A potential divider is an arrangement of resistors in series across a given potential difference to provide a known fraction of the potential difference. A resistor with a sliding contact can be used as a potential divider. The arrangement provides a continuously variable potential difference from zRead more
We know, Gravitational force=GM1×M2/r^2…..(1) Let,m be mass of meteor Me= mass of earth. Using Newton’s Second Law of motion we can rewrite (1) as Gravitational force(F)=ma…..(2) Comparing1) and (2) we get, a=FG/m a=(GMe)/(r^2) Inserting various values we get a={(6.67408×10^−11)×(6×10^24)}/{(10^7)^2Read more
We know, Gravitational force=GM1×M2/r^2…..(1) Let,m be mass of meteor Me= mass of earth. Using Newton’s Second Law of motion we can rewrite (1) as Gravitational force(F)=ma…..(2) Comparing1) and (2) we get, a=FG/m a=(GMe)/(r^2) Inserting various values we get a={(6.67408×10^−11)×(6×10^24)}/{(10^7)^2Read more
Since we know, W=mg Considering W=600N and g= 9.81m/s^2 in case 1 Mass of the object will be 61.16 kg Again considering W=300N and M=61.16 kg Then g=4.905m/s^2 (Using W=mg) Now we know g1/g2= (R2/R1)^2 9.81/4.905 = (6400+x/6400)^2 Solving we get, X=2648.20 km( from the surface of the earth)
reasoning question
Samikshya Thakuri
When the earth stops rotating, the value of g increases at the equator.
Resistance(Current electricity)
Samikshya Thakuri
Given, Current(I)=20mA=20/1000=0.02 A Voltage of battery (V) =12V Output voltage (V2) =3 V We know, IR2=V2 or, R2=3/0.02 Therefore, R2= 150 ohm Again, VR2/(R1+R2) =V2 or, (12×150)/150+R1=3 or,1800=3R1+450 or,R1=1350/3 Therefore,R1=450 ohm Next, Rt(R1+R2) = 150+450=600 ohm
Resistance(Current electricity)
Samikshya Thakuri
A potential divider is an arrangement of resistors in series across a given potential difference to provide a known fraction of the potential difference. A resistor with a sliding contact can be used as a potential divider. The arrangement provides a continuously variable potential difference from zRead more
Quiz
Samikshya Thakuri
Deer(Cervidae) can breathe through both nose and eyes.
Quiz
Samikshya Thakuri
Butterfly (Papilionoidea) have largest number of eyes in the world.
Fact
Samikshya Thakuri
The three words mostly used in English language are : 1)Hello 2)Stop 3)Taxi
Social
Samikshya Thakuri
Balbhadra Kunwar (30 January 1789 – 13 March 1823) is known as "HERO OF KHALANGA".
Force numerical
Samikshya Thakuri
We know, Gravitational force=GM1×M2/r^2…..(1) Let,m be mass of meteor Me= mass of earth. Using Newton’s Second Law of motion we can rewrite (1) as Gravitational force(F)=ma…..(2) Comparing1) and (2) we get, a=FG/m a=(GMe)/(r^2) Inserting various values we get a={(6.67408×10^−11)×(6×10^24)}/{(10^7)^2Read more
Force numerical
Samikshya Thakuri
We know, Gravitational force=GM1×M2/r^2…..(1) Let,m be mass of meteor Me= mass of earth. Using Newton’s Second Law of motion we can rewrite (1) as Gravitational force(F)=ma…..(2) Comparing1) and (2) we get, a=FG/m a=(GMe)/(r^2) Inserting various values we get a={(6.67408×10^−11)×(6×10^24)}/{(10^7)^2Read more
force Numerical
Samikshya Thakuri
Since we know, W=mg Considering W=600N and g= 9.81m/s^2 in case 1 Mass of the object will be 61.16 kg Again considering W=300N and M=61.16 kg Then g=4.905m/s^2 (Using W=mg) Now we know g1/g2= (R2/R1)^2 9.81/4.905 = (6400+x/6400)^2 Solving we get, X=2648.20 km( from the surface of the earth)