a) What is the magnitude and direction of the wind?b) How far and in what direction must the ship now sail to reach its original destination?P4. The two vectors a? and b?in the figure have equal magnitudes of 10 m and the angles are ??301?and??1052?. Finda) x and y components for each vector,b) x and y components of their vector sum R?,c) the magnitude of R?,d) the angle R?makes with the positive direction of the x axis.
Hoist yar sails laddie.... your in for a gale!
5000 km east on the equator is approximately 40 5 stages (a million/8 of the circumference) of longitude. 5000 km due north from the equator is likewise approximately a million/8 of the circumference - at 40 5 stages of variety. At this variety, 40 5 stages of longitude covers approximately 3555 km. on account that all of those are approximations, we can proceed with an approximation of a proper triangle with facets of 5000 km and 3555 km. Which comes out to an approximate entire distance now of 6135 km out of your unique trip spot.
5000 km east on the equator is approximately 40 5 stages (a million/8 of the circumference) of longitude. 5000 km due north from the equator is likewise approximately a million/8 of the circumference - at 40 5 stages of variety. At this variety, 40 5 stages of longitude covers approximately 3555 km. when you consider that all of those are approximations, we are in a position to proceed with an approximation of a suitable triangle with components of 5000 km and 3555 km. Which comes out to an approximate entire distance now of 6135 km out of your unique trip spot.
First of all, u must draw the event. b) Use phytagoras, to find the original distance that the ship must sail Let a is the original distance, so a^2 = (120)^2 + (100)^2 a^2 = 14400 + 10000 a = sqrtroot [24400] a = 156.2 km To find the direction of the ship, use again phytagoras. Let @ the direction of the ship. tan@ = 100/120 @ = 39.8 a) The direction of the wind is usually -- {east} I'm sorry I think the data incomplete.
