Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
Answer:
bounce up and down
Explanation:
Buoys are used for two main reasons, one is to let the people on land know of a big incoming wave, while the second reason is to generate electricity. When a big wave is approaching the buoy starts to bounce up and down with the strength of the smalled previous waves and then bounce very strongly up as the bigger wave passes by. This movement is combined with pistons within the buoy in order to conduct electricity.
Metallic bonding<span> is the force of attraction between valence electrons and the metal ions. It is the sharing of many detached electrons between many positive ions,
Hopefully this can help you understand
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Answer:
e.26m/s
Explanation:
Vf=Vi+at (1)
Vf=9j+(2i-4j)t
X= X₀+at
now, in the i direction
15=O+2t or t=7.5 when x position is 15
Lets put that into the (1) equation, solve for Vf.
Vf=9j+(2i-4j)7.5
Vf= 15i - 21j
Speed= 
Vf= 25.8 m/s
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math
at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
(1/2)mv^2 = umgh x cot(a)
u = (1/2)v^2 x tan (a) / gh
