Answer:
mas of water displaced = 41.4 g
Explanation:
Weight in air = True weight = 45 g
Apparent weight = 3.6 g
Apparent weight = True weight - Buoyant force
Buoyant force = 45 g - 3.6 g = 41.4 g
Weight of water displaced = Buoyant force
Weight of water displaced = 41.4 g dyne
mas of water displaced = 41.4 g
Answer: 3.4s
Explanation:
There are three stages in the motion of the ball, so you have to calculate the times for every stage.
1) Ball dropping from 9.5m: free fall
d = Vo + gt² / 2
Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²
⇒ t = √ (1.94 s²) = 1.39s
2) Ball rising 5.7m (vertical rise)
i) Determine the initial speed:
Vf² = Vo² - 2gd
Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²
⇒ Vo = 10.6 m/s
ii) time rising
Vf = Vo - gt
Vf = 0 ⇒ Vo = gt ⇒
t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s
3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)
i) d = 5.7m - 1.20m = 4.5m
ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²
⇒ t = √ (0.92 s²) = 0.96s
4) Total time
t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s
Answer:
The pack has fallen 275.6 m
Explanation:
The pack is in free fall, so it is moving by uniform accelerated motion, so we can use the following suvat equation:
where, choosing upward as positive direction:
v is the final velocity
u is the initial velocity
is the acceleration of gravity
s is the vertical displacement
For the pack in this problem,
u = 0
v = -73.5 m/s
Solving for s, we find how far the packet has fallen:
And the negative sign means the direction of the displacement is downward.
Distance= speed * time
So multiply speed and time to get the value of distance
20*9.8 =196 meters
Answer:
Explanation:
charge on the capacitor = capacitance x potential
= 1.588 x 3.4
= 5.4 C
Energy of capacitor = 1 / 2 C V ² , C is capacitance , V is potential
= .5 x 3.4 x 1.588²
= 4.29 J
If I be maximum current
energy of inductor = 1/2 L I² , L is inductance of inductor .
energy of inductance = Energy of capacitor
1/2 L I² = 4.29
I² = 107.25
I = 10.35 A
Time period of oscillation
T = 2π √ LC
=2π √ .08 X 3.4
= 3.275 s
current in the inductor will be maximum in T / 4 time
= 3.275 / 4
= .819 s.
Total energy of the system
= initial energy of the capacitor
= 4.29 J
