Answer:
2156J
Explanation:
Given parameters:
Height of lift = 10m
Mass = 22kg
Unknown:
Work done by the machine = ?
Solution:
Work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Here;
Work done = mass x acceleration due to gravity x height
Work done = 22 x 9.8 x 10 = 2156J
Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
Earthquakes and volcanoes most commonly occur around plate boundaries because of the movement from the plate boundaries. The interactions between the plates by moving under, upon, or sliding against other boundaries may cause earthquakes and volcanoes.
Answer:
The reading of the scale during the acceleration is 446.94 N
Explanation:
Given;
the reading of the scale when the elevator is at rest = your weight, w = 600 N
downward acceleration the elevator, a = 2.5 m/s²
The reading of the scale can be found by applying Newton's second law of motion;
the reading of the scale = net force acting on your body
R = mg + m(-a)
The negative sign indicates downward acceleration
R = m(g - a)
where;
R is the reading of the scale which is your apparent weight
m is the mass of your body
g is acceleration due to gravity, = 9.8 m/s²
m = w/g
m = 600 / 9.8
m = 61.225 kg
The reading of the scale is now calculated as;
R = m(g-a)
R = 61.225(9.8 - 2.5)
R = 446.94 N
Therefore, the reading of the scale during the acceleration is 446.94 N
