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lina2011 [118]
4 years ago
11

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

Physics
1 answer:
VARVARA [1.3K]4 years ago
8 0

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

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Physics is a branch of science which deals with........... in relation to.................
dybincka [34]

Answer:

Physics is the branch of science that deals with the structure of matter and how the fundamental constituents of the universe interact.

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3 0
3 years ago
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A 780g, 50-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short but
Semmy [17]

Answer:

9.6 rad/s

Explanation:

L = length of the metal rod = 50 cm = 0.50 m

M = Mass of the long metal rod = 780 g = 0.780 kg

Moment of inertia of the rod about one end is given as

I = \frac{ML^{2}}{3} = \frac{(0.780)(0.50)^{2}}{3} = 0.065 kgm^{2}

F = force applied by the hammer blow = 1000 N

Torque produced due to the hammer blow is given as

\tau = \frac{FL}{2}

\tau = \frac{(1000)(0.50)}{2}

\tau = 250 Nm

t = time of blow = 2.5 ms = 0.0025 s

w = Angular velocity after the blow

Using Impulse-change in angular momentum, we have

I w = \tau t\\(0.065) w = (250) (0.0025)\\w = 9.6 rads^{-1}

5 0
3 years ago
To give an idea of sensitivity of the platypus's electric sense, how far from a 80nC n C point charge does the field have this m
SVETLANKA909090 [29]

The question is incomplete. The complete question is :

A platypus foraging for prey can detect an electric field as small as 0.002 N/C.

-To give an idea of sensitivity of the platypus's electric sense, how far from a +80nC point charge does the field have this magnitude?

Solution :

Given electric field,  E = 0.002 N/C

Charge, Q = + 80 nC

$\therefore E = \frac{kQ}{R^2} $

or $R^2=\frac{kQ}{E}$

    $R^2=\frac{9\times 10^9 \times 80 \times 10^{-9}}{0.002}$

   R = 600 m

This is the distance of the charge from the point of observations.

5 0
3 years ago
Which is the equivalent resistance of the circuit<br><br> shown below?
Misha Larkins [42]
1/Rt=1/R1+1/R2+1/R3, 1/Rt=1/3+1/12+1/4=2/3, Rt=equivalent resistance= 1.5 ohms
8 0
3 years ago
A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin
erastovalidia [21]

Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's  = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫   dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

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B = 3.44 x 10⁻⁵ x 105.5  / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

8 0
3 years ago
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