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Stells [14]
4 years ago
9

The drawing shows an end-on view of three wires. They are long, straight, and perpendicular to the plane of the paper. Their cro

ss-sections lie at the corners of a square. The currents in wires 1 and 2 are ????1 = ????2 = ???? and are directed into the paper. What is the direction of the current in wire 3, and what is the ratio ????3⁄????, so that the net magnetic field at the empty corner is zero?
Physics
1 answer:
notka56 [123]4 years ago
8 0

Answer:

l3/l = 2

Explanation:

Suppose B1 is the magnetic field due to cable 1 and B2 is the field due to cable 2. Therefore, the magnitude of the field at a distance from the cable is equal to:

B = (uo*I)/(2*pi*r)

B1-2 = (B1^2 + B2^2)^1/2 = ((((uo*I)/(2*pi*r)^2 + (uo*I)/(2*pi*r)^2))))))^1/2 = (2^1/2)*(uo*l)/(2*pi*r)

From this equation we can say that the direction of the field will be from the corner that is empty and cable 3:

B3 = B1-2

(uo*l3)/(2*pi*(2^1/2)*r = (2^1/2)*(uo*l)/(2*pi*r)

From here we have that the relationship between both currents will be equal to:

l3/l = 2

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The driver of a stationary car hears a siren of an approaching police car at a frequency of 280Hz. If the actual frequency of th
deff fn [24]

Answer:

The speed of the police car is 294 m/s

Explanation:

Given;

frequency of the siren in air, f = 280 Hz

speed of sound in air, v = 343 m/s

Determine the wavelength of the sound in air to the stationary car:

v = fλ

where;

λ is wavelength of the sound

λ = v/f

λ = 343 / 280

λ = 1.225 m

Now, determine the speed at which the police car is approaching the stationary car;

The actual frequency of the police car, F = 240 Hz

V = Fλ

Where;

V is speed of the police car

λ is the distance between the police car and the stationary car, (wavelength)

V = 240 x 1.225

V = 294 m/s

Therefore, the speed of the police car is 294 m/s

5 0
3 years ago
A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the
jasenka [17]

Answer:

1.8 cm

Explanation:

m = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

q = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

\Delta V =Potential difference through which the ion is accelerated = 215 V

v = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}

r = Radius of the path followed by ion

B = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm

5 0
3 years ago
(a) Check all of the following that are correct statements, where E stands for γmc2. Read each statement very carefully to make
Reil [10]

Answer:

V=9.60m/s

Attached is the solution

5 0
3 years ago
Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)
Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

P_{1}=\rho g h_{1}

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

P_{1}=\rho gh_{1}.....(I)

Pressure for second pipe,

P_{2}=\rho gh_{2}.....(II)

From equation (I) and (II)

P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

Put the value of P₁ and P₂

\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)

2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2....(III)

We know that,

The continuity equation

v_{1}A_{1}=v_{2}A_{2}

v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})

Put the value of v₂ in equation (III)

2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2

2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2

Here, \dfrac{A_{1}}{A_{2}}=\gamma

So, 2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)

v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Hence, (a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

8 0
3 years ago
The absolute temperature of a gas is t. in order to double the rms speed of its molecules, what should be the new absloute tempe
jenyasd209 [6]

The new absloute temperature should be 4t.

<h3>Temperature </h3>

The hotness of matter or radiation is expressed by the physical quantity known as temperature.

There are three different types of temperature scales: those, like the SI scale, that are defined in terms of the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron in a body; those that solely depend on strictly macroscopic properties and thermodynamic principles, like Kelvin's original definition; and those that are not defined by theoretical principles but rather by useful empirical properties of particula.

Using a thermometer, one can gauge temperature. It is calibrated using different temperature scales, each of which historically defined itself using a different set of reference points and thermometric materials.

Learn more about temperature here:

brainly.com/question/15267055

#SPJ4

6 0
2 years ago
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