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xz_007 [3.2K]
3 years ago
11

You charge a capacitor and then discharges it through a resistor. You notice that, after four time constants, the voltage across

the capacitor has decreased to _____ of its value just prior to the initiation of the discharge.
Physics
1 answer:
ollegr [7]3 years ago
5 0

Answer: half of it's value?

Explanation:

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What is the rate of 12 liters of water moving through a water hose in 4.0 minutes?
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3 0
3 years ago
In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas
Anni [7]

Answer:

The phase difference is  \Delta \phi  = 180^o

Explanation:

From the question we are told that

     The distance between the slits is d = 0.2 \ mm = \frac{0.2}{1000}  = 0.2 *10^{-3} \ m

     The distance to the screen is D = 100 cm = \frac{100}{100} = 1 \ m

      The wavelength is  \lambda = 400nm

 The distance of the wave from the  central maximum is L =  5mm = 5*10^{-3} m

   

Generally the path difference of this  waves is mathematically represented as

              y = d sin \theta

Here \theta is the angle between the the line connecting the mid-point of the slits with  the screen and the line  connecting the mid-point of the slits to the central maximum

  This implies that

              tan \theta  = \frac{L}{D}

     =>     \theta = tan ^{-1} \frac{L}{D}

             \theta = tan ^{-1} [\frac{5*10^{-3}}{1}]

           \theta =0.2865

Substituting values into the formula for path difference

       y = 0.2 *10^{-3} sin(0.2864)  

       y = 9.997*10^{-7} \ m  

The phase difference is mathematically represented as

          \Delta \phi = \frac{2 \pi }{\lambda }  * y

Substituting values        

         \Delta \phi = \frac{2 \pi }{400 *10^{-9} }  \ * 9.997*10^{-7}

         \Delta \phi =5 \pi

Converting to degree

         \Delta \phi =5 \pi radians = 5 (180^o)  =  180^o  

the solution is subtracted by 360° in order to get the actual angle

 

             

4 0
3 years ago
While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta
Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to:  brainly.com/question/12550364

#SPJ4

8 0
2 years ago
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