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stiv31 [10]
3 years ago
12

A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he

at is lost from the water. The temperature rise of water is : (a) 21.50 (b) 12.0C (c)31.1C (d) 50.0C
Engineering
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

                                =900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

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Answer:

h=0.46m

Explanation:

From the question we are told that:

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Generally, the equation for Water Velocity is mathematically given by

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Therefore Height h is given as

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What are the dimensions of the base of the pyramid?
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b

Explanation:

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For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time
max2010maxim [7]

Answer with Explanation:

The general equation of simple harmonic motion is

x(t)=Asin(\omega t+\phi)

where,

A is the amplitude of motion

\omega is the angular frequency of the motion

\phi is known as initial phase

part 1)

Now by definition of velocity we have

v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )

part 2)

Now by definition of acceleration we have

a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )

part 3)

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where

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Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is A\omega ^{2} and A\omega respectively.

4 0
3 years ago
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

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rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0
3 years ago
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