Some unit 2 questions

[dekoyl]

I hope someone can help me out with these questions that has cropped up in practice exams.

Below is a circuit. 

-----[67 Ω]------------------
|                  |                  |
|                 /\ diode        [180Ω]
[6V]             |                  |
|                  |                  |
| ___(A)___ |__________|

If the diode(/\ pointing up) were to be removed from the circuit, why would the ammeter read the same current as when it's attached? I thought if it were removed, it would allow current flow down the path as well.

------------------------------------------------------------------------------------------------------------------------

Shaun releases the cork from a fresh bottle of champagne vertically into the air from the same height as his head. 3.5 seconds later, it hits his head again. (Acceleration is 10ms^-2 and air resistance is neglible.)

Determine the maximum height of cork.

v = 0; t = 1.75; a = -10
which is correct.

Next question asks: Determine the speed of the cork when it hits his head.
Wouldn't v = 0, as for the question above v = 0?
These are the solutions.

v = u + at
v = -v + at
2v = at
v = -17.5 17.5 m/s downwards.

Thankeee

[Flaming_Arrow]

Next question asks: Determine the speed of the cork when it hits his head.
Wouldn't v = 0, as for the question above v = 0?
These are the solutions.

v = u + at
v = -v + at
2v = at
v = -17.5 17.5 m/s downwards.

Thankeee

v isnt = 0 because its coming down, notice how it says after 3.5s it hits his head. so we are trying to find velocity in the downward journey so u = 0

[dekoyl]

^Ah thank you for that. Your explanation makes a lot more sense.

[huynhvien00]

Solution:

Question 1: - The ammeter will have the same value even if the diode is removed. The diode is reverse biased so there is no current or voltage drop across the diode ( assume the diode is ideal). Hence, the diode has no effect to the  180 ohms resistor and the current will be be the same in both cases.

Question 2: - It will be much easier to  break the motion of the cork into 2 parts to do calculations. Each part has the same acceleration and time.

a. The cork is projected from the head to the highest point:
- We have: a = -10 m/s^2
                v=0 ( the cork stops at the highest point)
                t=3.5/2 = 1.75 s
Initial velocity is unknown so we have to work it out by using the formula  v = u  + at ( u is the initial velocity)
From the formula, we have u= v - at
                                    u = 0 - (-10x1.75)= 17.5 m/s

                                    d=((u+v)/2) x t
                                  d = ((17.5+0)/2)x1.75 = 15.3125 m (highest point)

b. The cork returns from the highest point to the head.

u=0
t= 1.75 s
d= 15.3125 m
a= 10 m/s^2
u=0

[/b][/size]

v^2 = u^2 + 2ad
v^2 = 2 x 10 x 15.3125 = 306.25
v= - 17.5 ( the cork is going down)