Which of the following leads most directly to the production of igneous rocks?
1 answer:
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Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.
The distance of the earth from the sun is .
<h3>What is Kepler's third law?</h3>
Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.
Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.
By using Kepler's third law, this can be written as,
Substituting the values, we get the value of constant k for mars.
The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.
Hence we can conclude that the distance of the earth from the sun is .
To know more about Kepler's third law, follow the link given below.
Answer:
The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west
Explanation:
Hi there!
Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:
The vector for the first displacement is:
First displacement = (20 mi, 0)
The second displacement:
Second displacement = (0, 6.0 mi)
The resultant displacement will be:
R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)
The magnitude of this vector will be:
The magnitude of the vector displacement is 21 mi.
To find the direction of the vector R, we have to apply trigonometry:
In a right triangle the following trigonometric rule applies:
cos θ = adjacent side to the angle/ hypotenuse
In this case:
cos θ = 20 mi / magnitude of R
θ = 16.7°
The direction of the vector is 16.7° north of west.
The question is incomplete. Here is the complete question.
Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.
Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²
Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:
= -
-
The is a x-component of force due to gravity (W) and, in this case, is given by:
= W.sin(14)
W is described as: W = m.g
Force due to friction () is given by:
= μs.N
N is the normal force and, in the system, is equivalent of , so:
= m.g.cos(14)
Therefore, the formula will be:
= -
-
m.a = - (m.g.sin14) - (μs.mg.cos14)
a = - g (sin14 + μscos 14)
a) For dry concrete, μs = 1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 1.cos14)
a = - 11.05 m/s²
b) For wet concrete, μs = 0.7:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin 14 + 0.7.cos14)
a = - 10.64 m/s²
c) For ice, μs = 0.1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 0.1cos14)
a = - 9.84 m/s²
