An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi
1 answer:
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Answer:
I = 11.26 mA
Explanation:
given,
V = 0.7 V length = 80 m
diameter = 0.2 mm = 0.02 cm
radius = 0.01 × 10⁻² m
ρ for gold wire = 2.44 × 10⁻⁸ ohm-m at 20 °C
A = cross sectional area = π r² = π (0.01 × 10⁻² )²
= 31.4× 10⁻⁹ m²
R = 62.165 Ω
I = 11.26 mA
Answer:
the diameter of the outside edge of the receiver is
Explanation:
From the schematic free body diagram illustrating what the question is all about below;
Let represent A to be the vertex where the receiver is being placed
S to be the focus
BP to be equal to r (i.e radius of the outer edge)
BC to be 2 r (i.e the diameter)
Given that AS = 4 in and AP is 18 in
Let AP be x- axis and AY be y -axis
A=(0,0)
S=(4,0) = (0,0)
So that the equation of the parabolic path of the receiver will be:
B = (AP, BP)
B = (18, r)
B lies y² = 16 x
r² = 16 x
r² 16 × 18
Diameter BC = 2r
