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3 years ago

What is Parabolic flight

Physics

2 answers:

andreyandreev [35.5K]3 years ago

6 0

Answer:

It's on graphs

Explanation:

During such parabolic flight an aircraft flies a trajectory that provides freefall for up to 40 seconds. Parabolic flight generates freefall by following a trajectory wherein the acceleration of the aircraft cancels the acceleration due to gravity along the aircraft vertical (z) axis.

blagie [28]3 years ago

3 0

This aircraft is used to train astronauts in zero-g maneuvers, giving them about 25 seconds of weightlessness out of 65 seconds of flight in each parabola. During such training, the airplane typically flies about 40–60 parabolic maneuvers

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what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current

Sophie [7]

One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

R = 480 Ohms

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.

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3 years ago

Suppose that a constant force is applied to an object. Newton's Second Law of Motion states that the acceleration of the object

omeli [17]

(9 kg)(5 m/s^2) = M(3 m/s^2)
that the acceleration of the object varies inversely with its mass.

4 0

3 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

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3 years ago

g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and

cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

= 20(11.8)

= 236N

4 0

3 years ago

What value do we use to describe acceleration due to gravity

andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

7 0

3 years ago