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LenKa [72]
2 years ago
13

How did Thomson show that cathode rays are different from light?

Chemistry
1 answer:
Flura [38]2 years ago
5 0

Answer:

They had a negative charge

Explanation:

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Which type of wave is a sound wave?
Vesna [10]

Answer:

longitudinal wave

Explanation:

hope it helps you and Mark me down as brainlist

3 0
2 years ago
Using de Broglie's equation, λ = h/mv which would have a larger wavelength, a slow-moving proton or a fast-moving golf ball? Exp
tatuchka [14]

Answer: Proton will have larger wavelength

Explanation:

\lambda =\frac{h}{mv}  (de-Broglie's equation)

\lambda =\text{wavelenght of the particle}

h= Planck constant

m= mass of the particle

v= velocity of the particle

As we can see from the de-Broglie's equation , that wavelength is inversely proportional to the product of mass into velocity of the object.

The wavelength of proton will be higher than that fast moving golf ball because mass of proton (1.67 \times 10^{-24}g) is very small than that of the golf ball (45.93 g). Proton is moving  at slow velocity and the golf ball is moving with fast velocity by which value of product of mass into velocity of proton will be lower than the value of product of mass into velocity of the golf ball which will result in larger value of wavelength of the proton.

6 0
3 years ago
What do we mean by basic MgO?
aliina [53]

Answer:

Magnesium oxide is a simple basic oxide, because it contains oxide ions. It reacts with water to form magnesium hydroxide which is a base.

5 0
2 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
2 years ago
What type of bond is formed between the oxygen atom of one water molecule and the hydrogen atom of another water molecule? what
Scilla [17]

Answer:

The bond formed is a polar covalent bond.

7 0
3 years ago
Read 2 more answers
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