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Advocard [28]
3 years ago
10

How much work is done in lifting a 1.4-kg book off the floor to put it on a desk that is 0.7 m high?

Mathematics
1 answer:
nalin [4]3 years ago
3 0

Answer:

Step-by-step explanation:

Work is said to be done when a force applied to an object cause the body to move in a specified direction.

Work-done = Force * Distance

Since Force = mass * acceleration due to gravity

Work-done =  mass * acceleration due to gravity * distance

Given mass = 1.4kg, distance = 0.7m and g = 9.8m/s²

Workdone in lifting the book off the floor = 1.4*0.7*9.8

Workdone = 9.604Joules

- Similarly, work done in lifting a 21-lb weight book 6 ft off the ground is expressed using the same formula as above;

Given mass = 21-lb, g = 32ft/s² and distance = 6ft

Workdone = 21 * 32 * 6

Workdone = 4,032 lb-ft²/s²

Hence, work-done in lifting a 21-lb weight book 6 ft off the ground is  4,032 lb-ft²/s²

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A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form
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Answer:

\vec{v}= \text{ or } \approx

Step-by-step explanation:

Component form of a vector is given by \vec{v}=, where i represents change in x-value and j represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector \vec{v}=, the magnitude is ||v||=\sqrt{i^2+j^2.

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being i, one leg being j, and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}

To find the other leg, i, we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}

Verify that (9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark

Therefore, the component form of this vector is \vec{v}=\boxed{}\approx \boxed{}

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