In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
Not positive but id use a bar graph<span />
If it is the video I am thinking of then Yes
Answer:
there are conversion scales used for doing that
to convert Celsius to Fahrenheit we use the formula,
T(°F) = T(°C) × 9/5 + 32
and to convert Fahrenheit to Celsius we use,
T(°F) = T(°C) × 5/9 + 32
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Explanation:
