the resultant of two forces at right angle is 5N.when the angle between them is 120°.The resultant is√13 then the forces are
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f1 + f2 = 5N when angle=90 , given
f1+f2= square root(f1^2 + f2^2 + 2f1f2cos90)
5 = square root( f1^2 +f2^2 )
squaring on both sides,
25 = f1^2 + f2^2 - equation 1
f1+f2=square root(13), when angle = 120
square root(13)=square root(f1^2 + f2^2 + 2f1f2cos120)
squaring on both sides,
13 = f1^2 + f2^2 + 2f1f2x(-1/2)
13 = 25 - f1f2 (since f1^2 + f2^2 = 25)
f1f2=13 - equation 2
on solving equations 1 and 2
the two forces are :
(f1=9N and f2=4N) or (f1=4N and f2=9N)
f1+f2= square root(f1^2 + f2^2 + 2f1f2cos90)
5 = square root( f1^2 +f2^2 )
squaring on both sides,
25 = f1^2 + f2^2 - equation 1
f1+f2=square root(13), when angle = 120
square root(13)=square root(f1^2 + f2^2 + 2f1f2cos120)
squaring on both sides,
13 = f1^2 + f2^2 + 2f1f2x(-1/2)
13 = 25 - f1f2 (since f1^2 + f2^2 = 25)
f1f2=13 - equation 2
on solving equations 1 and 2
the two forces are :
(f1=9N and f2=4N) or (f1=4N and f2=9N)
Answered by
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Answer:
Explanation:
1) According, to the third law of motion by newton, every action has an equal and opposite reaction.
1) According, to the third law of motion by newton, every action has an equal and opposite reaction.
2) so, if the bullet exerts a momentum when going forward, same amount of momentum will push the gun backward
3) Hence, the momentum of both would be equal
10 g * v = 1000 g * 5 m/s
v = 500 m/s
2) so, if the bullet exerts a momentum when going forward, same amount of momentum will push the gun backward
3) Hence, the momentum of both would be equal
10 g * v = 1000 g * 5 m/s
v = 500 m/s
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