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1 month ago

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A wire, of length L = 3.8 mm, on a circuit board carries a current of I = 2.54 μA in the j direction. A nearby circuit element g

enerates a magnetic field in the vicinity of the wire of B = Bxi + Byj + Bzk, where Bx = 6.9 G, By = 2.6 G, and Bz = 1.1 G.A) Calculate the i component of the magnetic force Fx, in newtons, exerted on the wire by the magnetic field due to the circuit element.
B) Calculate the k component of the magnetic force Fz, in newtons, exerted on the wire by the magnetic field due to the circuit element.
C) Calculate the magnitude of the magnetic force F, in newtons, exerted on the wire by the magnetic field due to the circuit element.

1 answer:

Keith_Richards [3.2K]1 month ago

5 0

It equals 5z

Explanation:

5z

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3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

serg [3582]

Response:

(a) $W=-19.25J$

(b) $W=-52.8J$

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

$W=nRTln(\frac{V_1}{V_2} )$

The moles are calculated based on the provided mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Consequently, the work amounts to:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Best regards.

4 0

1 month ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Softa [3030]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Reservoir pressure = 10 atm

$T_1$ = Reservoir temperature = 300 K

$P_2$ = Exit pressure = 1 atm

$T_2$ = Exit temperature

$R_s$ = Specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

Assuming isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

Flow density at exit measures 2.2721 kg/m³

4 0

2 months ago

A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

Given 52.92 g of metal in the sample, the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0

1 month ago

A suspension system is being designed for a 1300 kg vehicle. When the vehicle is empty, its static deflection is measured as 2.5

Keith_Richards [3271]

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

3 0

1 month ago

A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and

kicyunya [3294]

Answer:

$\displaystyle W=-m.g.H$

Explanation:

Transformation of Energy

Also known as energy conversion, this refers to the process in which energy shifts from one type to another. In this context, three energy forms are involved. When the object is stationary at the ramp's peak, it possesses gravitational potential energy, calculated as

$U=m.g.H$

As the object descends the frictionless ramp, it converts all its potential energy into kinetic energy, represented as

$\displaystyle K=\frac{m.v^2}{2}$

Thus,

$K=m.g.H$

Ultimately, when the object encounters a rough surface, all energy converts to thermal energy. The work performed by the friction force corresponds to the alteration in kinetic energy, as all velocity is lost in this process:

$\displaystyle W=\Delta E=K_f-K=0-K=-\frac{m.v^2}{2}$

Given the kinetic energy equals the initial potential energy:

$\boxed{\displaystyle W=-m.g.H}$

The negative sign indicates that the work acted against the direction of movement, meaning the force and displacement are 180° apart.

This outcome is independent of the distance D needed to halt the block or the kinetic friction coefficient.

7 0

2 months ago