Ball A was carried to the top of a hill in a straight line, while ball B was carried in a longer, zigzag path. At the top of the
2 answers:
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3,89,988 cm/min is the linear velocity
Given,
Diameter of CD = 12 cm
So, Radius of CD = 6 cm
CD is spinning at 10350 rev/min
Firstly , convert rev/min into rad/min
1 rev = 2π radians
10350 rev/min = 10350 × 2π
= 64998 rad/min
Formula used,
where,
is the Linear velocity
is the radius
is the angular velocity
= 6 cm × 64998rad/min
= 3,89,988 cm/min
Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.
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Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F =
F =
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹
θ = tan⁻¹ 0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East