The velocity is given by:
V = √(Vx²+Vy²)
V = velocity, Vx = horizontal velocity, Vy = vertical velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for V:
V = √(6²+12²)
V = 13.42m/s
Now find the direction:
θ = tan⁻¹(Vy/Vx)
θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for θ:
θ = tan⁻¹(12/6)
θ = 63.4°
The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.
Answer:
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Explanation:
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
Answer:
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Explanation:
