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Question:

I used to play a game on pc about 10 years ago. It was a turn based strategy game.?

The characters wears robotic suits and fires laser at long range and swords for melee attacks.

Answer:

7 Layers of OSI :every physique seem to choose records Processing to help you undergo in techniques. A utility - layer that interfaces directly to accomplish utility amenities for processing P Presentation - delivers a context between utility layers S consultation - manages connections between desktops(as I even have remembered ^_^) T delivery - delivers circulate of records between the top purchasers. controls the records circulate and blunder administration. N community - it performs community routing D records link - it detects blunders opportunities interior the actual layer P actual - it refers on your pins, voltages, cable specs, Hubs, repeaters, community adapters desire this help ^_^
Here's the diagram of your problem: . .._..???.._.. _-????_.. . _-??..?/_ Vy/./.↑ '.↑..//. ..|./.55? )/.|Y ..|/-→Vx_______/______↓_ .. ..|←X-→| The initial Velocity V 300m/s .V 300m/sec .. /↑ /| /I Vy /I /I /.55? ..I. /- Vx.. The components are: Vx Vcos55? 300(.574) 172.2 m/sec Vy Vsin55? 300((.8192) 245.76 m/sec To find the Y -coordinate, you may use the linear equation S Vit + ? at ? Where: S Y Vi Vy a -g (freefall acceleration due to gravity -9.8m/sec? t 42.0 secs (given) Y Vyt + ? (-g)t ? .. (Vsin55?)t - ? (9.8)t ? .. 300(.8192)(42) - 4.9(42)? .. 10321.32 - 8643.6 . Y 1677.72 m ? To find the X-coordinate, use the linear equation .._ S Vt Where _ V Vx in theory, Vx is not accelerated, it has a constant value S X t 42 secs (given) X Vx t (Vcos55? ) t 300(.574)(42) X 7232.4 m ? ≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡ ._??-___ ._??.??.↑ .._????.| Vi.._????..H1 ___♂_?)???_↓_ .▓▓▓▓▓▓█▄???..↑H2 ╘OO╛__╘O╛..???_↓__ |←-d→| Take note that distances are always reckoned from the point of origin of the trajectory. In cases of Firefighters whose nozzle of the fire hose is at some distance (H2) from the ground, the actual height of the building (h H1 + H2) H1 is what you will get from the projectile equations. Vx Vicos? Vy Visin? Since time is not given, you first solve it from d Vxt d (Vicos?)t t is the only unknown here Then use the equation for freefall motion. In this case, it takes the form (see previous problem above) Y Vyt + ? (-g)t ? H1 (Visin? )t - ? gt ? Then h H1 + H2 H2 must be given, the distance of the hose nozzle from the ground)?
Remember for projectile motions, there are two very important equations 1) x Vo cos (theta) t 2) y Vo sin (theta) t - (1/2)at^2 you can find the x and y coordinates in the given problem because u're given all unknowns. x (300) cos (55) (42) v (300) sin (55) (42) - (1/2)(-9.8)(42)^2

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