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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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C) During chemical reaction, total mass remains the same.

This is because due to the law of conservation of mass, stating that in a chemical reaction, neither mass will be created nor destroyed. So, the total mass will not change when chemical reactions occur.
7 0
3 years ago
An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre
irakobra [83]

Answer:

T = -282.33^o C

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)

here we know that

P_1 = 6.50 \times 10^4

P_o = 4.80 \times 10^4

T_1 = 100^o C

T_o = 0.01^o C

now we will have

P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)

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now if P = 0

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0 = 4.80 \times 10^4 + 170.02(T - 0.01)

T = -282.33^o C

7 0
3 years ago
s A horizontal insulating rod of length 11.0-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge
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Answer:

F = 2.26 ×  10⁻³ N

Explanation:

given,

length of rod = 11 cm

charge  = 19 nC

linear charge density = 3.9 x 10⁻⁷ C/m

electric force at 2 cm away.

E(r) = \dfrac{2K\lambda}{r}

F = E q

F= \dfrac{2K\lambda\ q}{L}\int \dfrac{dr}{r}

integrating from 0.02 to 0.02 + L

F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]

F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]

F = 2.26 ×  10⁻³ N

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Answer:

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