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algol [13]
2 years ago
14

A 1200-N crate rests on the floor. How much work is required to move it at constant speed (a) 5.0 m along the floor against a fr

iction force of 230 N, and (b) 5.0 m vertically?
Physics
1 answer:
Bogdan [553]2 years ago
6 0

Answer:

(a) The work required to move the crate = 7150 J

(b) The Work required to move the crate vertically = 6000 J

Explanation:

(a) Frictional Force: These is the force that tend to oppose the motion of  two bodies in contact.

Work required to move the crate = work done against friction + work required to move it through a distance of 5.0 m.

Wr = Wf + Wd

Where Wr = work required to move the crate, Wf = work done against friction, Wd = work required to move the crate through 5.0 m

Given: Wf = friction force × distance = 230×5 = 1150 J

and Wd = force × distance = 1200×5.0 =6000 J.

Substituting these values into equation 1

Wr = 6000 + 1150

Wr = 7150 J

Therefore the work required to move the crate = 7150 J

(b)

Work required to move the crate vertically = mgh

or

Work required to move the crate vertically =  Wh................ Equation 2

Where W = weight of the block, h = vertical height

<em>Given:W = 1200 N, h= 5.0 m</em>

Substituting these values into equation 2

Work required to move the crate vertically = 1200×5

Work required to move the crate vertically = 6000 J

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Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

s = ut +\frac{1}{2}at^{2}

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0
3 years ago
Carpet can keep a room quiet by:
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Hard surfaces reflect sound back into the room, while carpets help to absorb the sound so it reflects less
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A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?
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Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

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b = 0.000255


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2 years ago
Exactly one pound of bread dough is placed in a baking tin. the dough is cooked in an oven at 350 °f releasing a wonderful aroma
Sladkaya [172]
The mass of the baked loaf will be less than the original dough. In making dough for bread, we have ingredients that are liquid such as water, melted butter, food flavoring, etc. All of this liquid ingredients mixed on the dough will definitely turn into vapor. This vapor is responsible for releasing of the aroma of the freshly baked bread.
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3 years ago
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
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Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

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