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Ede4ka [16]
3 years ago
15

A wire carrying a 32.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's

field, and there is a 2.15 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?
Physics
1 answer:
marin [14]3 years ago
7 0

Answer:

2.24 T

Explanation:

From Electromagnetic Field,

F = BILsin∅................ Equation 1

Where F = Force on the wire, B = Field strength, I = current flowing in the conductor, L = length of the conductor, ∅ = The angle the conductor makes with the magnetic field.

Making B the subject of the equation,

B = F/ILsin∅..................... Equation 2

Given: F = 2.15 N, I = 32 A, L = 3.00 cm = 0.03 m, ∅ = 90° ( the wire is perpendicular to the magnetic field)

Substitute into equation 2

B = 2.15/(32×0.03×sin90°)

B = 2.15/0.96

B = 2.24 T.

Hence the Field strength = 2.24 T

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what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s
kotegsom [21]
Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.
7 0
3 years ago
I would like to know why this is the correct answer
Marta_Voda [28]

Answer:

see below

Explanation:

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I hope this helps!

4 0
3 years ago
In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
Twenty is the _________________ of potassium
Mumz [18]
Twenty is the atomic number of potassium.
7 0
3 years ago
Radiation in the ultraviolet region of the electromagnetic spectrum is quite energetic. It is this radiation that causes dyes to
N76 [4]

The  energy of photon in kJ/mol is 329kJ/mol.

Wavelength of radiation is 370nm. The frequency of given wavelength is

            ν = c / λ

            ν = 3×10^8 / 370×10^-9

            ν = 8.11 × 10^14 s^-1

      Now the energy of photon is:

             E = hν

             E = 6.63×10^-34  J.s/photon × 8.11×10^14s^-1

             E = 5.41× 10^-19  J/photon

         To find in mole

              E = 5.41× 10^-19 × 6.022×10^23

               E = 3.29 ×10^ 5 J/mol

   So, the energy of mole of photon is equal to 329 kJ/mol.

        Learn more about radiation here:

                brainly.com/question/18650102

                     #SPJ4  

5 0
1 year ago
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