If a weather system is moving from the Pacific coast to North Carolina,

assuming that the seafloor is spreading at a rate of 1.3cm/yr, calculate how long it took africa's west coast to move 2400 kn aw

Verdich [7]

Answer: 184,615,384.6 years

Explanation:

This problem can be solve by the following equation:

$V=\frac{d}{t}$ (1)

Where:

$V=1.3 \frac{cm}{year} \frac{1 km}{100000 cm}=0.000013 \frac{km}{year}$ is the spreading rate of the seafloor, its velocity

$d=2400 km$ is the distance the Africa's west coast moved at this rate

$t$ is the time it took to the coast to move the descibed distance

Isolating $t$ from (1):

$t=\frac{d}{V}$ (2)

$t=\frac{2400 km}{0.000013 \frac{km}{year}}$ (3)

Finally:

$t=184,615,384.6 years$ This is the time it took to the Africa's west coastto move away from the Mid atlantic ridge.

3 0

2 years ago

When magnesium fluoride reacts with iodine, the products are fluorine and magnesium iodide. What would the equation be?

likoan [24]

Answer:

Explanation:

Every chemical reaction can be represented by simple chemical equations that shows how compounds are combining to give some products.

Such reactions are usually made up of:

Reactants on the left hand side

Products on the right hand side

Reactants → Products

In this given problem;

Reactants are : Magnesium fluoride = MgF₂

Iodine in form of Iodide = I₂

Product : Magnesium iodide = MgI₂

MgF₂ + I₂ → MgI₂ + F₂

7 0

3 years ago

One horsepower is a unit of power equal to 746 W. How much energy can a 150 horsepower engine transform in 10.0 seconds

11111nata11111 [884]

1,119,000 basically 746 times 150 then multiply that answer by 10

7 0

3 years ago

A block has two strings attached to it on opposite ends. One string has a force of 5 N,

juin [17]

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = m (1 m/s²)

where m is the mass of the block. Solve for m :

7 N = m (1 m/s²)

m = (7 N) / (1 m/s²)

m = 7 kg

(B) The friction force is proportional to the normal force, so that if f is the mag. of friction and n is the mag. of the normal force, then f = µ n where µ is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. w) cancel out:

n + (-w) = 0

n = w

n = m g

where g = 9.8 m/s² is the mag. of the acceleration due to gravity.

n = (7 kg) (9.8 m/s²)

n = 68.6 N

Then

3 N = µ (68.6 N)

µ = (3 N) / (68.6 N)

µ ≈ 0.044

4 0

3 years ago

How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

4 0

3 years ago