Answer:Time constant gets doubled
Explanation:
Given
L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by
![i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%20%7D%7D%5Cright%20%5D)
where 
is maximum current
i=current at any time
thus if inductance is doubled then time constant also gets doubled or twice to its original value.
This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:
D = 1/2 A T²
Distance = (1/2) (acceleration) (time²)
The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)
The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .
In 5 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (5 sec)²
D = (4.9 m/s²) (25 sec²)
D = 122.5 meters
In 6 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (6 sec)²
D = (4.9 m/s²) (36 sec²)
D = 176 meters
Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
d=vi*t+(1/2)gt²
d=11 m
g=9.8 m/s²
vi=0 m/s
11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s
Answer: the time the sone is in flight is 1.5 s
