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vlabodo [156]
3 years ago
12

A 3600 kg rocket traveling at 2900 m/s is moving freely through space on a journey to the moon. The ground controllers find that

the rocket has drifted off course and that it must change direction by 11◦ if it is to hit the moon. By radio control the rocket’s engines are fired instantaneously (i.e., as a single pellet) in a direction perpendicular to that of the rocket’s motion. The gases are expelled (i.e., the pellet) at a speed of 4300 m/s (relative to the rocket). What mass of gas must be expelled to make the needed course correction? Answer in units of kg.
Physics
1 answer:
Nana76 [90]3 years ago
7 0

Answer:

m=417.24 kg

Explanation:  

Given Data

Initial mass of rocket  M = 3600 Kg

Initial velocity of rocket vi = 2900 m/s  

velocity of gas vg = 4300  m/s

Θ = 11° angle in degrees

To find

m = mass of gas  

Solution

Let m = mass of gas    

first to find Initial speed with angle given

So

Vi=vi×tanΘ...............tan angle

Vi= 2900m/s × tan (11°)

Vi=563.7 m/s

Now to find mass

m = (M ×vi ×tanΘ)/( vg + vi tanΘ)

put the values as we have already solve vi ×tanΘ

m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)

m=417.24 kg

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Mechanical advantage allows you to apply a force over a short distance to increase the distance and object moves.

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When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance b
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When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

<h3>Newton's law of universal gravitation</h3>

Newton's law of universal gravitation states that the force of attraction between two masses in the universe is directly proportional to the product of the masses and inversely proportional to the the square of the distance between them.

The mathematical interpretation of the above law is

  • F ∝Mm/r²

Removing the proportionality sign,

  • F = GMm/r².

Where:

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  • G = Gravitational constant
  • M = Bigger mass
  • m = Smaller mass
  • r = Distance between the masses.

From the above, When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

Learn more about Newton's law of universal gravitation here: brainly.com/question/9373839

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7 0
2 years ago
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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Answer:

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An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same
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Answer:

v_2=\sqrt{2}v_1

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

v_1^{2} =v_0^{2} +2gh

Since the object is initially at rest, v₁ becomes:

v_1=\sqrt{2gh}

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

v_2^{2}=v_1^{2} +2gh

Substituting v₁ in this expression and solving for v₂, we get:

v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}

Now, dividing v₂ over v₁, we get the expression:

\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\   \\\implies v_2=\sqrt{2}v_1

It means that v₂ is √2 times v₁.

4 0
3 years ago
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