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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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2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compres

lutik1710 [3]

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3 years ago

Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that

Ainat [17]

Answer:

(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

Explanation:

Solution

(a) Let Mn denotes the number of maintenance visits after the nth journey

Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3

We therefore, notice that M follows an arithmetic sequence

So,

Mn = M₁ + (n-1) (M₂ -M₁)

= 1 + (n- 1) 1 = n

or Mn =n

(b) For what value of n we will get fro Mn > 10

Thus,

n > 10 (exceed 10) or n =11

(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused

So,

n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

7 0

3 years ago

Write a Python program that does the following. Create a string that is a long series of words separated by spaces. The string i

dybincka [34]

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

$\\$ Then I used methods append(), insert() and extend() for adding elements to the list.

$\\$ Finally I converted list into a string using join() and adding space in between the elements of the list.

4 0

4 years ago

A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2

lora16 [44]

(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

Z = √[R²+(2πfL)²]........................ Equation 1

Where:

R = resistance of the coil
f = Frequency
L = Inductance of the coil
Z = Overall impedance of the circuit

From the question,

Given:

R = 7 ohms
L = 31.8 mH = 0.0318 H
f = 50 Hz
π = 3.14

Substitute these values into equation 1

Z = √[7²+(2×3.14×50×0.0318)²]
Z = √(49+99.7)
Z = √(148.7)
Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

I = V/Z.................. Equation 2

Where:

V = Voltage
I = current in the circuit.

Given:

V = 230 V.

Substitute into equation 2

I = 230/12.19
I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

∅ = Phase angle.

Substitute into equation 3

∅ = tan⁻¹(2×3.14×50×0.0318/7)
∅ = tan⁻¹(9.9852/7)
∅ = tan⁻¹(1.426)
∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

pf = cos∅............ Equation 4

Where:

pf = power factor.

Substitute the value of ∅ into equation 4

pf = cos(54.97°)
pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

P = V×I×pf................ Equation 5

Where:

P = The power consumed

Substitute the values into equation 5

P = 230(18.87)(0.574)
P = 2491.22 W

Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

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2 years ago

Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil

Karo-lina-s [1.5K]

Answer:

$E = \frac{3Q}{2A\epsilon_0}$

Explanation:

By Gauss Law for electric field:

$E = \frac{\sigma}{2\epsilon_0}$

Where $\sigma$ is the charge density Q/A. Since we have 2 parallel plates with different charge, the electric field at point P in the gap would be the sum of 2 field

$E = E_1 + E_2$

$E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}$

$E = \frac{3Q}{2A\epsilon_0}$

5 0

3 years ago