A 65-kg bungee jumper, who is attached to one end of an 85-m long bungee cord that has its other end tied to a bridge, jumps off
the bridge and toward the river below. When stretched, the bungee cord acts like a spring and provides a force on the jumper that increases linearly as the cord is stretched. When the bungee cord is stretched to its maximum length, it exerts a 2.2 kN force on the jumper. If the bungee cord is stretched beyond its equilibrium length for a duration of 1.2 seconds, what is the impulse delivered to the bungee jumper
1 answer:
Answer:
The impulse delivered to the bungee jumper is 1.32 kN.s
Explanation:
The situation can be shown graphically as shown in the figure.
Impulse delivered to the bungee jumper = Area under the curve.
The curve represents a triangle and the area of traiangle = (1/2)base×height
The base of the triangle from the graph = 1.2 seconds.
The height of the triangle from the graph = 2.2 kN
Thus,
<u>Impulse = (1/2)×(1.2 seconds)×(2.2 kN) = 1.32 kN.s</u>
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Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb