To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
0.173 m.
Explanation:
The fundamental frequency of a closed pipe is given as
fc = v/4l .................. Equation 1
Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.
Making l the subject of the equation,
l = v/4fc ................ Equation 2
also
v = 331.5×0.6T ................. Equation 3
Where T = temperature in °C, T = 18.0 °c
Substitute into equation 3
v = 331.5+0.6(18)
v = 331.5+10.8
v = 342.3 m/s.
Also given: fc = 494 Hz,
Substitute into equation 2
l = 342.3/(4×494)
l = 342.3/1976
l =0.173 m.
Hence the length of the organ pipe = 0.173 m.
I think it might be A. I’m sorry if I’m wrong
The Primary Colors are Red Yellow and Blue
Explanation:
Calculating this according to the m/s rate
Now solving the question
If the car goes 20 m/s mathematically we can generate it as
15 × 60 = 600 secs
If the car goes 20 meters per every second it means
The car will go 20 meters for 900 secs
Which is 900 ×20 = 18000m/s
=18000÷1000 = 1km
Therefore the answer is 18km
