Answer:
solved
Explanation:
a) F_net = (F2 - F3)i - F1 j
b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)
= sqrt( (9- 5)^2 + 1^2)
= 4.123 N
c) θ = tan^-1( (Fnet_y/Fnet_x)
= tan^-1( -1/(9-5) )
= -14.036°
Transverse Waves: Displacement of the medium is perpendicular to the direction of propagation of the wave.
Longitudinal Waves: Displacement of the medium is parallel to the direction of propagation of the wave.
Answer:
the maximum height reached by the object from where it was launched is 0.4591 m
Explanation:
initial speed of the object, u= 3 m/s
The velocity at the maximum height will always be 0.
Therefore, final velocity, v= 0 m/s
Using the Newton's equation of motion,
v^2 - u^2 = 2*g*h(max)
0 - u^2= 2*g*h(max)
h(max) = -u^2 /2* g
where g is the gravitational acceleration.
g= - 9.8 m/s^2
substituting the values in equation,
h(max)= - (3*3) / 2*(-9.8)
h(max) = 0.4591 m
the maximum height reached by the object from where it was launched is 0.4591 m
learn more about maximum height reached here:
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