Question 1 (1 point)
1 answer:
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to find i1, i2, and i3 we need to find the total current.
to find the total current, you need to find the total resistance
you're already given the total voltage, Vs
to find Rtotal, start from the resistors furthest from the voltage source.
R3 and R4 are in series so
Rtotal= R3+R4 = 6+3 = 9 ohms
9 ohms is now in parallel with R2 so,
Rtotal= + (\frac{1}{R2}) ^{-1})^-1= (1/18)^-1 +( 1/9)^-1 = 6 ohms
6 ohms is in series with R1 so
Rtotal= 4+6=10 ohms
itotal= (\frac{Vtotal}{Rtotal})
= 120 v/10 ohms = 12 A
i total = i1 because all the current flows through it
i1= 12A
so the current splits into i2 and i3 and the amount of current that flows through a branch depends on the total resistance in each branch.
we already calculated the resistance in the R3+R4 & R2 branch as 6 ohms
since r3 and r4 are in series, the same current will flow through them
r3+r4 = 9 ohms
r2= 18 ohms
so the current in r2 will be half that of R3 & R4 (V=IR)
using the current divider rule
i2= 12A x (6 ohms/18 ohms)= 4 A
i3= 12A x (6 ohms/9 ohms) = 8 ohms
Answer:
a) The maximum height reached by the projectile is 558 m.
b) The projectile was 21.3 s in the air.
Explanation:
The position and velocity of the projectile at any time "t" is given by the following vectors:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).
v = velocity vector at time t
a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:
vy = v0 · sin α + g · t
0 = 175 m/s · sin 36.7° - 9.80 m/s² · t
- 175 m/s · sin 36.7° / - 9.80 m/s² = t
t = 10.7 s
Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).
Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²
y = 558 m
The maximum height reached by the projectile is 558 m
b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.
We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²
0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)
0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t
- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t
t = 21.3 s
The projectile was 21.3 s in the air.
