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Alona [7]
4 years ago
7

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

ischarge completely. Part A How much energy is dissipated by the 55Ω resistor?
Physics
1 answer:
Ipatiy [6.2K]4 years ago
6 0

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

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2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
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a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

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#LearnwithBrainly

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3 years ago
A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume that the pressure of the gas
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Answer:

The  pressure is  P =   1652 \  Pa

Explanation:

From the question we are told that

    The  volume of the container is  V  =  1.83  \ L =  1.83 *10^{-3 } \  m^3

     The mass of  N_2 is  m_n  =  0.246 \ g =  0.246 *10^{-3} \ kg

     The root-mean-square velocity is  v =  192 \ m/s

The  root -mean square velocity is mathematically represented as

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Now the ideal gas law is mathematically represented as

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Where  M  is the molar mass of  N_2

So  

        RT  =  \frac{PVM_n }{m _n  }

=>    v =  \sqrt{ \frac{3 \frac{P* V  *  M_n }{m_n } }{M_n  } }

=>    v =  \sqrt{  \frac{ 3 *  P* V  }{m_n } } }

=>   P =   \frac{v^2   *  m_n}{3 *    V  }

substituting values

    =>    P =   \frac{( 192)^2   *  0.246 *10^{-3}}{3 *    1.83 *10^{-3} }

=>         P =   1652 \  Pa

       

     

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