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3 years ago

The main difference between turbojets and rocket engines is the fact that

Physics

2 answers:

mafiozo [28]3 years ago

5 0

Answer;

D. rocket engines are not dependent on oxygen from the air.

Explanation;

-Jet engines and rockets work on the same principle. They produce thrust through an internal pressure difference and, as explained by Newton’s Third Law of Motion, eject exhaust gases in an equal and opposite direction.

-The main difference between them is that jets get the oxygen to burn fuel from the air and rockets carry their own oxygen, which allows them to operate in space.

Additionally, Jet engines have two openings (an intake and an exhaust nozzle). Rocket engines only have one opening (an exhaust nozzle).

o-na [289]3 years ago

3 0

The correct answer is D. rocket engines are not dependent on oxygen from the air. Rockets carry their own oxygen.

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Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.

Nataly_w [17]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

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4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

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3 years ago

What type of clouds usually accompany cold fronts?

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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).