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Citrus2011 [14]
3 years ago
10

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

is possible to calculate ____________________
Physics
1 answer:
Tasya [4]3 years ago
3 0

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

                                                     \Phi=B\ A\ cos\theta

We are told the values of \Phi and B, then we can calculate the magnitude

                                                      \frac{\Phi}{A}=B\ cos\theta

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A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a
DanielleElmas [232]

Answer:

a) f_o=454.11Hz

b)f_o=425.89Hz

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

f_o=\frac{v\pm v_o}{v\pm v_s} *f_s

where:

f_o=Observed\hspace{3}frequency

f_s=Actual\hspace{3}frequency

v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves

v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source

v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer

Now let's consider the next cases:

+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source

-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source

-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer

+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer

The data provided by the problem is:

f_s=440Hz\\v_o=11m/s

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

v=343m/s

Now, in the first case the observer alone is in motion towards to the source, hence:

f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz

Finally, in the second case the observer alone is in motion away from the source, so:

f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz

6 0
4 years ago
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s
Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

P=\frac{W}{t}

The work is the force the is applicated in a distance so

W=F*d

replacing:

P=F*\frac{d}{t} and \frac{d}{t} determinate distance in time is velocity so

a).

P=F*v

F=m*a\\F=m*g*sin(33.5)

P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}

The maximum force is when the car is accelerating so

Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N

so the maximum force is the maximum force by the maximum speed

Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

4 0
4 years ago
in an experiment the following readings were observed volume of alcohol flowing per minute equals to 10 raise to power - 5 cube
tekilochka [14]

Answer:

The viscosity is 1.30 x 10^-3 deca poise.

Explanation:

Volume per minute, V = 10^-5 m^3

Volume per second, V = 1.67 x 10^-7 m^3

density, d = 800 kg/m^3

radius, r = 0.05 cm

Length, L = 0.5 m

Height, h = 60 cm

Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa

Use the formula  of rate of flow

V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 =  3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise

6 0
3 years ago
Peter and Fred are skateboarding in a large semicircular halfpipe. Peter starts out from rest at a height h and collides with Fr
jolli1 [7]

Answer:

The answer is h/4

Explanation:

When Peter collides with Fred, the collision is inelastic & they both proceed with a velocity of V/2.

let m represent the masses for Peter & Fred

v represent the initial velocity of Peter

V represent final velocity of both of them

mv + 3m × 0 = (m+m)V

V = v/2

Using the expression; H = v² / 2g .............Eqn 1

Upon substitution of V/2 into Eqn 1 above,

H = (V/2)² / 2g

H = (V²/4) ÷ 2g

Therefore height will be h/4

6 0
3 years ago
Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and b
Alona [7]

Answer:

E. The force on A is exactly equal to the force on B.

Explanation:

The force between two charges is given by

F=\dfrac{kq_1q_2}{r^2}

where

q_1 = Charge on particle 1

q_2 = Charge on particle 2

r = Distance between the charges

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N

This force will be exerted on both the charges equally.

So, The force on A is exactly equal to the force on B

6 0
3 years ago
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