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Alex Ar [27]
3 years ago
11

When an electric current is passed through water during the process of electrolysis two gases are formed one gas has a boiling p

oint of -183 c and the other has a boiling point of -253 change or a chemical change occurred explain had a physical change or a chemicial change occurred explain?
Physics
2 answers:
Olin [163]3 years ago
4 0

Answer:hi

Explanation:

vagabundo [1.1K]3 years ago
3 0

There is a chemical change because the liquid water, after passing an electric current, produces hydrogen and oxygen gases. This is a reaction. The molecular component of eater has been broken down into two. These two gases have different structures compared to water.

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Gwar [14]
No becuase a round room has no corners so no
3 0
3 years ago
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Earth’s surface is subjected to unusually high rates of weathering. The Moon and some of the inner planets of our solar system a
Komok [63]
<h2>Answer:</h2>

<em><u>Running water on the Surface of Earth.</u></em>

<h2>Explanation:</h2>

Weathering can be defined as the phenomenon of breaking or disintegration of the rocks, minerals, soil etc. due to coming in contact with the Earth's atmosphere such as the Water, Wind, Organisms etc.

In this process the rocks, minerals disintegrate into lot of small particles.

But the rate of weathering on Earth is at a very unusual or unique rate because Earth's surface has the presence of the running water that leads to the high rate of surface weathering.

<em><u>Therefore, the reason of Earth's rapid weathering rates is Running water on the Surface of Earth.</u></em>

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3 0
3 years ago
When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the thir
nirvana33 [79]

Answer:

- Contact 1 with 3 ,  initial charge of 1.8 C.

- contact 1 with 2 and then 1 with 3 , first body should have 3.6 C

Explanation:

The excess charge on a body is distributed evenly throughout the body.

We can have two different configurations:

- Contact 1 with 3

When the third body was touched with the first, the initial charge was distributed between the two, so that when each one separated, it had half the charge, in this configuration the first body should have an initial charge of 1.8 C.

- contact 1 with 2 and then 1 with 3

Another possible configuration of the exercise is that the first body touches the second and the charge decrease to the half and then touches the third where it again decreases by half, so that the first body only gives it every ¼ of its initial load.

Therefore in this configuration if the third body has a load of 0.9C the first body should have 3.6 C

8 0
3 years ago
A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
3 years ago
Pls someone help me with this :(
RUDIKE [14]

4.0 is gonna be your answer






6 0
2 years ago
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