When 525 J of heat is added to 7.6 g of metal at 26°C, the temperature increases to 80°C. What is the specific heat of the metal
1 answer:
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Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone,
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
To solve this problem, we use the Law of Universal Gravitation:
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons