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sergiy2304 [10]
2 years ago
5

One component of a magnetic has a magnitude of 0.045T and points along the +r axis, while the other component has a magnitude of

0.065T and points along the -y axis. A 1- gram particle carrying a charge of +2.0 ൈ 10-ହC Cis moving along the +Z and at a speed of 4.2 ൈ 10ଷ/. Determine the angle that the net force makes with respect to the +r axis,
Physics
1 answer:
jek_recluse [69]2 years ago
4 0

The angle that the net force makes with respect to the x- axis, is determined as 55.3⁰.

<h3>Angle of the net force</h3>

The angle that the net force makes with respect to the +x axis, is determined as follows;

F = qvBsinθ

<h3>with respect to y-axis</h3>

Fy =  qv(By)sinθ

<h3>with respect to x-axis</h3>

Fx =  qv(Bx)sinθ

The angle that the net force makes with respect to the +x axis;

tanθ = Fy/Fx

tanθ = qv(By)sinθ / qv(Bx)sinθ

tanθ = By/Bx

tanθ = 0.065/0.045

tanθ = 1.444

θ = tan⁻¹(1.444)

θ = 55.3⁰

Thus, the angle that the net force makes with respect to the x- axis, is determined as 55.3⁰.

Learn more about net force here; brainly.com/question/14361879
#SPJ1

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The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids
BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0
3 years ago
Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

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S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

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Answer:

Explanation:

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(0.9 x 10⁻²)² x .35 = ( .45 x 10⁻² )² x v₂

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