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2 years ago

How deep would you have to drill to reach the center of earth?

1 answer:

4 0

The radius of the Earth is about 3,960 miles.

That's how far the center of the Earth is straight down under your feet.
It's farther than the distance all the way across the USA, from New York
to Los Angeles. It's like the distance from London (England) to Chicago.

If you drill absolutely straight and aim straight down, that's how far
you need to drill to reach the center. Unfortunately, your drill will melt
long before you get there.

The longest hole ever drilled into the Earth (so far) is the Sakhalin-I Oduptu Well,
drilled by Russia into the sea-bottom in 2011.

That hole is 40,502 feet ... roughly 7.67 miles, or about 0.0019 of the distance
to the center of the Earth. You can see that we still have quite a long way to go !

You might be interested in

Please help.....Which of the following is most likely the distance between the Milky Way and another galaxy?

bagirrra123 [75]

Answer:

The answer would be 10.8 billion light years

6 0

3 years ago

Give two mathematical examples of Newton's third law and how you get the solution

bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

and R is the normal reaction of the weight of the particle on the

surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

horizontal plane.

- When an external force P acts on the particle, then its equation

of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

and R is the normal reaction of the weight of the particle on the

surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0

3 years ago

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

kenny6666 [7]

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

4 0

3 years ago

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

5 0

3 years ago