Ask question

Ask question

All categories

GenaCL600 [577]

4 years ago

11

If the density of a diamond is 3.5 g/cm", what would be the mass of a diamond whose

1 answer:

alexandr1967 [171]4 years ago

7 0

Answer:

Explanation:

You might be interested in

every computer consists of physical components and non physical components the non physical component of a computer that underst

sp2606 [1]

Answer:

C. software

Explanation:

software, is a collection of data or computer instructions that tell the computer how to work. This is in contrast to physical hardware, from which the system is built and actually performs the work.

4 0

3 years ago

Water is completely filling black metallic vessel having cubic form and thin walls. The mass of water is 1 kg and initial temper

Brilliant_brown [7]

Answer:

Check the attached image

Explanation:

To solve the problem for time you will have to use the formula for time, t = d/s which means time equals distance divided by speed.

Kindly check the attached image below for the step by step explanation to the question.

5 0

3 years ago

Which layer of the earth includes the ozone layer

Rus_ich [418]

Troposphere is the answer

7 0

3 years ago

A car battery has a rating of 250 ampere-hours. This rating is one indication of the total charge that the battery can provide t

Serga [27]

Answer:

Total charge provided by the battery could be 900000 C.

Maximum current provided by the battery for 37 minutes could be 405.405 A

Explanation:

Rating= 250 A-h

a. Total charge:

$Rating=Q/t\\Q=Rating.t\\$

Suppose t=1h

$Q=250 A(1h)\\Q=250 A(3600 sec)\\Q= 900000 A.sec$

We konw that $Ampere=\frac{Coulomb}{sec}$ , replacing:

$Q=900000(\frac{Coulomb}{sec})(sec)\\Q=900000 Coulomb$

Total charge provided by the battery could be 900000 C.

b. Maximum current for 37 minutes

$I=\frac{Q}{t} \\I=\frac{900000 C}{37*60 sec}\\I=405.405 A$

Maximum current provided by the battery for 37 minutes could be 405.405 A

4 0

3 years ago

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t

postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^{\circ}$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

$V_1=1.6198\times 10^{-3}\ m^3$

Final Volume $V_2=4V_1$

$V_2=4\times (1.6198\times 10^{-3})$

$V_2=6.4792\times 10^{-3}\ m^3$

Specific volume at this stage

$\nu _2=\frac{V_2}{m}$

$\nu _2=\frac{6.4792\times 10^{-3}}{1.4}$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})$

$T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)$

$T_2=370.7^{\circ} C$

4 0

3 years ago