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3 years ago

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A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d

irection of its velocity?
FIND THE DIRECTION I WILL MARK YOU BRAINLIEST FOR THE CORRECT ANSWER

1 answer:

QveST [7]3 years ago

3 0

Answer:

The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

(target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

(10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

(target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

-94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

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A marathon runner completes a 42.238 km course in 2 h, 31 min, and 46 s . There is an uncertainty of 29 m in the distance run an

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Answer:

The percentage uncertainty in the average speed is 0.10% (2 sig. fig.)

Explanation:

Consider the formula for average speed $\bar{v}$ .

$\displaystyle \bar{v} = \frac{s}{t}$ ,

where

$s$ is the total distance, and
$t$ is the time taken.

The percentage uncertainty of a fraction is the sum of percentage uncertainties in

the numerator, and
the denominator.

What are the percentage uncertainties in $s$ and $t$ in this question?

The unit of the absolute uncertainty in $s$ is meters. Thus, convert the unit of $s$ to meters:

$s = \rm 42.238\;km = 42.238\times 10^{3}\;m$ .

$\begin{aligned}\displaystyle \text{Percentage Uncertainty in }s &= \frac{\text{Absolute Uncertainty in } s}{\text{Measured Value of }s}\times 100\% \\ &=\rm\frac{29\; m}{42.238\times 10^{3}\;m}\times 100\%\\ &= 0.0687\%\end{aligned}$ .

The unit of the absolute uncertainty in $t$ is seconds. Convert the unit of $t$ to seconds:

$t = \rm 2\times 3600 + 31\times 60 + 46 = 9106\;s$

Similarly,

$\begin{aligned}\displaystyle \rm \text{Percentage Uncertainty in }t &= \frac{\text{Absolute Uncertainty in }t}{\text{Measured Value of }t}\times 100\% \\ &=\rm\frac{46\; s}{9106\;s}\times 100\%\\ &= 0.0329\%\end{aligned}$ .

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Generally, keep

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The percentage uncertainty in $\bar{v}$ here is less than 2%. Thus, keep two significant figures. However, keep more significant figures than that in calculations to make sure that the final result is accurate.

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4 years ago

A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

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Answer:

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

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From Newton's second law of motion

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Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity

Making $u_{i}$ the subject we obtain $u_{i}=\frac {u_{y}}{sin \theta}$

Since $u_{y}$ and $\theta$ are known as $46.43ms^{-1}$ and $80^{o}$ respectively, then $u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}$

Therefore, the velocity of potato is $47.15ms^{-1}$

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation $v^{2}=u^{2}+2sa$

The final velocity v is answer in part a which is $47.15ms^{-1}$ , initial velocity u is $0ms^{-1}$ hence the equation is re-written as

$v^{2}=2sa$ and making a the subject we obtain

$a=\frac {v^{2}}{2s}$

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Therefore, average acceleration is $2470.13ms^{-2}$

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is $2470.13ms^{-2}$

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

$R=\frac {F}{W}=\frac {F}{mg}$ and since F=1235.06, m=0.500kg and g=9.8

$R=\frac {1235.06}{0.500*9.8}$ =252.05

Therefore, ratio of average force to weight is 252.05

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4 years ago

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3 years ago

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

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Answer:

Q = 47.06 degrees

Explanation:

Given:

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