The period of the wave is determined as 0.083 seconds.
<h3>What is period of a wave?</h3>
The period of a wave is the time taken by a particle of the medium to complete one vibration.
<h3>Period of the wave</h3>
The period of the wave is calculated as follows;
T = 1/f
where;
- T is the period of the wave
- f is frequency of the wave
T = 1/12
T = 0.083 seconds
Thus, the period of the wave is determined as 0.083 seconds.
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here we will use the concept of Newton's III law
as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat
So here we will say
impulse gain by the ball = impulse lost by the bat
given that
For ball the change in speed will be
now from above equation
so speed of bat will decrease by 6.72 mph
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
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The actual answer is B) Chlorine
According to the Bohr Model diagram, the atom has seventeen electrons. This makes it Chlorine.
Sorry if i'm late!!
Metals are not brittle so it can’t be the first one or the third one, both metalloids and metals are shiny so it can’t be the second one. Therefore, it would be the last one because both metalloids and metals are shiny and both are solids at room temperature because it is not a high enough melting point.
ANSWER: Both are shiny and are solid at room temperature.
