A 4.0 kg mass has a velocity of 10 m/s to the EAST. The 4.0 kg mass is subjected to a constant net force of 16 N to the WEST for
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Answer:
distance of 2nd team from 1st team will be: 58.2
Direction of 2nd team from 1st team will be: 14.90 deg North of east
Explanation:
ASSUME Vector is R and makes angle A with +x-axis,
therefore component of vector R is
From above relation
Assuming base camp as the origin, location of 1st team is
away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)
location of 2nd team is at
, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)
Now position of 2nd team with respect to 1st team will be given by:
Using above values:
distance of 2nd team from 1st team will be:
Direction of 2nd team from 1st team will be:
Direction = 14.90 deg North of east
Answer:
141.56 N.
Explanation:
Data given:
Weight of the box= 200.2 N
Angle with the horizontal= 37.1°
Solution;
Gravitational force on the box, = weight of the box
= 200.2 N
Component of gravitational force along plane = ( ∅ )
= W * (sin∅)
= (200.1) * sin (37.1°)
= 141.56 N