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3 years ago

What does it take to launch a rocket in space?

Engineering

2 answers:

il63 [147K]3 years ago

5 0

You need more than enough fuel engine to blast you up in to space for if anything goes wrong you have enough

Dafna11 [192]3 years ago

3 0

To get rockets into orbit, they need much more thrust than the amount that will get them up to the required altitude. They also need sufficient thrust to allow them to travel with very high orbital speed. ... If speed is less than this, an object will fall back to the Earth

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Cast iron has about how much carbon content?

Aliun [14]

Answer:

Explanation:

7 0

3 years ago

A 0.50 m3 drum was filled with 0.49 m3 of liquid water at 25oC and the remaining volume was water vapor without any air. The dru

timurjin [86]

Answer:

There is not going to be pressure build up in the system,that is isobaric process.

Explanation:

Assumptions to be made

1. No mass is gained or lost during the heating process.

2. There are no friction losses,so work is transmitted efficiently.

3. It was started the water in the drum and its surrounding have same temperature.

4. This system is closed,so there is no mass transfer across its boundaries.

5 0

3 years ago

) In a disk test performed on a specimen 32-mm in diameter and 7 mm thick, the specimen fractures at a stress of 680 MPa. What w

Radda [10]

Answer:

Sorry it doesnt tall me anythikng

Explanation:

6 0

2 years ago

I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going to tr

Alina [70]

Answer:

Δx = 25 ft.

Explanation:

Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:

1 ) Egg falling:

If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:

yf-y₀ = 1/2* g* t²

If we take the up direction as positive, we can solve for t as follows:

0-100 ft = 1/2* (-32.15 ft/s²)* t²

⇒ t = $\sqrt{(100*2)/32.15}$ = 2.5 sec.

2) Person on the ground running away:

In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.

We can get the distance at which he can reach, applying the definition of velocity:

v = (xf-x₀) / (tfi-t₀)

If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:

xf = v*t = 20 ft/sec*1.25 sec = 25 ft.

8 0

2 years ago

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago