)\\T = 33000Nm$

6 0

3 years ago

Imagine a particle that has three times the mass of the electron. All other quantities given above remain the same. What is the

melamori03 [73]

Answer:

The only parameter that changes is mass m

It is only necessary to calculate the ratio Eh/Ee

$m_{h}=3m_{e}\\E_{h}=\frac{3m_{e}v^{2}}{2}\\E_{e}=\frac{m_{e}v^{2}}{2}\\\frac{E{h}}{E{e}}=3$

The kinetic energy of the heavy paricle is three times the kinetic energy of an electron

5 0

3 years ago

An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +32.6°. When the potentia

Reil [10]

Answer:

V=-8.35v

Explanation:

Pls see attached file

6 0

3 years ago

A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi

krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

(m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

k = [(m)×(v^2)]/(x^2)

k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

v^2 = [(k)(x^2)]/m

v = \sqrt{ [(k)(x^2)]/m}

v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

= 0.491 m/s

8 0

3 years ago

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