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Nesterboy [21]
3 years ago
6

Find the work done lifting a 290-pound weight 28 feet in the air.

Mathematics
2 answers:
adell [148]3 years ago
4 0

Answer: The correct answer is 8120 foot-pound.

Step-by-step explanation:

The expression of work done in terms of mass and height is as follows;

W= mgh

Here, m is the mass of the object, g is the acceleration due to gravity and h is the height.

It is given in the problem that the work is done in lifting a 290-pound weight 28 feet in the air. Here, the work done is equal to the potential energy.

Calculate the work done.

W= mgh

Put mg=290-pound and h= 28 feet.

W= (290)(28)

W= 8120 foot-pound

Therefore, the work done is  8120 foot-pound.

IgorLugansk [536]3 years ago
3 0
The work may be calcualted as the differential of potencial energy or  as the product of the force x the distance. The result is the same.

T = weight x height

T = 290 pound * 28 feet = 8120 foot-pound
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Which of these triangles appears not to be congruent to any others shown
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Is (-3,-9) a solution of the equation y = 3x ?<br> O YES<br> O NO<br> O NOT A SOLUTION
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If sinA+cosecA=3 find the value of sin2A+cosec2A​
Irina18 [472]

Answer:

\sin 2A + \csc 2A = 2.122

Step-by-step explanation:

Let f(A) = \sin A + \csc A, we proceed to transform the expression into an equivalent form of sines and cosines by means of the following trigonometrical identity:

\csc A = \frac{1}{\sin A} (1)

\sin^{2}A +\cos^{2}A = 1 (2)

Now we perform the operations: f(A) = 3

\sin A + \csc A = 3

\sin A + \frac{1}{\sin A} = 3

\sin ^{2}A + 1 = 3\cdot \sin A

\sin^{2}A -3\cdot \sin A +1 = 0 (3)

By the quadratic formula, we find the following solutions:

\sin A_{1} \approx 2.618 and \sin A_{2} \approx 0.382

Since sine is a bounded function between -1 and 1, the only solution that is mathematically reasonable is:

\sin A \approx 0.382

By means of inverse trigonometrical function, we get the value associate of the function in sexagesimal degrees:

A \approx 22.457^{\circ}

Then, the values of the cosine associated with that angle is:

\cos A \approx 0.924

Now, we have that f(A) = \sin 2A +\csc2A, we proceed to transform the expression into an equivalent form with sines and cosines. The following trignometrical identities are used:

\sin 2A = 2\cdot \sin A\cdot \cos A (4)

\csc 2A = \frac{1}{\sin 2A} (5)

f(A) = \sin 2A + \csc 2A

f(A) = \sin 2A +  \frac{1}{\sin 2A}

f(A) = \frac{\sin^{2} 2A+1}{\sin 2A}

f(A) = \frac{4\cdot \sin^{2}A\cdot \cos^{2}A+1}{2\cdot \sin A \cdot \cos A}

If we know that \sin A \approx 0.382 and \cos A \approx 0.924, then the value of the function is:

f(A) = \frac{4\cdot (0.382)^{2}\cdot (0.924)^{2}+1}{2\cdot (0.382)\cdot (0.924)}

f(A) = 2.122

8 0
3 years ago
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