A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40
1 answer:
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Answer:
A ≈ 40.3 cm²
Step-by-step explanation:
The area of right Δ ACD is calculated as
A = bh ( b is the base and h the perpendicular height )
Here A = 50.5, b = CD and h = AC , then
× 14 × AC = 50.5
7 AC = 50.5 ( divide both sides by 7 )
AC ≈ 7.214
The area of Δ ABC is calculated as
A = × BC × AC × sin36°
= 0.5 × 19 × 7.214 × sin36°
≈ 40.3 cm² ( to 3 sf )
Answer:
The value of the missing side is
Step-by-step explanation:
In this problem i assume that the length side x is tangent to the circle
so
the length side x is perpendicular to the radius
Applying the Pythagoras Theorem in the right triangle
we have that
Solve for x