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snow_lady [41]
2 years ago
14

Hello people ~

Physics
2 answers:
skad [1K]2 years ago
5 0

Answer:

64q

Explanation:

Note that

  • Charges are addaptive quantities

Here

Capacitance C and potential V is given

We know

  • Q=CV

So charge is also same

Now

there are 64 drops so 64q charge

Harlamova29_29 [7]2 years ago
3 0

Answer:

option d

  • <u> </u><u>6</u><u>4</u><u> </u><u>q </u>

Explanation:

<u> </u><u>C </u><u> </u><u>IS </u><u>A </u><u>CAPACITOR</u>

<u> </u><u> </u><u> </u><u>SO </u><u>,</u><u>,</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>64 drops each having the capacity C and potential V are combined to form a big drop. If the charge on the small drop is q, then the charge on the big drop will be

q = 64

q = 64 q

v are combined in big drop

so ,,,

ans is d 64 q

hope it's helpful to you

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Answer:

34.51

Explanation:

k=1/2mv² is the kenetic energy equation to fill is in

k=[1/2(0.235)×50]²

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Which illustration represents high accuracy but low precision?<br><br><br> plz help
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A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistance
Nookie1986 [14]
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
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by sub.
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4 0
3 years ago
A 5.0kg toolbox is raised from the ground by a rope. If the upward acceleration of the bucket is 2.5 m/s^2, find the force exert
swat32

Answer:

62 N

Explanation:

Sum of the forces on the toolbox:

∑F = ma

T − mg = ma

T = mg + ma

T = m (g + a)

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T = 61.5 N

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8 0
4 years ago
Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc
oee [108]

Answer:

The value is \Delta  \phi   =   4.12 \ rad

Explanation:

From the question we are told that

    The frequency of each sound is  f_1 = f_2 = f =  540 \  Hz

      The speed of the sounds is  v = 330 \  m/s

       The  distance of the first source from the point considered is  a = 4.40 \  m

        The distance of the second source from the point considered is  b  = 4.00  \  m

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_a =  2 \pi [\frac{a}{\lambda}  + ft]

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_b =  2 \pi [\frac{b}{\lambda}  + ft]          

Here b is the distance o f the first wave from the considered point  

Gnerally the phase diffencence is mathematically represented as  

           \Delta \phi= \phi_a - \phi_b  =  2 \pi [\frac{ a}{\lambda}  + ft ] - 2 \pi [\frac{b}{\lambda}  + ft ]      

=>      \Delta  \phi   =   \frac{2\pi [ a - b]}{ \lambda }

Gnerally the wavelength is mathematically represented as

        \lambda  =  \frac{v}{f}

=>     \lambda  =  \frac{330}{540}

=>     \lambda  =  0.611 \ m

=>    \Delta  \phi   =   \frac{2* 3.142 [ 4.40 - 4.0 ]}{  0.611  }

=>    \Delta  \phi   =   4.12 \ rad

     

5 0
3 years ago
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