essay about mobile phone in kannada​

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Here is your answer :

C) Lizard

Saurology is the study of ___________.

(A) Mosquitoes

(B) Snake

(C) Lizards

(D) Cockroach

answer...c) Lizard

Answer:

1. व्यवहार की वह रीति जिससे अपना कल्याण हो और समाज को भो कोई बाधा न पहुँचे ।
2. क्षमताशाली । 2. काम करने के योग्य । कार्य में समर्थ ।
3. अमूल्य संस्कृत [विशेषण] 1. जिसका मूल्य न हो ; अनमोल 2. बहुमूल्य ; मूल्यवान ; बेशकीमती 3. श्रेष्ठ ; उत्तम।

Explanation:

Answer:

व्यवहार की वह रीति जिससे अपना कल्याण हो और समाज को भो कोई बाधा न पहुँचे ।

क्षमताशाली । 2. काम करने के योग्य । कार्य में समर्थ ।

अमूल्य संस्कृत [विशेषण] 1. जिसका मूल्य न हो ; अनमोल 2. बहुमूल्य ; मूल्यवान ; बेशकीमती 3. श्रेष्ठ ; उत्तम।

Answer:

Winter vacation is a time to relax and celebrate the time in the company of your family and friends. Usually, winter vacations comes right after the second term exams are over. Thus, it helps in providing much-needed respite to students after a month of rigorous hard work. Students usually await the winter vacations as it indulges them a lot of fun stuff during the break. The essay on a winter vacation is an insight into the minds of students activities during the break. Winter vacations usually consist of 15 days and include 2 important festivals which are Christmas and New year. So, the holiday season is full of festival flavors. Thus, many students look forward to Christmas as their parent brings gift flr them. Also, the entire home is illuminated with lights and other decorations. Additionally, people also shop for stars, small bells, candies, and reindeers to put it the Christmas and decorate it.

Also, kids look forward to the cake and sweets that are prepared by their mother on Christmas. Many families celebrate the festival by having a family dinner and thus the vibe around the festival is great. The new year is also special for kids as they either invite or are invited to some relatives or friends place. Thus, it is time to meet family, friends, and relatives.

Explanation:

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Hint: In order to solve this problem we need to use the formula of force and then we have to put all the values and get the value of the gravitational constant. The main thing here is to get the force in Newton since it is given in milligram weight so first we will convert the force and then put the values to get the right answer.

Complete answer:

It is said that a sphere of mass 40 kg is attracted by a second sphere of mass 15kg, when their centers are 20cm apart, with a force of 0.1 milligram weight. We need to calculate the value of G here.

The situation of this problem can be denoted in figure as:

The spheres are 20cm(0.2m) apart and the force between them is 0.1miligram weight.

We know that 1 mg weight = 10−3×980dyne.

Therefore 0.1 mg weight = 98 x 10−3dyne

We also know that 1 dyne = 10−5N

So, 98 × 10−3dyne = 98×10−8 = F.

The formula of gravitational force between two masses m1 and m2 and the distance between them as r is,

F = Gm1m2r2.

On putting the respective values we get the values of G as:

⇒G=F(r)2m1m2=98×10−8×0.2×0.240×15=6.53×10−11 .

Therefore the value of gravitational constant G here is 6.53×10−11Nm2kg−2.

Explanation:

Distance(r)= 0.2 cm= 0.002m

m1= 40kg , m2= 15kg

F= 9.8×10^-7 N

G= F×r^2/ m1×m2

G= 9.8×10^-7 × 0.002×0.002 / 40×15

G= 9.8×10^-7 × 4× 10^-6 / 40×15

G= 0.6533× 10^-12

G= 6.533×10^-13