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3 years ago

What is the difference between charles law and boyle's law?

1 answer:

3 0

Boyle's law talks about the relationship between pressure and volume (high pressure = low volume, and vice-versa), while Charles's law talks about the relationship between volume and temperature (high temperature = high volume, and vice-versa).

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When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *

Ganezh [65]

+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.

5 0

3 years ago

Read 2 more answers

1. The term that describes where the supply curve intersects the demand curve is known as

nadezda [96]

Equilibrium is the answer

6 0

2 years ago

Which evidence did Alfred Wegener’s original theory of continental drift have access to?

Sophie [7]

Answer:

Evidence for continental drift

Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces.

Explanation:

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2 years ago

The diagram below represents the rock cycle.

lara31 [8.8K]

Answer:

C. Heat and Pressure

Explanation:

The arrow which is labeled A points from igneous rock to metamorphic rock.

There are three types of Rocks:

1. Igneous Rock

2. Metamorphic Rock

3. Sedimentary Rock

Rock cycle:

Rock cycle is the process that describes the transition between these three types of rocks. Each type has its own form and its own equilibrium condition. The rock type alters when it is pushed out of its equilibrium conditions.

Transition of Igneous rock to Metamorphic rock:

Igneous rock forms when magma cools down. The transition of Igneous Rock to Metamorphic Rock is a result of a process called Metamorphism. Metamorphism is the alteration in the structure of rock as a result of certain heat and pressure conditions. Inside Earth heat comes from pressure. Heat with pressure does not melt the rock but it bakes the rock. Baking is not melting but it changes the shape of the rock while it is still solid. It actually forms crystals. Because the rock changes its structure, it is called Metamorphic Rock.

5 0

3 years ago

Read 2 more answers

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)

$g = G \cdot M/ r^{2}$ (equation 4)

where the universal gravitation constant $G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}$

Thus

$\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}$

3 0

2 years ago