Is the force of gravity stronger on a piece of crumpled paper or a normal piece of flat paper?
1 answer:
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Answer:
dr/dt = -2 cm/s.
Explanation:
The volume of a cone is given by:
(1)
- r is the radius
- h is the height
Let's take the derivative with respect to time in each side of (1).
(2)
We know that:
- dh/dt = 10 cm / s (rate increasing of height)
- dV/dt = 0 (constant volume means no variation with respect of time)
- r = 4 cm
- h = 10 cm
We can calculate how fast is the radius changing using the above information.
Therefore dr/dt will be:
The minus signs means that r is decreasing.
I hope it helps you!
The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:
Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)
where
Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height
Part A.
Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m
Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:
y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds
a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s
Part C.
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m
Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)
where
Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height
Part A.
Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m
Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:
y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds
a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s
Part C.
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m