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Nitella [24]
3 years ago
8

You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?​

Physics
2 answers:
77julia77 [94]3 years ago
5 0

Answer:

20 J of work is done by you, against gravity

Explanation:

Work done by gravity = Potential energy stored in the book after it is put ona shelf.

Potential energy stored:

U = mgh

U = wh

w= weight in Newton

m= mass of the book

g= acceleration due to gravity

h= height through which, it is raised.

Work is being done against gravity.

U = 10 \times 2

U = 20 J

Murljashka [212]3 years ago
3 0

Answer:

20 J

Explanation:

Given:

Weight of the book is, W=10\ N

Height or displacement of the book is, d=2\ m

The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.

So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.

Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.

Therefore, work done is given as:

Work=W\times d\\Work=10\times 2=20\ J

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.

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On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav
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Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is:  300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

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An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i
crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}

q=\frac{1}{0.047 cm^{-1}}=+21.3 cm

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

h_i = - h_o \frac{q}{p}

where  

h_i, h_o are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o

So we notice that:

|h_i| < |h_o| : this means that the image is smaller than the object

h_i < 0 : this means that the image is inverted

so, the correct option is:

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7 0
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