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Umnica [9.8K]
4 years ago
13

3.12 ** To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries 60% of its initial mas

s as fuel. (That is, the mass of fuel is 0.6m 0.) What is the rocket's final speed, accelerating from rest in free space, if it burns all its fuel in a single stage?
Physics
1 answer:
madreJ [45]4 years ago
8 0

Answer:

0.916v_{ex}

Explanation:

A multistage rock is also known as a step rocket and it is a form of vehicle that can use more than a rocket stage. The rocket stage typically contains propellant as well as its engine. The final speed of the rocket can be estimated using the equation below:

v = v_{ex} ln(\frac{m_{o}}{0.4m_{o}}) = v_{ex}*ln(2.5) = v_{ex}*0.916 = 0.916v_{ex}

Therefore, the maximum speed of the rocket is 0.916v_{ex}

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A motor has an internal resistance of 12.1 Ω. The motor is in a circuit with a current of
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Answer:

Explanation:

V = I * R

V = 4 * 12.1 = 48.4 v

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What shape is the Milky way galaxy
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What is 0.94kg divided by 2.4n
LuckyWell [14K]

Answer:

0.39166666666

Explanation:

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3 years ago
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
3 years ago
How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?​
miss Akunina [59]

Answer:

The answer to your question is    Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

           Pe = mgh

Substitution

           Pe = (50)(5)(9.81)

Simplification

           Pe = 2452.5 J

8 0
3 years ago
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