求问一道物理题
黑笔电路中:左边变成右边情况时,外电阻增大,导致外电路电压增大为什么红笔电路中:左边情况变成右边情况时,按我的理解说上支路阻值增大了,那么外路总电阻增大,外路电压之增大才...
黑笔电路中:左边变成右边情况时,外电阻增大,导致外电路电压增大
为什么红笔电路中:左边情况变成右边情况时,按我的理解说上支路阻值增大了,那么外路总电阻增大,外路电压之增大才对啊,为什么题目答案里说不变呢外电压 展开
为什么红笔电路中:左边情况变成右边情况时,按我的理解说上支路阻值增大了,那么外路总电阻增大,外路电压之增大才对啊,为什么题目答案里说不变呢外电压 展开
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这题没给出波长的条件,以下假设两个波波长是一样的。
公式:
The equation for the fundamental frequency of an ideal taut string is:
f = (1/2L)*√(T/μ)
where
f is the frequency in hertz (Hz) or cycles per second
T is the string tension in gm-cm/s2
L is the length of the string in centimeters (cm)
μ is the linear density or mass per unit length of the string in gm/cm
思路:
As Sting A and B have the exact same length and tension, so the only factor that drives a different frequency of wave is μ. Suppose the mass of B is m, A has the mass of 4m. By comparing f_A and f_B, we found that f_A / f_B = 1:2 (f_A : f_B = 4m^(-1/2) : 1m^(-1/2) = 1:2) due to the equation above .
Due to the equation of wave : speed= wave length * frequency, the wave lengths are identical. and f is directly proportional to speed of wave, easily found that:
S_A:S_B = 1:2
So answer is B. one half the speed of a wave on string B.
有错误请指出,望能帮到你
公式:
The equation for the fundamental frequency of an ideal taut string is:
f = (1/2L)*√(T/μ)
where
f is the frequency in hertz (Hz) or cycles per second
T is the string tension in gm-cm/s2
L is the length of the string in centimeters (cm)
μ is the linear density or mass per unit length of the string in gm/cm
思路:
As Sting A and B have the exact same length and tension, so the only factor that drives a different frequency of wave is μ. Suppose the mass of B is m, A has the mass of 4m. By comparing f_A and f_B, we found that f_A / f_B = 1:2 (f_A : f_B = 4m^(-1/2) : 1m^(-1/2) = 1:2) due to the equation above .
Due to the equation of wave : speed= wave length * frequency, the wave lengths are identical. and f is directly proportional to speed of wave, easily found that:
S_A:S_B = 1:2
So answer is B. one half the speed of a wave on string B.
有错误请指出,望能帮到你
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