Category: Algebra
"Published in Newark, California, USA"
Alberto would like to distribute a bagful of marbles equally among his friends. He figures out that if he were to give them 7 marbles each, he will need 4 marbles more but if he were to give them 6, he will have 3 marbles left. How many marbles does Alberto have?
Solution:
For this type of word problem, we will use the principles of linear equation where the partition or distribution of items is involved. Let's analyze the given word problem as follows
Let x be the number of marbles
y be the number of friends to distribute the marbles
If the statement says "if he were to give them 7 marbles each, he will need 4 marbles more", then the working equation will be
or
If the statement says "if he were to give them 6, he will have 3 marbles left", then the working equation will be
or
Equate x = x, we have
Therefore,
Substitute the value of y to either one of the two equations, we have
Therefore, Alberto has 294 marbles that he distributed to his 46 friends.
This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)
Thursday, February 28, 2013
Wednesday, February 27, 2013
Escape Velocity - Solar System, Planet
Category: Differential Equations, Physics
"Published in Newark, California, USA"
The radius of the moon is roughly 1080 miles. The acceleration of gravity at the surface of the moon is about 0.165g, where g is the acceleration of gravity at the surface of the earth. Determine the velocity of escape for the moon. (Hint: g = 32.16 ft/sec2 and 1 mile = 5280 ft)
Solution:
Consider the problem of determining the velocity of a particle projected in a radial direction outward from the earth and acted upon by only one force, the gravitational attraction of the earth.
We shall assume an initial velocity in a radial direction so that the motion of the particle takes place entirely on a line through the center of the earth.
According to the Newtonian Law of Gravitation, the acceleration of the particle will be inversely proportional to the square of the distance from the particle to the center of the earth. Let r be that variable distance, and let R be the radius of the earth. If t represents time, v is the velocity of the particle, a its acceleration, and k is the constant of proportionality in the Newtonian Law, then
The acceleration is negative because the velocity is decreasing. Hence the constant k is negative. When r = R, then a = - g, the acceleration of gravity at the surface of the earth. Thus
or
from which
We wish to express the acceleration in terms of the velocity and the distance. We have a = dv/dt and v = dr/dt. Hence
so the differential equation for the velocity is now seen to be
The method of separation of variables applies to equation above and leads at once to the set of solutions
or
Suppose that the particle leaves the earth's surface with the velocity v0. Then v = v0 when r = R, from which the constant C is easily determined to be
Thus, a particle projected in a radial direction outward from the earth's surface with an initial velocity v0 will travel with a velocity v given by the equation
If v becomes zero, the particle would stop, the velocity would change from positive to negative, and the particle would return to the earth.
If r becomes very large, then the first term will be very small and even equal to zero.
To avoid that v becomes zero, then the next two terms in the equation above must be
If
then we can solve for the value of v0 as follows
or
The above equation is called the velocity of escape. This is applicable at any planets and moons in a solar system including earth.
Now, let's go back to the given problem that if the radius of a moon is 1080 miles and the acceleration of gravity is 0.165g, then the velocity of escape at the moon will be equal to
or
"Published in Newark, California, USA"
The radius of the moon is roughly 1080 miles. The acceleration of gravity at the surface of the moon is about 0.165g, where g is the acceleration of gravity at the surface of the earth. Determine the velocity of escape for the moon. (Hint: g = 32.16 ft/sec2 and 1 mile = 5280 ft)
Solution:
Consider the problem of determining the velocity of a particle projected in a radial direction outward from the earth and acted upon by only one force, the gravitational attraction of the earth.
We shall assume an initial velocity in a radial direction so that the motion of the particle takes place entirely on a line through the center of the earth.
According to the Newtonian Law of Gravitation, the acceleration of the particle will be inversely proportional to the square of the distance from the particle to the center of the earth. Let r be that variable distance, and let R be the radius of the earth. If t represents time, v is the velocity of the particle, a its acceleration, and k is the constant of proportionality in the Newtonian Law, then
The acceleration is negative because the velocity is decreasing. Hence the constant k is negative. When r = R, then a = - g, the acceleration of gravity at the surface of the earth. Thus
or
from which
We wish to express the acceleration in terms of the velocity and the distance. We have a = dv/dt and v = dr/dt. Hence
so the differential equation for the velocity is now seen to be
The method of separation of variables applies to equation above and leads at once to the set of solutions
or
Suppose that the particle leaves the earth's surface with the velocity v0. Then v = v0 when r = R, from which the constant C is easily determined to be
Thus, a particle projected in a radial direction outward from the earth's surface with an initial velocity v0 will travel with a velocity v given by the equation
If v becomes zero, the particle would stop, the velocity would change from positive to negative, and the particle would return to the earth.
If r becomes very large, then the first term will be very small and even equal to zero.
To avoid that v becomes zero, then the next two terms in the equation above must be
If
then we can solve for the value of v0 as follows
or
The above equation is called the velocity of escape. This is applicable at any planets and moons in a solar system including earth.
Now, let's go back to the given problem that if the radius of a moon is 1080 miles and the acceleration of gravity is 0.165g, then the velocity of escape at the moon will be equal to
or
Tuesday, February 26, 2013
Angle - Depression Problem
Category: Trigonometry
"Published in Newark, California, USA"
A surveyor on a mountain peak observes below him two vessels lying at anchor 1 mile apart and in the same vertical plane with his position. He finds the angles of depression of the ships to be 18° and 10° respectively. How high does the peak rise above the water? (1 mile = 5280 feet)
Solution:
This is also a great application of angle of depression especially for the surveyors and Civil Engineers. Let's analyze and label further the above figure as follows
but
The above equation becomes
but
Therefore, the height of a mountain peak from the water is
"Published in Newark, California, USA"
A surveyor on a mountain peak observes below him two vessels lying at anchor 1 mile apart and in the same vertical plane with his position. He finds the angles of depression of the ships to be 18° and 10° respectively. How high does the peak rise above the water? (1 mile = 5280 feet)
Photo by Math Principles in Everyday Life |
Solution:
This is also a great application of angle of depression especially for the surveyors and Civil Engineers. Let's analyze and label further the above figure as follows
Photo by Math Principles in Everyday Life |
but
The above equation becomes
but
Therefore, the height of a mountain peak from the water is
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