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ikadub [295]
3 years ago
8

how mobile phones make use of chemical , lightning and magnatic effects? lease give a simple answer !

Physics
1 answer:
Basile [38]3 years ago
3 0

Answer:  By a mechanism that is not fully understood, charges in a thundercloud are separated such that the lower portion of a cloud takes on a negative charge while the upper portion is positive.  In addition to the negative charge on the base of the cloud, there are small pockets of positive charge on the base.

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Suppose you sound a 1056-hertz tuning fork at the same time you strike a note on the piano and hear 2 beats/second. You tighten
Sindrei [870]

Answer:

The frequency of the piano string is either 1053 HZ or 1059 HZ.

Explanation:

Here we know that frequency of beats is equal to the difference between the frequencies between two waves .

     Given that frequency of tuning fork is 1056 HZ  .

     Let the frequency of the piano be ' f ' .

   Given that number of beats = 3.

    We know that   | 1056 - f | = 3 ;

                      ⇒       1056- f = ±3,

         Upon solving this we get  

                        f = 1056-3 and 1056 + 3

                   ⇒ f = 1053 or 1059 .

                     

6 0
3 years ago
Which of the following statements describe the transfer of energy a. Collision of atoms causing nuclear reactions B. A car speed
oksano4ka [1.4K]
I think the answer would be A.
8 0
3 years ago
A 5.17 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The othe
Andreyy89

Answer:

v=0.816 m/s

Explanation:

The force of the spring and the motion of the block are in equilibrium so without any force of friction the motion is

E_{s}=E_{k}

\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2

First determinate the constant of the spring that produce the kinetic energy of the bloc

k=\frac{m*v^2}{d_{s}^2}

k=\frac{5.17kg*(1.30\frac{m}{s})^2}{0.102^2m}

k=839.8 \frac{kg}{s^2}

Now the motion with the force of friction in the kinetic

E_{s}=E_{k}-W_{k}

\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2-u*m*g

Resolve to v

v^2=\frac{(k*d_{s}^2)+(2*g*u)}{m}

v=\sqrt{\frac{839.8*(0.102m)^2-2*9.8*0.270}{5.17}}

v=0.81 \frac{m}{s}

3 0
3 years ago
What must you do before you make a hypothesis?a)Run the experiment
insens350 [35]
B Is your answer research info about your question
Reasoning is because c) has nothing to do with the scientific method a) is where you text your hypothesis and d) is a conclusion which is the final results of your experiment
4 0
3 years ago
What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change
sammy [17]

Answer:

0.1 m

Explanation:

The closest distance the electrodes used in an NCV test in oerder to measure

the voltage change as a response to the stimulus is 0.1 m.

This is because the shortest observable time period is not less than the action-potential time response of 1 mili second the length traveled by the sensation during this time is 1 m sec x 100 m / s =0.1 m, which is the shortest distance the electrodes could be positioned on the nerve.

6 0
4 years ago
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