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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leva [86]

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3 years ago

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

Over [174]

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:

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