(b) 56 g/cm3 to kg/m3
Write vectors equations for each diagram below.
(a) | (b) | ||
(c) | (d) | ||
(e) | (f) |
(a) | (b) | ||
(c) | (d) | ||
(e) | (f) |
First we get the magnitude of the momenta: and . Next we neatly sketch the problem and its solution.
Vector problems are solved by breaking the given vectors into
their i and j components.
Using Pythagoras' Theorem, . We use
trigonometry to find the direction. The angle is .
The total or net momentum is thus 0.141 kg m/s at 40.3° north
of east.
First we neatly sketch the problem and its solution.
Vector problems are solved by breaking the given vectors into
their i and j components.
Using Pythagoras' Theorem, . We use
trigonometry to find the direction. The angle is .
The total or net force is thus 260 N at 70.1° south of east.
It would be more common to state this as 260 N at 19.9° east of
south.
First, we must realize that the velocity that the observer sees is the sum of the velocity of the boat and the velocity of the current, i.e. . Since we are looking for , we are dealing with the subtraction of vectors which is the addition of the negative of a vector, i.e. . To solve this, or any other vector problem, we sketch the solution first.
Vector problems are solved by breaking the given vectors into
their i and j components.
Using Pythagoras' Theorem, . We use
trigonometry to find the direction. The angle is .
The velocity of the current is thus 1.36 m/s at 56.2° north of
east.
Since we are told the total force and one of the forces, the other force is given by . To solve this, or any other vector problem, we sketch the solution first.
Vector problems are solved by breaking the given vectors into
their i and j components.
Using Pythagoras' Theorem, . We use
trigonometry to find the direction. The angle is .
The second force is thus 2200 N at 40.9° south of west.
Questions? mike.coombes@kpu.ca