Answer:
Explanation:
radius of circle r = 0.9 m.
(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .
( b )
Tension in string T = m ω²r
Putting the values
60 = .072 x ω² x 0.9
ω² = 926
ω = 30.4 rad /s
angle made in 20 revolutions θ = 20 x 2π = 126.6 rad
time taken = θ / ω
= 126.6 / 30.4
= 4.16 s .
Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
By law of refraction we know that image position and object positions are related to each other by following relation
here we know that
now by above formula
so apparent depth of the bottom is seen by the observer as h = 3.39 cm
Answer:
sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Explanation:
We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis
So x component of the vector 
y component of the vector 
So vector will be 6.06i+3.5j
Now other vector of length of 7 units and makes an angle of 120° with positive x-axis
So x component of vector 
y component of the vector 
Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
-- In order to achieve constant verlocity, the net force on the mass must be zero. So if there ARE any forces acting on it, they must be balanced.
-- There is already a force on the mass that can't be eliminated . . . the force of gravity.
-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = <em><u>245N</u></em> in the <u><em>downward</em></u> direction.
-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of <em>245N</em>, all in the <em>upward</em> direction.
-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.
