All categories

2 years ago

Of the following systems, which contains the most heat?

Physics

1 answer:

densk [106]2 years ago

8 0

Answer:d

Explanation:

Given systems are state of matter and do not contain any heat instead Heat is required to change Phase or raise the temperature of the particular system.

For example 600 kg of ice at $0^{\circ}C$

Heat Required to convert it to water at $0^{\circ}C$ is

$Q=m\times L$

Where L=latent heat of Fusion $=334 J/gm$

$Q=600\times 334\times 1000$

$Q=200.4 MJ$

You might be interested in

A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the

Elis [28]

Answer:

a) $R = 171$ μΩ

b) $E = 1.7 *10^{-4} V/m$

c) $R_{2} = 1.16 *10^{-4}$ Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder = 1.1 cm

resistivity depends on the distance x from the left

$p(x)=a+bx^2$ ............(i)

using equation

$R = \frac{pl}{a}$

let dR is the resistance of thickness dx

$dR =\frac{p(x)dx}{a}$

where p(x) is resistivity l is length

a is area

$\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\$ .........................(2)

after integration

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$ ...............(3)

it is given $p(0) = a = 2.25 * 10 ^{-8}$ Ωm

$p(L) = a + b(L)^2$ = $8.5 * 10 ^{-8}$ Ωm

$8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\$

(here * stand for multiplication )

on solving we get

$b = 2.78* 10^{-8}$ Ωm

put each value of a and b and r value in equation 3rd we get

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$

$R = 1.71 * 10^{-4}$ Ω

$R = 171$ μΩ

FOR (b)

for mid point x = L/2

E = p(x)L

for x = L/2

$p(L/2) = a+b(L/2)^2$

for given current I = 1.75 A

so electric field

$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

by substitute the values

we get;

$E = 1.7 *10^{-4} V/m$

(here * stand for multiplication )

c ).

75 cm means length will be half

that is x = L/2

integrate the second equation with upper limit L/2

Let resistance is $R_{1}$

so after integration we get

$R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}$

substitute the value of a , b and L we get

$R_{1} = 5.47 * 10 ^{-5}$ Ω

for second half resistance

$R_{2} = R- R_{1}$

$R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}$

$R_{2} = 1.16 *10^{-4}$ Ω

(here * stand for multiplication )

5 0

2 years ago

A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc

Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s, u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest), = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0

2 years ago

A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv

zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 $KV_{LL}$

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0

2 years ago

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined

MrRa [10]

Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

$F_{B}=-5755N$

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

$\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N$