(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14
1 answer:
Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules
You might be interested in
Answer:
Explanation:
Torque, is given by
where F is force and r is perpendicular distance
where
is the angle of inclination
Torque, can also be found by
where I is moment of inertia and a is angular acceleration
Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity
Making a the subject, and already I is given as
hence
Taking g as 9.81, is given as 37 and L is 1.2
Answer:
The current is
Explanation:
From the question we are told that
The radius of the loop is
The earth's magnetic field is
The number of turns is
Generally the magnetic field generated by the current in the loop is mathematically represented as
Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth
=>
Where is the permeability of free space with value
=>