By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
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Answer:
13.95
Explanation:
Given :
Vector A polar coordinates = ( 7, 70° )
Vector B polar coordinates = ( 4, 130° )
To find A . B we will
A ( r , ∅ ) = ( 7, 70 )
A = rcos∅ + rsin∅
therefore ; A = 2.394i + 6.57j
B ( r , ∅ ) = ( 4, 130° )
B = rcos∅ + rsin∅
therefore ; B = -2.57i + 3.06j
Hence ; A .B
( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95
The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.
Define work. Explain the rate of doing work.
Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.
Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.
Solution Explained:
Given,
Weight = 1000N and distance = 5m
A/Q, the work here is done in lifting then
Work = (weight) × (distance moved)
= 1000 X 5
= 5000Nm or 5000J = 5kJ
Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.
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Answer:
False
Explanation:
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