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yanalaym [24]
3 years ago
5

You pick up a 3.4 kg can of paint from the ground and lift it to a height of 1.8 m. how much work do you do on the can of paint?

you hold the can stationary for half a minute, waiting for a friend on a ladder to take it. how much work do you do during this time? you friend decides against the paint, so you lower it back to the ground. how much work do you do on the can as you lower it?
Physics
1 answer:
zepelin [54]3 years ago
5 0

1) +60.0 J

The work done when lifting an object is equal to the change in gravitational potential energy of the object:

W=\Delta U= m g \Delta h

where m is the mass of the object, g is the gravitational acceleration, \Delta h is the variation of height of the object.

In this problem, m = 3.4 kg, g = 9.8 m/s^2 and \Delta h = +1.8 m, so the work done is:

W=(3.4 kg)(9.8 m/s^2)(1.8 m)=+60.0 J

And the work done is positive, since the object has gained potential energy.


2) 0 J

In this case, there is no variation of the heigth of the object:

\Delta h =0

Therefore, the variation of potential energy is zero:

\Delta U = 0

And so, the work done is zero as well:

W=0


3) -60.0 J

In this case, the variation of height of the object is negative, since the object has been lowered to the ground:

\Delta h = -1.8 m

So, the object has lost potential energy:

W=(3.4 kg)(9.8 m/s^2)(-1.8 m)=-60.0 J

And so the work done is negative:

W=-60.0 J

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Answer:Simple Covalent substance

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Therefore, when simple covalent substance melts,only the intermolecular forces are broken leaving only the covalent bond in the substance.

(2) poor conductivity: for a substance to conduct electricity,it must have charged particles which are free to move to and fro.

But in the simple covalent substance,there are no charged particles that can be separated due to the covalent bond present in simple covalent substance.

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Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

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