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7nadin3 [17]
2 years ago
5

A missile is fired from a jet flying horizontally at Mach 1 (1100 ft/s). The missile has a horizontal acceleration of 1000 ft/s

2 . Calculate its horizontal velocity at 10.0 seconds after it is fired. (Assume constant acceleration of the missile throughout its flight.)
Physics
2 answers:
Nata [24]2 years ago
8 0
Vo = 1100 ft/s 
<span>a = 1000 ft/s^2 </span>
<span>t = 10 s </span>
<span>Use velocity formula </span>
<span>Vh = Vo + at </span>
<span>Vh = 1100 ft/s + 1000 ft/s^2 *10 s </span>
<span>Vh = 11,100 ft/s </span>

<span>Explanation: In this case, the missile continues travelling at the plane's initial velocity, Vo, (due to Newton's 1st law) with an added velocity component, V = a*t, due to the missile's acceleration. The two velocity components are vectors which act in the same direction, so they may be summed together. </span>

<span>Hope this helps:D
Have a great rest of a brainly day!</span>
gayaneshka [121]2 years ago
5 0
V = u + a*t = 1100ft/s + (1000*10) ft/s = 11100 ft/s 
Answer is <span>11,100 ft/s  </span>
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Complete Question

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Answer:

a

The torque acting on the particle is  \tau = 48t \r k

b

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Explanation:

From the equation we are told that

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        The time is  t

   

The torque acting on  the particle is mathematically represented as

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where \r l is change in angular momentum which is mathematically represented as

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Where X mean cross- product

   \r v is the velocity which is mathematically represented as

           \r v = \frac{d \r r }{dt}

Substituting for  \r r

           \r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]

           \r v =  8t \r i - (2 + 12 t) \r j

Now the cross product of \r r \ and \ \r v is  mathematically evaluated as    

          \r r  \  \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]

                       = 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k

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So the angular momentum becomes

       \r l = m (8t^2 \r k)

Substituting for m

      \r l = 3 *  (8t^2 \r k)

      \r l =24t^2  \r k

Substituting into equation for torque

       \tau = \frac{d}{dt} [24t^2 \r k]

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The magnitude of the angular momentum can be evaluated mathematically as

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Answer:

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b) We repeat in calculation for a conducting sphere.

For r = 4 cm

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