We have: A= F.s
So, A = 22 x 16
A= 352J
Só A is correct answer
(89000/102000)×100
=87.25%
(92000/104000)×100
=88.46%
efficiency is (output/input)×100
if u get confused which way input and output should go, remember the smaller value is always output and it's above in the fraction, then only it's possible to get a efficiency lower than 100.
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1
No, because the distance-time would show a constant velocity but the velocity-time graph shows an increasing velocity.
