३) टायफून वादळे कोठे निर्माण होतात ?​

Answer:

YAC - Yeast Artificial Chromosome

It is extracted from yeast and used for manipulation in large fragments of DNA like human genome.

hope this will help you

Answer:

YAC - Yeast Artificial Chromosome

Explanation:

Answer:

अभिधा शब्द शक्ति का पहला प्रकार है जो शब्दों के शब्दकोशीय अर्थ का बोध कराती है। इसमें किसी शब्द का सामान्य अर्थ में प्रयोग किया जाता है। जैसे 'सिर पर चढ़ाना' का अर्थ किसी चीज को किसी स्थान से उठाकर सिर पर रखना होगा। साक्षात् सांकेतित अर्थ (मुख्यार्थ या वाच्यार्थ) को प्रकट करने वाली शब्दशक्ति अभिधा शब्दशक्ति कहलाती है।

Answer:

ICM=25MR2 , where ICM is the moment of inertia of the solid sphere

Hope it helps!!!

Explanation:

The moment of inertia of a solid sphere about an axis passing through its center​ is [tex]\frac{2}{5} MR^{2}[/tex].

Explanation:

The moment of inertia (I) of a solid sphere about an axis passing through its center​ is

normally represented as :

Here, R = radius of the solid sphere

         M = mass of the solid sphere.

The moment of inertia is (I) = [tex]\frac{2}{5} MR^{2}[/tex].

The moment of inertia (I) can be also calculated for the tangent of the solid sphere.

This is achieved by using the parallel axis theorem.

The moment of inertia of a solid sphere about an axis passing through its tangent is [tex]\frac{7}{5} MR^{2}[/tex].

Explanation:

processes P1,P2,P3 and P4 with Arrival times 0, 1, 2

and 4 respectively. The Burst times of P1,P2,P3 and P4 are 5, 4,

2 and 1 respectively. Calculate the average turnaround time and

average waiting time

Given are the arrival times and the burst times of four processes, Find their average turnaround and average waiting times using round robin

Explanation:

• In the round-robin scheduling algorithm each process is given a fixed amount of time (time quantum) to run and then the next process is given the same time to run and so on in a cyclic way.
• Meanwhile a queue known as the ready queue is maintained to follow which process is to be scheduled next based on the arriving order.  
• We get the order of processes scheduled and time for which they each run as,  [tex]P1_{0-2}\ \ \ P2_{2-4}\ \ \ P3_{4-6}\ \ \ P1_{6-8}\ \ \ P4_{8-9}\ \ \ P2_{9-11}\ \ \ P1_{11-12}[/tex]  
• Given below is the table for turnaround and waiting times for each process, [tex]Process\ no. \ \ \ \ AT\ \ \ \ BT\ \ \ \ \ CT \ \ \ \ \ TAT(=CT-AT)\ \ \ \ \ WT(=TAT-BT) \\P1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ 5\ \ \ \ \ \ \ \ 12\ \ \ \ \ \ \ \ \ \ 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7 \\P2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ 4\ \ \ \ \ \ \ \ 11\ \ \ \ \ \ \ \ \ \ 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\\[/tex]  [tex]P3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ \ \ \ \ \ 2\ \ \ \ \ \ \ \ 6\ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\\P4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ 9\ \ \ \ \ \ \ \ \ \ 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4[/tex]              
• Hence we get the average waiting and turnaround times as,  [tex]Av._{TAT}=\frac{12+10+4+5}{4} \ \ \ \ \ \ ->Av._{TAT} =7.75 \ units\\Av._{WT}=\frac{7+6+2+4}{4} \ \ \ \ \ \ ->Av._{TAT} =4.75 \ units[/tex]