Don't break or crush mercury-containing lamps because mercury powder may be released.

The diagram illustrates a method of producing plastics called

hodyreva [135]

Answer:

polymerisation,

Explanation:

6 0

1 year ago

3.94 x 105) + (2.04 x 105)

Flura [38]

627.9 is the answer

6 0

2 years ago

A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s from a depth of 200 m. For a pump effi

nasty-shy [4]

Answer:

Input power of the geothermal power will be 686000 J

Explanation:

We have given density of brine $\rho =1050kg/m^3$

Rate at which brine is pumped $V=0.3m^3/sec$

So mass of the pumped per second

Mass = volume × density = $1050\times 0.3=315$ kg/sec

Acceleration due to gravity $g=9.8m/sec^2$

Depth h = 200 m

So work done $W=mgh=315\times 9.8\times 200=617400J$

Efficiency is given $\eta =0.9$

We have to fond the input power

So input power $=\frac{617400}{0.9}=686000J$

So input power of the geothermal power will be 686000 J

5 0

2 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

2 years ago

Can a heat engine operating with only one thermal reservoir?

Harlamova29_29 [7]

Answer:

No

Explanation:

Heat engine is a device which operates between two thermal reservoir.One thermal reservoir supply the thermal energy to the heat engine and another thermal reservoir absorb heat rejected by heat engine.These thermal reservoir are infinite amount of heat capacity system.Temperature of thermal reservoir remain constant always.

So we can say that heat engine can not operate by using only one thermal reservoir.

5 0

2 years ago