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weqwewe [10]
3 years ago
7

Please help me fast and i brainliest!! Two balls are released from the same height. Ball A is released on the surface of Earth,

and ball B is released on the surface of the moon. The mass and net force on each ball are recorded in the table. Net Force (N) Mass (kg) Ball A 19.6 2 Ball B 9.6 6 Ball C 6.6 ? Which statement is correct about the mass of ball C, which is released on the surface of the moon from the same height? The mass of ball C is less than the masses of ball A and ball B. The mass of ball C cannot be determined from the information provided. The mass of ball C is greater than the masses of ball A and ball B. The mass of ball C is greater than the mass of ball A but less than the mass of ball B
Physics
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

If they both start at the same height then most likely that means that each has an intical velocity  and a net force sooooo ball A 19.6

Explanation:

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5 0
3 years ago
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
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