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svetlana [45]
2 years ago
14

What do you mean by Parallel Universe ?​

Physics
2 answers:
elena-14-01-66 [18.8K]2 years ago
7 0

Answer:

Parallel Universe is a hypothetical self-contained plane of existence, co-existing with one's own

BartSMP [9]2 years ago
3 0

Answer:

Parallel universe, or alternate reality, is a hypothetical self-contained plane of existence, co-existing with one's own

You might be interested in
An experiment is carried out to measure the extension of a rubber band for different loads.
PtichkaEL [24]

Complete question is;

An experiment is carried out to measure the extension of a rubber band for different loads.

The results are shown in the image attached.

What figure is missing from the table?

Answer:

17.3 cm

Explanation:

The image attached showed values for load, extension and initial length.

Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.

The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.

This gives: 16.2 - 15.2 = 1 cm

This corresponds to what is given in the table.

For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.

Thus;

2.1 cm = Final length - 15.2 cm

Final length = 15.2 + 2.1

Final length = 17.3 cm

3 0
2 years ago
A rocket sled is tested at "5 g" (5 times the acceleration due to gravity). If the sled starts from rest at position do= 0.00, h
stealth61 [152]
D=at²
441m=(5*9.81m/s²)(t²)
t²=441/(5*9.81)
t≈√8.99
t≈3 sec
8 0
2 years ago
Read 2 more answers
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
Nuclear fusion is when two atoms of __________________ join together to form _____________.
solong [7]

Answer:

1. Hydrogen
2. Helium

Explanation:

Nuclear fusion is when two atoms of Hydrogen join together to form one Helium atom.

3 0
2 years ago
if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a
Whitepunk [10]

Answer:

r=0.41m

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

\tau=r\times F

Due to the definition of cross product, the magnitude of the torque is given by:

\tau=rFsin\theta

Where \theta is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when sin\theta is equal to one, solving for r:

r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m

7 0
2 years ago
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