1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ohaa [14]
3 years ago
13

The flash on a high-tech camera contains a capacitor of 750 μ F. The battery in the camera supplies 330 V. (a) Determine the ene

rgy that is used to produce the flash. (b) Assuming that the flash lasts for 5 ms, find the power of the flash.
Physics
1 answer:
Arada [10]3 years ago
3 0
Two equations for this one: Q=CV  & PE=(1/2)QV. 
C=750 * 10^-6 F , V=330 V . 
so if you use Q=CV=(<span>750 * 10^-6)(330). Q=0.2475

Then solve for PE. PE=(1/2)QV= (1/2)(0.2475)(330)= 40.8 J
So PE=40.8 J</span>
You might be interested in
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
Kamir and Alexis are studying the properties of water. They conducted a variety of experiments to determine its physical and che
solniwko [45]

The answer is D) neutral water reacts with carbon dioxide to form an acid solution

4 0
3 years ago
Read 2 more answers
. The Soviet Yuri Gagarin was the first human to orbit Earth true or false
Dmitrij [34]

Answer:

true , I searched and got u the answer

3 0
2 years ago
In order to be _______________ forces, their effects must cancel each other out and not cause a change in the object's motion
Tju [1.3M]
Balancing.

When the same force is applied from both sides,the forces cancel out each other.
4 0
3 years ago
What is the magnitude b of the magnetic field at the location of the charge due to the current-carrying wire?
elena-14-01-66 [18.8K]

Answer:

k

Explanation:

3 0
3 years ago
Other questions:
  • slader A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut fo
    7·1 answer
  • A student, standing on a scale in an elevator at rest, sees that his weight is 840 n. as the elevator rises, his weight increase
    7·1 answer
  • You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h
    6·1 answer
  • How does technology make sustainability more difficult?
    7·2 answers
  • DO NOT ANSWER IF YOU DON'T KNOW
    8·1 answer
  • How can the pilot determine, for an ILS runway equipped with MALSR, that there may be a penetration of the obstacle identificati
    12·1 answer
  • Help please:))))))))))))))))))))
    12·1 answer
  • Example of kinetic energey
    11·2 answers
  • The table below shows the measurements you took in an experiment. Trial Length ( miles) 1.9 4.2 N 3 5.9 4 What is the longest me
    13·1 answer
  • The Ancient Roman economy did not make use of
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!