S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
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Answer:
b) 338 N
Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.
the force required to accelerate the gymnast is given by:
F = m×a
= (45.0)×(7.50)
= 337.5 N
Therefore, the force a trampoline has to apply is 138 N.
The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.
<h3>What is friction force?</h3>
The friction force is the opposing force which acts on the object which is in relative motion.
The driving force is equal and opposite to the friction force acting between road and bicycle.
Friction force = 400N
The friction force between rider and bike is zero.
So the magnitude of sum of friction force = 400N +0 = 400N
Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.
Learn more about friction force.
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Yes that looks correct to me. good luck!!