Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law



so voltage across the capacitor + voltage across resistor = V


Now we know that

here rate of change in energy of the capacitor is given as



We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y
Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m
displacement = √(116² + 333²)
= 353 m
Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.
Answer:
the acceleration is reduced by gravity
a = (15 / .35) - [9.8 * sin(65º)]
Explanation:
break the launch vector into two components, vertical and horizontal
Force Net Vertical=-9.8*.350+15cos65 N
force net horizonal=15sin65
initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i
using i,j vectors..
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