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Greeley [361]
3 years ago
12

g An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of It carries a current of 0.300 A. (a) Wha

t is the energy stored in the magnetic field
Physics
1 answer:
valkas [14]3 years ago
4 0

Answer:

Explanation:

Energy of an inductor = 1/2 L i²

L is inductance , i is current .

= 1/2 x 12 x .3²

= .54 J

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A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
Advocard [28]

Answer:

a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

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3 years ago
What is the advantage of nuclear plants?
DerKrebs [107]

They emit low leavls of carbon dioxide (CO2) and the emissions of green house gases and therefore the contribution of nuclear power plants to global warming is therefore pretty little

8 0
3 years ago
Read 2 more answers
A cylinder with a piston contains 0.300 mol of oxygen at 2.50×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The
alukav5142 [94]

Answer:

a) W =  900   J.  b) Q =  3142.8   J . c) ΔU =  2242.8   J. d) W = 0. e) Q =   2244.78   J.  g) Δ U  =  0.

Explanation:

(a) Work done by the gas during the initial expansion:

The work done W for a thermodynamic constant pressure process is given as;

W  =  p Δ V

where  

p  is the pressure and  Δ V  is the change in volume.

Here, Given;

P 1 = i n i t i a l  p r e s s u r e  =  2.5 × 10^ 5   P a

T 1 = i n i t i a l   t e m p e r a t u r e  =  360   K

n = n u m b er   o f   m o l e s  =  0.300  m o l  

The ideal gas equation is given by  

P V = nRT

where ,

p  =  absolute pressure of the gas  

V =  volume of the gas  

n  =  number of moles of the gas  

R  =  universal gas constant  =  8.314   K J / m o l   K

T  =  absolute temperature of the gas  

Now we will Calculate the initial volume of the gas using the above equation as follows;

PV  =  n R T

2.5 × 10 ^5 × V 1  =  0.3 × 8.314 × 360

V1 = 897.91 / 250000

V 1  =  0.0036   m ^3  = 3.6×10^-3 m^3

We are also given that

V 2  =  2× V 1

V2 =  2 × 0.0036

V2 =  0.0072   m^3  

Thus, work done is calculated as;

W  =  p Δ V  = p×(V2 - V1)

W =  ( 2.5 × 10 ^5 ) ×( 0.0072  −  0.0036 )

W =  900   J.

(b) Heat added to the gas during the initial expansion:

For a diatomic gas,

C p  =  7 /2 ×R

Cp =  7 /2 × 8.314

Cp =  29.1  J / mo l K  

For a constant pressure process,  

T 2 /T 1  =  V 2 /V 1

T 2  =  V 2 /V 1 × T 1

T 2  =  2 × T 1  = 2×360

T 2  =  720  K

Heat added (Q) can be calculated as;  

Q  =  n C p Δ T  = nC×(T2 - T1)

Q =  0.3 × 29.1 × ( 720  −  360 )

Q =  3142.8   J .

(c) Internal-energy change of the gas during the initial expansion:

From first law of thermodynamics ;

Q  =  Δ U + W

where ,

Q is the heat added or extracted,

Δ U  is the change in internal energy,

W is the work done on or by the system.

Put the previously calculated values of Q and W in the above formula to calculate  Δ U  as;

Δ U  =  Q  −  W

ΔU =  3142.8  −  900

ΔU =  2242.8   J.

(d) The work done during the final cooling:

The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.

(e) Heat added during the final cooling:

The final process is a isochoric process and for this, the first law equation becomes ,

Q  =  Δ U  

The molar specific heat at constant volume is given as;

C v  =  5 /2 ×R

Cv =  5 /2 × 8.314

Cv =  20.785  J / m o l   K

The change in internal energy and thus the heat added can be calculated as;  

Q  = Δ U  =  n C v Δ T

Q =  0.3 × 20.785 × ( 720 - 360 )

Q =   2244.78   J.

(f) Internal-energy change during the final cooling:

Internal-energy change during the final cooling  is equal to the heat added during the final cooling Q  =  Δ U  .

(g) The internal-energy change during the isothermal compression:

For isothermal compression,

Δ U  =  n C v Δ T

As their is no change in temperature for isothermal compression,  

Δ T = 0 ,  then,

Δ U  =  0.

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Please help with number 18
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