A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
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Answer:
4.96 × 10⁵ Pa
Explanation:
F = mg
This force is evenly distributed on the three leg
radius, r = d/2
= 2.8 / 2
= 1.4 cm = 0.014 m
total cross sectional area of the three legs, A = 3*pi*r^2
Pressure due to weight,
P = Weight/A
P = 4.96 × 10⁵ Pa
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
where F is the force applied, and
is the stretch of the spring with respect to its equilibrium position. Using the data, we find
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
where F is the force applied, and
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring: