The resultant vector is 5.2 cm at a direction of 12⁰ west of north.
<h3>
Resultant of the two vectors</h3>
The resultant of the two vectors is calculated as follows;
R = a² + b² - 2ab cos(θ)
where;
- θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
- a is the first vector
- b is the second vector
R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)
R² = 27.02
R = 5.2 cm
<h3>Direction of the vector</h3>
θ = 90 - 78⁰
θ = 12⁰
Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.
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Answer:
161.86 N
Explanation:
mass of box m= 55.0 kg
weight of the box, mg= 55×9.81
g here is acceleration due to gravity =9.81 m/sec^2
coefficient of friction between the box and the surface μ= 0.3
the friction force F_s= μmg= 0.3×55×9.81
=161.86 N
to move the ball horizontal force required is 161.86 N
Answer:
i think it is iron
Explanation:
its the only one that makes sense to me
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
Answer:
C) For every action, there is an equal and opposite reaction
Explanation:
Newton's Third Law
For every force, there is an equal and opposite reaction time. Explains why forces act in pairs.
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