-- Class I lever
The fulcrum is between the effort and the load.
The Mechanical Advantage can be anything, more or less than 1 .
Example: a see-saw
-- Class II lever
The load is between the fulcrum and the effort.
The Mechanical Advantage is always greater than 1 .
Example: a nut-cracker, a garlic press
-- Class III lever
The effort is between the fulcrum and the load.
The Mechanical Advantage is always less than 1 .
I can't think of an example right now.
Answer:
a) x = (0.0114 ± 0.0001) in
, b) the number of decks is 5
Explanation:
a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated
x = d / 52
x = 0.590 / 52
x = 0.011346 in
Let's look for uncertainty
Δx = dx /dd Δd
Δx = 1/52 Δd
Δx = 1/52 0.005
Δx = 0.0001 in
The result of the calculation is
x = (0.0114 ± 0.0001) in
b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is
#_cards = n 52
The formula for thickness is
x = d / n 52
Uncertainty
Δx = 1 / n 52 Δd
n = 1/52 Δd / Δx
n = 1/52 0.005 / 0.00002
n = 4.8
Since the number of decks must be an integer the number of decks is 5
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
Answer:
Explanation:
v = Velocidad final = 
u = Velocidad inicial = 0
t = Tiempo empleado = 15 s
a = Aceleración
De las ecuaciones cinemáticas tenemos
La aceleración del camión en el primer intervalo de tiempo es .
