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Umnica [9.8K]
2 years ago
5

4. With respect to a mass of 1kg. (a) How much change in elevation must it undergo to change its potential energy by 1 kJ? (g=9.

8 ms²) ​
Physics
1 answer:
mojhsa [17]2 years ago
4 0

Answer:

102 m upwards.

Explanation:

Just from a qualitative analysis we can tell the mass it needs to go upwards. How much we determine with the fact that the increase will be - in absolute value - equal to the work gravity does on it to go down that same distance.

Fixed that work being 1 kJ, we get

\vec F \cdot \vec{ \Delta h} = |F| |\Delta h| cos 0 = ( 9.8 N )\Delta h \cdot 1 = 10^3 J\\\Delta h = \frac {10^3}{9,8} m \approx 102 m

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Ksivusya [100]

The situation (heat going through the ceiling) describes
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4 years ago
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
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With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

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21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

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\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

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\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

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which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

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4 years ago
P*V = n*R*T where P = pressure V = volume n = number of moles R = the universal gas constant T = temperature in degrees Kelvin T
seraphim [82]

The Gay-Lussac's law or Amonton's law states that the pressure of a given amount of a gas is directly propotional to its temperature if its volume is kept constant  .

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3 0
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Answer:

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3 years ago
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Answer:

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