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3 years ago

How would cars fly what would keep them flying

Physics

2 answers:

BartSMP [9]3 years ago

7 0

Answer:

if people somehow try puting airplane parts into cars than they wud use the same ways to fly etc.

Explanation:

diamong [38]3 years ago

6 0

Two weeks ago I would have said flying cars were still firmly in the realm of techno-utopian fantasy, as they have been for decades. Now I’m not quite so sure.

In the coming few years nearly 20 small airborne vehicles are supposedly hitting the market (see table below). Some are drone-like, with anywhere from four to 18 rotors keeping them aloft. Most are fixed-wing craft with propellers that point upwards for vertical takeoff and landing (VTOL), and tilt forward for flight.

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Kayla drew two ray diagrams to compare the behavior of light rays.

Naddika [18.5K]

The answer is the last option

Explanation:

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3 years ago

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Butoxors [25]

Solutions are basically a release from a problem. This is more than helpful.

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3 years ago

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Balok diam diatas bidang miring pada sudut kemiringan 40° balok mulai bergerak,tentukan koefisien gesek statis antara balok dan

Pachacha [2.7K]

Answer:

0.84

Explanation:

m = Massa balok

g = Percepatan gravitasi

$\theta$ = Sudut kemiringan

$\mu$ = Koefisien gesekan statik antara balok dan bidang miring

Gaya balok karena beratnya diberikan oleh

$F=mg\sin\theta$

Gaya gesekan diberikan oleh

$f=\mu mg\cos\theta$

Kondisi dimana balok mulai bergerak adalah ketika gaya balok akibat beratnya sama dengan gaya gesek pada balok.

$mg\sin\theta=\mu mg\cos\theta\\\Rightarrow \mu=\dfrac{mg\sin\theta}{mg\cos\theta}\\\Rightarrow \mu=\tan\theta\\\Rightarrow \mu=\tan40^{\circ}\\\Rightarrow \mu=0.84$

Koefisien gesekan statik antara balok dan bidang miring adalah 0.84.

7 0

3 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

3 years ago

A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust i

Lerok [7]

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

5 0

2 years ago