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kotegsom [21]
3 years ago
11

Neutralization is an important part of digestion. Why?

Physics
1 answer:
Anarel [89]3 years ago
5 0
It's Breaks Away The Proteins And Nutrients That Your Body Would Use
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wich type of friction can best be described as the force between objects that are moving (static friction)(rolling friction )(ki
Yuri [45]

The answer is kinetic friction.

7 0
3 years ago
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What is the difference between V(peak voltage) and Vrms (root-mean-square) of AC voltage source?
Cerrena [4.2K]

Answer:

V(peak voltage) is the highest voltage that the waveform will ever attain and the Vrms(root-mean-square) is the effective voltage of the total waveform representing the AC source.

8 0
4 years ago
What Current is drawn by a vacuum cleaner from 120 V circuit having a resistance of 28
Ostrovityanka [42]

Answer:

4.3A

Explanation:

V=IR

120=I×28

I=120/28

=4.3

hope this helps

please like and Mark as brainliest

5 0
3 years ago
Copy the diagram. add a voltmeter to show how you would measure the voltage of the cell
Kruka [31]

Answer: the answer is 23voltage

Explanation: because the voltage and time put together is 23

7 0
3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
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