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2 years ago

11

A cargo spacecraft was orbiting the earth. It is equipped with rocket and has an initial mass of 30 metric ton. It wants to catc

h up with the international space station to deliver the cargo. So its rocket engine was fired for 1 minute. The engine ejects mass at a rate of 30 kg/s with an exhaust velocity of 3.1 km/s. The pressure at the nozzle exit is 5 kPa and the diameter of exit area is 94.4 cm. What is the thrust of the engine in a vacuum

1 answer:

Burka [1]2 years ago

5 0

Answer:

F = 96.5 kN

Explanation:

F = dm/dt v + PA

F = 30 kg/s(3100 m/s) + 5000(0.994/2)²π

F = 93000 + 3,499.482 N

F = 96.5 kN

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What determines the state of matter for any substance

Andrews [41]

Amount of molecules present in the substance

Explanation:

In general state of matter is classified into three types based on the amount of molecules present in the matter. They are widely classified as

SOLID
LIQUID
GAS

SOLID

In solid the molecules are tightly packed without any free space to move inside. Therefore Solids have strong Inter-Molecular Force of attraction and have definite shape and volume.

LIQUID

In Liquid the molecules are not tightly packed. So that molecules move freely inside the matter. Therefore Liquids have Weaker Inter-Molecular Force of attraction and have no definite shape and no definite volume.

GAS

In Gas the molecules are loosely packed with more free space to move inside. Therefore Gas have the weakest Inter-Molecular Force of attraction and don't have definite shape and definite volume.

7 0

2 years ago

Financial planning begins with

suter [353]

<h3><u>Answer;</u></h3>

C. a solid foundation

<h3><u>Explanation</u>;</h3>

<em><u>Financial planning is the process of determining the objectives; policies, procedures, program and budgets to deal with the financial activities of an enterprise.</u></em>
Financial planning is an important part of financial planning. It reflects the needs of the business and is integrated with the overall business planning.
it involves taking certain important decisions so that finances are continuously available to the company and are used efficiently.

3 0

2 years ago

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.

iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g

8 0

3 years ago

Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in

Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0

2 years ago

How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?

Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations. We are going to use vf = vi + at. Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.

5 0

3 years ago