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Marrrta [24]
3 years ago
5

Which of the following is NEITHER an acid nor a base?

Physics
1 answer:
vagabundo [1.1K]3 years ago
4 0

CaCl2 is neither an acid nor a base.

<u>Explanation</u>:

  • CaCl2 is neutral.  It does not ionize to make hydrogen ions, nor dissociate to give ions to hydroxide.  
  • Calcium chloride, CaCl2is a salt. At room temperature, it is white coloured crystalline solid and is extremely water-soluble. It can be formed by neutralizing the hydrochloric acid with calcium hydroxide.
  • These compounds are used essentially to de-icing and monitor dust.  It is used as a desiccant since the anhydrous salt is hygroscopic.
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Which of the following is an example of a properly written testable hypothesis? *
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Answer:

D. if it is dark, then an owl will find a mouse by the sound the mouse makes

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What can alter the motion of an object
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Force can alter its direction,slow or stop it you could say it can change its velocity

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A block lies on a plane raised an angle θ from the horizontal. Three forces act upon the block: F⃗ w, the force of gravity; F⃗ n
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Answer:

YFy = 0 = Ffsinθ + Fncosθ - Fw

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Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.

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A string is wound tightly around a fixed pulley having a radius of 5.0 cm. as the string is pulled, the pulley rotates without a
lubasha [3.4K]
The angular speed can be solve using the formula:
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8 0
3 years ago
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What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?
Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

V_{e}=\sqrt{\frac{2GM}{R}}

Where:

V_{e} is the escape velocity

G=6.67(10)^{-11} Nm^{2}/kg^{2} is the Universal Gravitational constant

M=5.976(10)^{24}kg is the mass of the Earth

R=6371 km=6371000 m is the Earth's radius

However, in this situation the craft would be launched at a height h=54000 km=54000000 m over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

V_{e}=\sqrt{\frac{2GM}{R+h}}

V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}

Finally:

V_{e}=3633.86 m/s \approx 3.63 km/s

7 0
2 years ago
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