Answer:
time=4s
Explanation:
we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion
, where
max=
Given that, at i(2)=
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
Now substitute 
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
now subs. 
⇒
also 
⇒
⇒
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
As per above given data
initial velocity = 19.3 km/s
final velocity = - 18.8 km/s
now in order to find the change in velocity




Part b)
Now we need to find acceleration
acceleration is given by formula

given that


now the acceleration is given as


so above is the acceleration
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:

where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,

<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:

(c)
Now we will consider the downward motion and use the third equation of motion:

where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,

<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:

(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>