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We have that the see-saw will be balance if a weight of 
is added to Nural's side of the see-saw
From the Question we are told that
Wan’s mass is 
Nurul’s mass is 
Generally
The Will be balance when the weight on both sides of the see-saw are equal
In conclusion
The see-saw will be balance if a weight of is added to Nural's side of the see-saw
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Answer:
A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.
B, is the answer to the question
Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
