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Hatshy [7]
3 years ago
10

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find

the range and hI e final velocity
Physics
1 answer:
True [87]3 years ago
7 0

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

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Under which condition does Ohm's law apply? a. The current must be constant b. The power must be constant c. The temperature mus
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option c

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2 years ago
If a pedestrian walked 900 meters forward and 300 meters back
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B

Explanation:

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2 years ago
A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (a) At this instant, wha
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Answer:

(A) L = 115.3kgm²/s

(B) dL/dt = 94.1kgm²/s²

Explanation:

The magnitude of the angular momentum of the rock is given by the foemula

L = mvrSinθ

We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.

Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =

115.3 kgm²/s

(B) The magnitude of the rate of angular change in momentum is given by

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Answer:

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E = 16471

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g = 10 m/s²

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W = 73.5×10×22.41 = 16471.35

W = 16.5 Kj

(ii) Power = workdone/time

P = 16471.35/330

P = 49.9 Watt

(iii) The energy burnt in this process = 16471

4 0
3 years ago
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