Question

please answer both questions ty so much

Two forces are acting on a 0.250 kg hockey puck as it slides along the ice. The first force has a magnitude of 0.400 N and points 35.0 north of east. The second force has a magnitude of 0.580 N and points 65,0° north of east. If these are the only two forces acting on the puck, what will be the magnitude and direction of the puck's acceleration? Enter the direction as an angle measured in degrees counterclockwise from due east. magnitude of acceleration: direction of acceleration:

Answer 1

two forces are acting on......

solution

F1 = 0.58* ( cos 65 i + sin 65 j)

F1 = 0.2451 i + 0.5256 j

other force

F2 = 0.4* ( cos 35 i + sin 35 j)

F2 = 0.3277 i + 0.229 j

net force

F = F1 + F2

0.25 a = 0.5728 i + 0.7546 j

a = 2.6 i + 3.43 j

====

a) magnitude of acceleration

a = 4.304 m/s^2

b)

direction

x = arctan ( 3.43 / 2.6)

x = 52.837 north of east

=======

plz post other problem separately

Comment before rate in case any doubt, will reply for sure.. goodluck