Hydroplaning is when ______________

(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45

Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m$

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0

3 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago

Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.

Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
 = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·g·sinθ

Therefore:
-(1/2)·m·g·sinθ + m·g·sinθ = μ·m·g·cosθ
(1/2)·m·g·sinθ = μ·m·g·cosθ
μ = (1/2)·m·g·sinθ / m·g·cosθ
 = (1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
 = tan⁻¹(2·0.9)
 = 61°

Therefore, the maximum angle you could ride down without worrying about skidding is 61°.

5 0

2 years ago

How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov

sveta [45]

Answer:

x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / v₀

cos θ = v₀ₓ / v₀

v_{y} = v_{oy} sin θ

v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

x = v₀ₓ t

x = v₀ cos θ t

vₓ = v₀ cos θ

Y axis

y = y₀ + $v_{oy}$ t - ½ g t2

y = y₀ + v₀ sin θ t - ½ g t2

v_{y} = v₀ - g t

v_{y} = v₀ sin θ - gt

$v_{y}^{2}$ = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

7 0

2 years ago

If you lived on Saturn, which planets would exhibit retrograde motion like that observed for Mars from Earth? (Select all that a

liberstina [14]

Answer:

a) Earth

b) Mercury

c) Neptune

Explanation:

All the planets move around the sun in eastward direction, but few planet have retrograde rotation i.e in westward direction. Retrograde motion is just an apparent change in the movement of planet which means it only seems as if the planet are rotating in opposite direction. Retrograde movement of planet like Saturn, Jupiter and mars is not real. Hence, if a person lives on Saturn, then following planets will exhibit retrograde motion

a) Earth

b) Mercury

c) Neptune

4 0

2 years ago

Hydroplaning is when _______________.