1) The equivalent resistance of two resistors in parallel is given by:

so in our problem we have

and the equivalent resistance is

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
Answer:
t₁ = 3 s
Explanation:
In this exercise, the vertical displacement equation is not given
y = 240 t + 16 t²
Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration
Let's replace
864 = 240 t + 16 t²
Let's solve the second degree equation
16 t² + 240 t - 864 = 0
Let's divide by 16
t² + 15 t - 54 = 0
The solution of this equation is
t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1
t = [-15 ±√(225 +216)] / 2
t = [-15 + - 21] / 2
We have two solutions.
t₁ = [-15 +21] / 2
t₁ = 3 s
t₂ = -18 s
Since time cannot have negative values, the correct t₁ = 3s
The solution would be like
this for this specific problem:
V^2 = 2AS = 2FS/M
V = sqrt(2FS/M) =
sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps
So the speed of the arrow as it leaves the bow
is 42.5 mps.
I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.
Explanation:
The given data is as follows.
mass = 0.20 kg
displacement = 2.6 cm
Kinetic energy = 1.4 J
Spring potential energy = 2.2 J
Now, we will calculate the total energy present present as follows.
Total energy = Kinetic energy + spring potential energy
= 1.4 J + 2.2 J
= 3.6 Joules
As maximum kinetic energy of the object will be equal to the total energy.
So, K.E = Total energy
= 3.6 J
Also, we know that
K.E = 
or, v = 
= 
= 
= 6 m/s
thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.