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nataly862011 [7]

2 years ago

7

A 2.0 m conductor is formed into a square and placed in the horizontal xy-plane. Magnetic field is oriented 30.0° above the hori

zontal with a strength of 1.0 T. What is the magnetic flux through the conductor

1 answer:

Sloan [31]2 years ago

6 0

Answer:

The magnetic flux through the conductor is $0.125\, T.m^{2}$

Explanation:

Given

Perimeter of square = 2.0 m

Magnetic field strength , B = 1.0 T at 30 degree with the horizontal

Therefore angle of magnetic field with the normal , $\Theta =(90-30)^{\circ}=60^{\circ}$

Let side of square be a

$\therefore 4a= 2.0\, m=>a=0.5\, m$

Therefore area of square is given by

$A=a^{2}=0.5^{2}\, m^{2}=0.25\, m^{2}$

Magnetic flux through the conductor is given by

$\phi =BA\cos \Theta =1.0\times 0.25\times \cos (60^{\circ})\, T.m^{2}$

=> $\phi =0.125\, T.m^{2}$

Thus the magnetic flux through the conductor is $0.125\, T.m^{2}$

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3 years ago

PLEASE PROVIDE AN EXPLANATION.<br><br> THANKS!!!

ziro4ka [17]

Answer:

(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz

(b) 250 m/s

(c) 1250 N

(d) Positive x-direction

(e) 6.00 m/s

(f) 0.0365 m

Explanation:

(a) The standard form of the wave is:

y = A cos ((2πf) t ± (2π/λ) x)

where A is the amplitude, f is the frequency, and λ is the wavelength.

If the x term has a positive coefficient, the wave moves to the left.

If the x term has a negative coefficient, the wave moves to the right.

Therefore:

A = 0.0800 m

2π/λ = 0.300 m⁻¹

λ = 20.9 m

2πf = 75.0 rad/s

f = 11.9 Hz

(b) Velocity is wavelength times frequency.

v = λf

v = (20.9 m) (11.9 Hz)

v = 250 m/s

(c) The tension is:

T = v²ρ

where ρ is the mass per unit length.

T = (250 m/s)² (0.0200 kg/m)

T = 1250 N

(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).

(e) The maximum transverse speed is Aω.

(0.0800 m) (75.0 rad/s)

6.00 m/s

(f) Plug in the values and find y.

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8 0

2 years ago

Read 2 more answers

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

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3 years ago

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GREYUIT [131]

<h2>Right answer: Comets have very elliptical orbits that usually take them far beyond the orbit of Pluto, but also take them closer to the Sun than Earth</h2>

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One of the main characteristics of a comet is that it travels quite fast, on its way around the Sun and has a long tail. It should be noted that the tails of comets always go in the opposite direction to the Sun (due to the radiation pressure of sunlight).

Therefore, the correct option is C.

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3 years ago

Read 2 more answers

The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

MaRussiya [10]

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

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Well depp $h=40.m$

a.

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

b.

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

4 0

3 years ago