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Mandarinka [93]
3 years ago
6

David lifts a book 1.2 meters from the floor to the top of his desk. If the book weighed 0.50 kg, what is the gravitational pote

ntial energy gained by the book? Estimate g as 9.81.
Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

5.886 J

Explanation:

Given:

The mass of the book is, m=0.5 kg

Height of lift is, \Delta h=1.2 m

Acceleration due to gravity is, g=9.81 m/s²

Now, gain in gravitational potential energy is a function of change in position and is given as:

\Delta PE=mg\Delta h

Here, \Delta PE is the change in gravitational potential energy.

Plug in 0.5 kg for m, 9.81 for g and 1.2 for \Delta h. Solve for \Delta PE

\Delta PE=0.50\times 9.81\times 1.2=5.886\ J

Therefore, the gain in gravitational energy of the book is 5.886 J.

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A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as t
salantis [7]

Answer:

2.07

Explanation:

Since you didn't supply the drawing, here is what I assumed:

A is the corner opposite the axis of rotation

B is one of the remaining two corners

L1 is the side between A & B

Centripetal acceleration is given by:

ac = v^2 / r = (v / r) * (v / r) * r…………1

Also angular speed is

w = v / r,………….2

Substituting (2) in (1) gives:

ac = (v / r) * (v / r) * r……….3

= (v / r)^2 * r

= w^2 * r

Therefore, the angular acceleration at A and at B are given by:

acA = w^2 * rA……..4

acB = w^2 * rB……..5

It is given that:

acA = n * acB…………6

Substituting (4) and (5) into (6) gives:

w^2 * rA = n * w^2 * rB ……….7==>

rA = n * rB……..8

In terms of the sides L1 and L2:

rA = sqrt (L1^2 + L2^2)…….9

and

rB = L2…………10

Considering (8):

n * L2 = sqrt (L1^2 + L2^2)………11

Squaring both sides:

n^2 * L2^2 = L1^2 + L2^2……….12

Dividing by L2^2:

n^2 = L1^2 / L2^2 + L2^2 / L2^2…….13

= (L1 / L2)^2 + 1 ==>

n^2 - 1 = (L1 / L2)^2 ………14==>

L1 / L2 = sqrt (n^2 - 1) ………15

= sqrt (2.30^2 - 1)

= 2.07. . . . . . <<<=== the value of the ratio L1 / L2 when n = 2.30

8 0
3 years ago
A 16 foot ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 feet per second, how fa
JulsSmile [24]

Answer:

11.625

Explanation:

L = length of the ladder = 16 ft

v_{y} = rate at which top of ladder slides down = - 3 ft/s

v_{x} = rate at which bottom of ladder slides

y = distance of the top of ladder from the ground

x = distance of bottom of ladder from wall = 4 ft

Using Pythagorean theorem

L² = x² + y²

16² = 4² + y²

y = 15.5 ft

Also using Pythagorean theorem

L² = x² + y²

Taking derivative both side relative to "t"

0 = 2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t}

0 = x v_{x} + y v_{y}

0 = 4 v_{x} + (15.5) (- 3)

v_{x} = 11.625 ft/s

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