Answer:
a) 6076 m
b) 43.33 m/s
c) 68 m/s
Explanation:
(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius
s = 2r = 2* 3.38km = 6.76 km or 6760 m
(b) The average velocity would be displacement over unit of time
v = s/t = 6760 / 156 = 43.33 m/s
(c) The length of the chord it's swept in 156s is half of the circle perimeter
c = πr = π3.38 = 10.62 km or 10620 m
The airplane average speed is its chord length over a unit of time
c / t = c / 156 = 68 m/s
Answer:
The mechanical advantage of the system is equal to 19.62
Explanation:
The ratio of the force produced by a machine to the force applied to it is called mechanical advantage. In other words it is the ratio of output force to the input force.
In this problem mass=200kg
applied force=100N
input force=100N
output force=
mechanical advantage 
It gives an idea about the efficiency of a mechanical device. It is indeed a measure of force amplification. In block and tackle system an assembly of ropes and pulleys is used to lift loads. When the moving block is supported by a greater number of rope sections the force amplification will be more.
Area is calculated as length times width.
In this case, the width is 5 m and the length is 3 m.
5 • 3 = 15
So the area of Nicole's office is 15 square meters.
Square meters because only length and width is 2 dimensional.
Hope this helps!
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.