All categories

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.

brainly.com/question/14277327

#SPJ4

You might be interested in

What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?

N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

S = v × t

= 660 miles/hour ×3.5 hours

= 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0

3 years ago

2.)

Oduvanchick [21]

Answer:

$-22.7 m/s^2$

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2$

4 0

3 years ago

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago

Which of the following explains how electrical charges flow? From areas of high-potential to areas of low-potential From areas o

skad [1K]

From areas of high potential to low potential. That is, high voltage to low voltage.

8 0

3 years ago

Read 2 more answers

A bumper car with a mass of 200 kilograms traveling at a velocity of 3.5 meters/second collides with a stationary bumper car wit

Ainat [17]

Answer: fourth option 1.2 × 10³ joules

Explanation:

1) kinetic energy is calculated as the half of the product of the mass times the square of the speed:

=> KE = (1/2) m × (v)²

2) you know the mass after the collision and the speed so you have all the data fo find the numerical value of the expression for KE:

m = m1 + m2 = 200 kg + 250 kg = 450 kg

KE = (1/2) × 450 kg × (2.3 m/s)² = 1,190.25 joules ≈ 1200 joules = 1.2 × 10³ joules, which is the last option of the list.

5 0

3 years ago