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Zigmanuir [339]
2 years ago
6

For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum ra

nge r of the incoming missile. find their relationship.

Physics
1 answer:
umka2103 [35]2 years ago
6 0
Refer to the diagram shown below.

The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.

Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is t_{f}, then
r=V_{o} \, t_{f}
where r = the range of the missile.

Also, the time, t, when the missile is at ground level is given by
0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}
where g = acceleration due to gravity.

t = 0 corresponds to when the missile is launched. Therefore
t_{f} =  \frac{2V_{o}sin\theta}{g}

Therefore
r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}

Typically, θ=45° to achieve maximum range, so that
r= \frac{V_{o}^{2}}{g}

This analysis is more applicable to a scud missile rather than a powered, guided missile.

Answer:
t_{f} =  \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}
Usually, θ=45°

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat
QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               j_{H_2O}=\frac{D}{l} \bigtriangleup c

where

  • j_{H_2O} is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as 0.25 cm^2/s
  • l is the thickness of the film which is 0.15 cm thick.
  • \bigtriangleup c is given as

                                \bigtriangleup c= \frac{P_{sat}-P_a}{RT}

In this

  • P_{sat} is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • P_a is the air pressure which is given as 0.5 times of P_{sat}
  • R is the universal gas constant as 8.314 kJ/kmol-K
  • T is the temperature in Kelvin scale which is 20+273= 293K

By substituting values in the equation

                                    \bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3

Converting \bigtriangleup c into cm^3/cm^3

As 1 mole of water 18 cm^3 so

                               \bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6}  cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6}  cm^3/cm^3

Putting this in the equation of mass flux equation gives

                            j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6}  cm/s

For calculation of water level drop in a day, converting mass flux as

                     j_{H_2O}=14.4 \times 10^{-6}  \times 24 \times 3600  cm/day\\ j_{H_2O}=1.24  cm/day

So the water level will drop by about 1.24 cm in 1 day.

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A highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is sm
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Answer:

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An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with
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Answer:

A) 0.957 J

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A cart moves along a track at a velocity of 3.5 cm/s. when a force is applied to the cart, its velocity increases to 8.2 cm/s. i
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Acceleration, in physics, is the rate of change of velocity of an object with respect to time. <span> Anytime an object's velocity is changing, the object is said to be </span><span>accelerating. It can be calculated as follows:

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Hope this answers the question.
5 0
2 years ago
Read 2 more answers
 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer
Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}

we know that y_{f}=0 and that v_{y0}=0 as well, so the formula is simplified to:

0=y_{0}+\frac{1}{2}at^{2}

we can now solve this for t, so we get:

-y_{0}=\frac{1}{2}at^{2}

-2y_{0}=at^{2}

\frac{-2y_{0}}{a}=t^{2}

t=\sqrt{\frac{-2y_{0}}{a}}

we know that y_{0}=1.20m and that a=g=-9.8m/s^{2}

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

V_{x}=\frac{x}{t}

so we can solve this for x, so we get:

x=V_{x}t

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

a=\frac{v_{f}-v_{0}}{t}

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

a=\frac{v_{f}}{t}

so we can solve this for the final velocity:

V_{yf}=at

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

V_{yf}=(-9.8m/s^{2})(0.495s)

which yields:

V_{yf}=-4.851m/s

so now we know the components of the final velocity, which are:

V_{xf}=1.5m/s and V_{yf]=-4.851m/s

so now we can find the speed by determining the magnitude of the vector, like this:

|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}

so we get:

|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

\theta = tan^{-1} (\frac{V_{y}}{V_{x}})

so we get:

\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})

which yields:

\theta = -72.82^{o}

4 0
2 years ago
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