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3 years ago

The planets in the solar system are believed to have formed as a result of a process known as

Physics

1 answer:

Nonamiya [84]3 years ago

8 0

The answer is the Big Bang

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Transformer is used in rectifier step up and step down voltage?

tensa zangetsu [6.8K]

Answer:

While the use of the type of transformer in a rectifier depends on the voltage requirement or to meet desired operating conditions, a step-down transformer is used mainly to reduce the voltage. It is used to bring the high AC voltage level to a reasonable value or the desired output voltage.

Explanation:

Hope it helps

Correct me if Im wrong

3 0

2 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

Answer: The final temperature of the solution is $80.14^oC$

Explanation:

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago

You use 35 J of energy to move an object 5 m. What is the weight of the object

Vlad [161]

Explanation:

35 ÷ 5 = 7kg.

=======================

7 0

2 years ago

To calculate the change in kinetic energy, you must know the force as a function of _______. the work done by the force causes t

trapecia [35]

Kinetic energy is a result of mass in motion at a certain velocity.
1 Joule = 1 kg • (m/s)2
the force as a function of mass of the object.

4 0

3 years ago

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

2 years ago