Answer
given,
weight of the oak board = 600 N
Weight of Joe = 844 N
length of board = 4 m
Joe is standing at 1 m from left side
vertical wire is supporting at the end.
Assuming the system is in equilibrium
T₁ and T₂ be the tension at the ends of the wire
equating all the vertical force
T₁ + T₂ = 600 + 844
T₁ + T₂ = 1444...........(1)
taking moment about T₂
T₁ x 4 - 844 x 3 - 600 x 2 = 0
T₁ x 4 = 3732
T₁ = 933 N
from equation (1)
T₂ = 1444 - 933
T₂ = 511 N
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(a)
KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J
(b)
The energy is entirely dissipated by the force of friction in the brake system.
(c)
W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J
(d)
Fd = delta KE
F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N
The magnitude of the friction force is 4800 N.
A) visible light because it just makes since
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector
.
The positive y-axis = the northern direction, with unit vector
.
The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
![\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B2%7D%20%3D40%28cos%2834%5E%7Bo%7D%29%5Chat%7Bi%7D%20-%20sin%2824%5E%7Bo%7D%29%5D%5Chat%7Bj%7D%29%20%3D%2033.1615%5Chat%7Bi%7D%20-22.3677%5Chat%7Bj%7D%29)
The plane's actual velocity is the vector sum of the two velocities. It is
The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h
The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°
Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.
