Answer:10842.33m/s
Explanation:
F=qvBsine
V=f/(qBsine)
V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)
V=10842.33m/s
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Answer:
force for start moving is 7.49 N
force for moving constant velocity 2.25 N
Explanation:
given data
mass = 7.65 kg
kinetic coefficient of friction = 0.030
static coefficient of friction = 0.10
solution
we get here first weight of block of ice that is
weight of block of ice = mass × g
weight of block of ice = 7.65 × 9.8 = 74.97 N
so here Ff = Fa
so for force for start moving is
Fa = weight × static coefficient of friction
Fa = 74.97 × 0.10
Fa = 7.49 N
and
force for moving constant velocity is
Fa = weight × kinetic coefficient of friction
Fa = 74.97 × 0.030
Fa = 2.25 N
Answer:
Explanation:
The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression
ω²R where ω is angular velocity and R is radius of the circular path .
angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.
ω = 2π / T
= 2π / 60
= .1047
angular acceleration = .1047² x .22
= 2.41 x 10⁻³ rad / s² .