Given,
the initial velocity = 0 m /s.
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
Answer:
a) b) d)
Explanation:
The question is incomplete. The Complete question might be
In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.
a)2 N; 2 N
b) 200 N; 200 N
c) 200 N; 201 N
d) 2 N; 2 N; 4 N
e) 2 N; 2 N; 2 N
f) 2 N; 2 N; 3 N
g) 2 N; 2 N; 5 N
h ) 200 N; 200 N; 5 N
For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces
a) 2+(-2)=0 here minus sign is to show the opposing firection of force
b) 200+(-200)=0
c) 200+(-201)
0
d) 2+2+(-4)=0
e) 2+2+(-2)
0
f) 2+2+(-3)
0; 2+(-2)+3
0
g) 2+2+(-5)
0; 2+(-2)+5
0
h)200 + 200 +(-5)
0; 200+(-200)+5
0
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