Water is taken from a well 11m deep in a bucket,the rope winding up to a drum 35cm in diameter.
A)Through what angle does the handle turn in winding up 1m of rope?
B)How many revolutions of the handle does it take to bring the bucket up from the bottom?
C)If the arm of the handle is 42cm long,how far does the hand of the winder travel when bringing the bucket up from the bottom?
My solution so far.
A)1m=100cm
100=tita/360*2*22/7*35/2
tita=327 3/11degree
B) 1rev=360 degree
xrev=327 3/11
=10rev.
Pls kindly help me with the part C
A=1/r=1//0.175
5.714
R=329degree
11/x=0.175/0.42
0.175x=11*0.42/0.175
=26.4m
I do not understand the c part please
thanks guys
C) So in effect, the radius of one revolution is 42 cm, and one rotation is 84pi.
Multiply 84pi by the number of rotations
10REV
Thank you so much the answer is correct.
i dont understand the b part pls
not understand
To find the distance the hand of the winder travels when bringing the bucket up from the bottom, we can use trigonometry.
First, let's consider the right triangle formed by the length of the arm of the handle (42cm), the radius of the drum (35cm/2 = 17.5cm), and the distance traveled by the hand of the winder (which we want to calculate).
Using the Pythagorean theorem, we can find the length of the hypotenuse, which represents the distance traveled by the hand of the winder:
hypotenuse^2 = arm_length^2 + radius^2
hypotenuse^2 = 42^2 + 17.5^2
hypotenuse^2 = 1764 + 306.25
hypotenuse^2 = 2070.25
Taking the square root of both sides, we get:
hypotenuse = √2070.25
hypotenuse ≈ 45.53 cm
So, the hand of the winder travels approximately 45.53 cm when bringing the bucket up from the bottom.