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Anarel [89]
3 years ago
15

Quick please and will give Brainliest!!!

Physics
2 answers:
zysi [14]3 years ago
8 0

Answer:

25Nc

Explanation:

masha68 [24]3 years ago
3 0

25 nC

That is the answer

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motion

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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant
atroni [7]

Answer:

The  time constant is  \tau  = 1.265 s

Explanation:

From the question we are told that

     the time take to charge is  t = 2.4 \  s

The mathematically representation for voltage potential of a capacitor at different time is

        V  =  V_o  - e^{-\frac{t}{\tau} }

Where  \tau  is the time constant  

           V_o is the potential of the capacitor when it is full

     So  the capacitor potential will be  100%  when it is full thus  V_o  =100%  =  1  

and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

      V  = 0.85

Hence

     0.85 =  1 -  e^{-\frac{2.4}{\tau } }

       - {\frac{2.4}{\tau } }  =  ln0.15

        \tau  = 1.265 s

     

7 0
3 years ago
Professional Application. A 96 kg football player catches a 0.900 kg ball with his feet off the ground with both of them moving
Zarrin [17]

To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

m_1u_1+m_2u_2 = (m_1+m_2)V_f

Where

m_{1,2} = Mass of each object

u_{1,2} = Initial Velocity of each object

V_f= Final velocity

Rearranging the equation to find the final velocity we have,

V_f = \frac{m_1u_1+m_2u_2}{(m_1+m_2)}

Our values are given as

m_1 = 96Kg\\m_2 = 0.9Kg\\u_1 = 6.3m/s\\u_2 = 27.4m/s

Replacing we have,

V_f = \frac{(96)(6.3)+(0.9)(27.4)}{(96+0.9)}

V_f = 6.4959m/s

Therefore the final velocity is 6.5m/s

3 0
3 years ago
A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to d
Ipatiy [6.2K]

Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1/(R1+R2) ×Vt

V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor

6 0
4 years ago
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