Grade 11Mechanics
the coefficint of friction is 0.75.if sin 37’=0.6,the angle of friction is?
the coefficint of friction is 0.75.if sin 37’=0.6,the angle of friction is?
4 Answers
Vikas TU
sin37 = 3/5
so cos37 = root(1 - (3/5)^2) = root( 1- 9/25) = root(16/25)
cos37 = 4/5
so tan37 = 3/4 = 0.75
angle of friction t = arc[tan(coefficient of friction)]
t = arc[tan(3/4)]
t = 37 degree
so cos37 = root(1 - (3/5)^2) = root( 1- 9/25) = root(16/25)
cos37 = 4/5
so tan37 = 3/4 = 0.75
angle of friction t = arc[tan(coefficient of friction)]
t = arc[tan(3/4)]
t = 37 degree
Khimraj
sin37 = 3/5
so cos37 = root(1 - (3/5)^2) = root( 1- 9/25) = root(16/25)
cos37 = 4/5
so tan37 = 3/4 = 0.75
angle of friction t = arc[tan(coefficient of friction)]
t = arc[tan(3/4)]
t = 37 degree
so cos37 = root(1 - (3/5)^2) = root( 1- 9/25) = root(16/25)
cos37 = 4/5
so tan37 = 3/4 = 0.75
angle of friction t = arc[tan(coefficient of friction)]
t = arc[tan(3/4)]
t = 37 degree
aswanth nayak
Dear Student,
Suppose the compression is x.The kinetic energy before colliding with spring is 12mv2=12×100×1032=15000 JThe potential energy after compression is 12kx2=250x2Frictional force is 0.75×100×10=750 N (taking g=10 ms-2)So energy loss due to friction is Ef=F.x=75xEquating total energy we have750x+150x2=15000⇒x=-12.8, 7.8Here negative sign means expansion. So the answer is 7.8 m. t = arc[tan(3/4)] ,t = 37 degree
aswanth nayak
Suppose the compression is x.
The kinetic energy before colliding with spring is 12mv2=12×100×1032=15000 J.
The potential energy after compression is 12kx 2 = 250 x 2.
Frictional force is 0.75×100×10=750 N (taking g=10 ms-2).
So energy loss due to friction is Ef=F.x=75x.
Equating total energy we have750x+150x2=15000⇒x=-12.8, 7.8.
Here negative sign means expansion.
So the answer is 7.8 m.
t = arc[tan(3/4)] ,
t = 37 degree
Regards
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