As the weight of the block increases the force needed to start it moving (N) increases because there is more friction between th
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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
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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°.
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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°
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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
Answer:
forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.
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Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion
h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.