Here,
Load distance (Ld) = 30 cm
Effort distance (Ed) = 60 cm
Load (L) = 200N
Effort (E) = ?
Now, By using formula,
or, E * Ed = L * Ld
or, E * 60 = 200 * 30
or, E = 6000/60
◆ E = 100N
This is a Right answer...
I hope you understand...
Answer:
71.4583 Hz
67.9064 N
Explanation:
L = Length of tube = 1.2 m
l = Length of wire = 0.35 m
m = Mass of wire = 9.5 g
v = Speed of sound in air = 343 m/s
The fundamental frequency of the tube (closed at one end) is given by

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz
The linear density of the wire is

The fundamental frequency of the wire is given by

The tension in the wire is 67.9064 N
Answer:
Try to chase it or try to put up missing posters if you cant him/her, i never had a dog but i would do that and ask neighbors if they know where the dog is at.
Explanation:
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s