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Dmitry_Shevchenko [17]
3 years ago
8

You have two identical capacitors and an external potential source.a) Compare the total energy stored in the capacitors when the

y are connected to the applied potential in series and in parallel.U(parallel)/U(series) =?b) Compare the maximum amount of charge stored in each case.Q(parallel)/Q(series) =?c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?E(parallel)/E(series) =?
Physics
1 answer:
AnnyKZ [126]3 years ago
8 0

Correct answer would be C.

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Que la tarcer ley de newton
slava [35]

Explanation:

If you write it in English so I can help u if you need it

3 0
3 years ago
Given a force of 100 N and a acceleration of 5 m/s, what is the mass
SIZIF [17.4K]

Answer:

force = mass \times acceleration \\ 100 = m \times 5 \\ m =  \frac{100}{5}  \\ m = 20 \:  kg

8 0
3 years ago
Read 2 more answers
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.
LiRa [457]

Answer:

         Time - taken = 2.5 s

          deceleration= -8 m/s²

Solution:

            Given:

                     speed, v = 8 m/s

                 distance, d = 20m

                     

              To Find:

                     deacceleration = ?

               

               As we know speed is defined as

                          v = d/t

                plugging in the values

                          t =  20/ 8

                          t = 2.5s

                Now from deceleration formula

                        a =  - v/ t

                        a = - 20/ 2.5

                        a = - 8 m/s²

          Thus, the time taken and acceleration is 2.5 s and -8 m/s²

          respectively.

          Learn more about deceleration here:

                brainly.com/question/13354629

                       #SPJ4  

               

                       

             

7 0
2 years ago
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