物理专业科技英语课后练习题含答案

物理专业科技英语课后练习题含答案
第一部分：力学
1. 题目
A 2.5 kg mass attached to a spring with a spring constant of 250 N/m is subject to a harmonic motion. If the maximum displacement is 0.1 meters, calculate:
a)The maximum force exerted by the spring.
b)The maximum velocity of the mass.
c)The maximum acceleration of the mass.
2. 答案
a)The maximum force exerted by the spring can be calculated as:
Fmax = k * xmax = 250 N/m * 0.1 m = 25 N
So, the maximum force exerted by the spring is 25 N.
b)The maximum velocity of the mass can be calculated as:
vmax = √(k/m) * xmax = √(250 N/m / 2.5 kg) * 0.1 m = 1 m/s
So, the maximum velocity of the mass is 1 m/s.
c)The maximum acceleration of the mass can be calculated as:
amax = √k/m * xmax = √(250 N/m / 2.5 kg) * 0.1 m = 1 m/s^2
So, the maximum acceleration of the mass is 1 m/s^2.
第二部分：热力学
1. 题目
A gas at a pressure of 2 atm and a volume of 3 L is compressed to a volume of 1 L while its pressure increases to 4 atm. Calculate the work done on the gas during this process.
2. 答案
The work done on the gas during this process can be calculated as: W = -PΔV = -(4 atm - 2 atm) * (1 L - 3 L) = 4 atm * 2 L = 8 J
So, the work done on the gas during this process is 8 J.
第三部分：电学
1. 题目
A capacitor with a capacitance of 50 nF is charged to a potential difference of 100 V. If the capacitor is then discharged through a resistor with a resistance of 5 kΩ, calculate:
a)The time constant of the circuit.
b)The time required for the voltage on the capacitor to drop
to 50 V.
c)The current through the resistor when the voltage on the
capacitor is 50 V.
2. 答案
a)The time constant of the circuit can be calculated as:
τ = RC = 5 kΩ * 50 nF = 0.25 ms
So, the time constant of the circuit is 0.25 ms.
b)The time required for the voltage on the capacitor to drop
to 50 V can be calculated using the equation:
V = V0 * e^(-t/τ)
where V0 is the initial voltage on the capacitor, e is the mathematical constant approximately equal to 2.71828, and t is the time elapsed. Solving for t, we get:
t = -τ * ln(V/V0) = -0.25 ms * ln(50 V/100 V) ≈ 0.173 ms
So, the time required for the voltage on the capacitor to drop to 50 V is approximately 0.173 ms.
c)The current through the resistor when the voltage on the
capacitor is 50 V can be calculated using Ohm’s law:
I = V/R = 50 V / 5 kΩ = 0.01 A
So, the current through the resistor when the voltage on the capacitor is 50 V is 0.01 A.。