Ask question

Ask question

All categories

2 years ago

15

A "home-made" solid propellant rocket has an initial mass of 9 kg; 6.8 kg of this is fuel. The rocket is directed vertically upw

ard from rest, burns fuel at a constant rate of 0:225 kg=s, and ejects exhaust gas at a speed of 1980 m=s relative to the rocket. Assume that the pressure at the exit is atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 s and the distance traveled by the rocket in 20 s. Plot the rocket speed and the distance traveled as functions of time.

1 answer:

nlexa [21]2 years ago

6 0

Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg (Initial mass)

me = 0.225 Kg/s (Rate of fuel consumption)

ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)

⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒ y = 741192.997 m = 741.19 km

The graphs are shown in the pics.

You might be interested in

A flying mosquito hits the windshield of a moving car and gets smashed, but the car is intact. Which of the following statements

zhannawk [14.2K]

Answer:

Action and reaction are the same

Explanation:

Let us carefully analyze the situation in the context of Newton's third law, when the car hits the mosquito exerts a force that acts on the mosquito, ACTION and the mosquito responds to the force that is being applied with a force of equal magnitude and direction or opposite applied to the car (Reaction).

Consequently, the correct answer is: The force of the car on the mosquito is of the same magnitude as that of the mosquito on the car

6 0

2 years ago

A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma

Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

= g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x .50 x ( 10+6 )

= 78.4 N

5 0

2 years ago

A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o

AlekseyPX

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

$P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB$ (1)

so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

$P_{out} = 0.100 W = 100 mW$

3 0

2 years ago

Zamir and Talia raced through a maze. Zamir walked 2 m north, 2 m east, 4 m south, 2 m east, 4 m north, 2 m east, 3 m south, 4 m

Alja [10]

Answer : Zamir's displacement and Talia's displacement is equal.

Explanation :

Displacement is explained to be the changing position of an object.

Zamir covers total distance 27 m and Talia covers total distance 19 m but Zamir's initial and final position and Talia's initial and final position is same.

So, we can say that Zamir's displacement and Talia's displacement is equal.

6 0

1 year ago

Read 2 more answers

For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou

Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0

2 years ago