please solve this ques. 11 for me
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I am solving the 2nd
(i) No the mass of metallic bar doesn't change when it is taken to equator as mass of a body is fixed
(ii)Yes, its weight will change when taken to equator as weight of body is maximum at poles and minimum at equator.
(iii) No , because the gravity is not zero on earth.
(i) No the mass of metallic bar doesn't change when it is taken to equator as mass of a body is fixed
(ii)Yes, its weight will change when taken to equator as weight of body is maximum at poles and minimum at equator.
(iii) No , because the gravity is not zero on earth.
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Hey!
(I)When the mass of an one object is double, then
force will be ,
can we use the formula :F=Gm1m2/r^2......(I)
then force (f)2=G*2m'1*m'2/r^2........(2)
from equation (1) and(2)
F1/F2=Gm1m2/r^2×r^2/Gm'1*2m'
then F1/F2=1/2..Ans.
(ii) when the distance between the object is double.then,
F1=Gm1m2/r^2...........(1)
so, distance will be double.
F2=m'1.m'2/2r^2.....(2)
Frome equation(1) and(2)
then, F1/F2=4/1.
similarly distance will be triple .then
force=9 times.
(iii) when the mass of an both object is double.
then force =Gm1m2/r^2......(1)
mass double is ,then force=G2m'1.2m'2......(2)
form equation(1)and(2)
F1/F2=Gm1m2/r^2×G2m'1.2m'1/r^2
then f1/f2=4/1 Ans..
Hope it helps you
(I)When the mass of an one object is double, then
force will be ,
can we use the formula :F=Gm1m2/r^2......(I)
then force (f)2=G*2m'1*m'2/r^2........(2)
from equation (1) and(2)
F1/F2=Gm1m2/r^2×r^2/Gm'1*2m'
then F1/F2=1/2..Ans.
(ii) when the distance between the object is double.then,
F1=Gm1m2/r^2...........(1)
so, distance will be double.
F2=m'1.m'2/2r^2.....(2)
Frome equation(1) and(2)
then, F1/F2=4/1.
similarly distance will be triple .then
force=9 times.
(iii) when the mass of an both object is double.
then force =Gm1m2/r^2......(1)
mass double is ,then force=G2m'1.2m'2......(2)
form equation(1)and(2)
F1/F2=Gm1m2/r^2×G2m'1.2m'1/r^2
then f1/f2=4/1 Ans..
Hope it helps you
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