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2 years ago

A rocket is being launched straight up. Air resistance is not negligible.

2 answers:

7 0

A rocket is launching upwards. The force acting on the rocket to move upward is F. The forces acting opposite to thrust are weight and drag force. The drag force is due to the air resistance. In order to move upwards the thrust force must be greater than weight and drag force. Figure shows the free body diagram of rocket at the time of launching.There are only three forces: weight (always present), thrust (provided from the rocket’s engines to lift it up) and air resistance (said specifically to be not negligible). There are no horizontal forces.

Yanka [14]2 years ago

3 0

Answer:

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What scientist shot alpha particles into gold

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Bromium has two naturally occurring isotopes: 79br, with an atomic weight of 78.918 amu, and 81br, with an atomic weight of 80.9

saul85 [17]

The two different isotopes have weights :

w1 = 78.918 amu

w2 = 80.916 amu

average weight w3 = 79.903 amu

The mixing of two components can be modeled as

let the fraction of w1 be 'x'

hence $w1. x + w2.(1-x) = w3$

now this is a linear equation in 'x'. Substituting the values we get

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3 years ago

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

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Explanation:

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we only see wavelengths from 400–700 nanometers.

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