А______ is a unit of measurement for energy. (7 Letters)
1 answer:
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Answer:
1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s
Explanation:
1) First Scenario.
After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.
Before starting, let's reduce the magnitudes to the SI system
v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s
y = 2650 m
Let's start by looking for the time it takes for the load to reach the ground.
y = y₀ + v_{oy} t - ½ g t²
in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero
0 = y₀ + 0 - ½ g t2
t =
t = √(2 2650/9.8)
t = 23.26 s
this is the horizontal scrolling time
x = v₀ t
x = 366.67 23.26
x = 8527 m
the speed at the point of arrival is
v_y = v_{oy} - g t = 0 - gt
v_y = - 9.8 23.26
v_y = -227.95 m / s
Module and angle form
v =
v = √(366.67² + 227.95²)
v = 431.75 m / s
θ = tan⁻¹ (v_y / vₓ)
θ = tan⁻¹ (227.95 / 366.67)
θ = - 31.97º
measured clockwise from x axis
We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.
2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º
we look for the components of speed
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v₀ sin θ
we look for the time for the arrival point that has coordinates x = 0, y = 0
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + vo sin θ t - ½ g t²
0 = -50 + vo sin 50 t - ½ 9.8 t²
x = x₀ + v₀ₓ t
0 = x₀ + vo cos θ t
0 = -400 + vo cos 50 t
podemos ver que tenemos un sistema de dos ecuación con dos incógnitas
50 = 0,766 vo t – 4,9 t²
400 = 0,643 vo t
resolved
50 = 0,766 ( ) t – 4,9 t²
50 = 476,52 t – 4,9 t²
t² – 97,25 t + 10,2 = 0
we solve the quadratic equation
t = [97.25 ± ] / 2
t = 97.25 ±97.04] 2
t₁ = 97.145 s
t₂ = 0.1 s≈0
the correct time is t1 the other time is the time to the launch point,
t = 97.145 s
let's find the initial velocity
x = x₀ + v₀ cos 50 t
0 = -400 + v₀ cos 50 97.145
v₀ = 400 / 62.44
v₀ = 6.4 m / s
Explanation:
<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively. Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>
At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.
At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.
<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>
Energy is conserved:
PE = PE + KE + RE
mgH = mgR + ½mv² + ½Iω²
mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²
mgH = mgR + ½mv² + ⅕mv²
gH = gR + ⁷/₁₀ v²
v² = 10g(H−R)/7
v = √(10g(H−R)/7)
<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>
Sum of forces in the radial (-x) direction:
∑F = ma
N = mv²/R
N = m (10g(H−R)/7) / R
N = 10mg(H−R)/(7R)
<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>
RE / KE
= (½Iω²) / (½mv²)
= ½(⅖mr²)(v/r)² / (½mv²)
= (⅕mv²) / (½mv²)
= ⅕ / ½
= ⅖
<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin. The sphere is replaced with a hoop of the same mass and radius. Will the value of Hmin increase, decrease, or stay the same? Justify your answer.</em>
When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0. Sum of forces in the radial (-y) direction:
∑F = ma
mg = mv²/R
gR = v²
Applying conservation of energy:
PE = PE + KE + RE
mgH = mg(2R) + ½mv² + ½Iω²
mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²
mgH = 2mgR + ½mv² + ½kmv²
gH = 2gR + ½v² + ½kv²
gH = 2gR + ½v² (1 + k)
Substituting for v²:
gH = 2gR + ½(gR) (1 + k)
H = 2R + ½R (1 + k)
H = ½R (4 + 1 + k)
H = ½R (5 + k)
For a sphere, k = 2/5. For a hoop, k = 1. As k increases, H increases.
<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above. The track makes an angle of θ above the horizontal at point C. Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate. Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>
C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.
v₀ = √(10g(H−R)/7)
The vertical component of this initial velocity is v₀ sin θ. At the maximum height, the vertical velocity is 0. During this time, the sphere is in free fall. The maximum height reached is therefore:
v² = v₀² + 2aΔx
0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)
0 = 10g(H−R)/7 sin²θ − 2g(h − R)
2g(h − R) = 10g(H−R)/7 sin²θ
h − R = 5(H−R)/7 sin²θ
h = R + ⁵/₇(H−R)sin²θ
