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77julia77 [94]
3 years ago
5

One of relatively few reactions that takes place directly between two solids at roo temperature is

Physics
1 answer:
andreev551 [17]3 years ago
3 0
A reaction between two pure solids is very rare. For a reaction to occur, the molecules must interact with each other. Since molecules of solids are very compact, they don't have the liberty to collide with one another. However, this is made possible by hydrates of solids. An examples would be CuSO₄·5H₂O. In this case, they have a an intermediate liquid to react them together.
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If anyone anwers this question, i will give you 100 points and brainliest!
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Answer:

33. at the top?

34. to add more "energy" to the water and keep it moving

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Explanation:

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3 years ago
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1. What kind of force is the force of gravity? (1 Point) contact force
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3 years ago
A proton, an electron, and a helium nucleus all move at speed v. Rank their de Broglie wavelengths from largest to smallest. Wri
MatroZZZ [7]

Answer:

The correct option is 'c':electron,proton,helium nucleus

Explanation:

The De-Broglie's wavelength of particle is given by

\lambda =\frac{h}{mv}

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6 0
4 years ago
Which type of pollution occurs when small particles of soil are suspended in water?
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7 0
3 years ago
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At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until
Sav [38]

Answer:

a) \Delta \theta = 617.604\,rad, b) \Delta t = 10.392\,s, c) \alpha = -14.128\,\frac{rad}{s^{2}}

Explanation:

a) The final angular speed at the end of the acceleration stage is:

\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)

\omega = 111.5\,\frac{rad}{s}

The angular deceleration is:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}

\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}

\alpha = -14.128\,\frac{rad}{s^{2}}

The change in angular position during the acceleration stage is:

\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}

\theta = 177.604\,rad

Finally, the total change in angular position is:

\Delta \theta = 440\,rad + 177.604\,rad

\Delta \theta = 617.604\,rad

b) The time interval of the deceleration interval is:

\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }

\Delta t = 7.892\,s

The time required for the grinding wheel to stop is:

\Delta t = 2.50\,s + 7.892\,s

\Delta t = 10.392\,s

c) The angular deceleration of the grinding wheel is:

\alpha = -14.128\,\frac{rad}{s^{2}}

4 0
4 years ago
Read 2 more answers
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