Answer:
A) Three hole punch and either a layered plastic or paper
B) Identify the lengths involved ,
Length of input arm / length of output arm = L1/ L2
Explanation:
<u>a) Materials involved includes :</u>
Three hole punch and either a layered plastic or paper
Identify the forces acting on the three-hole punch which are Input and output forces
Identify the points where they act
<u>B) procedures involved </u>
The mechanical advantage = output force / input force
step one: Identify the lengths involved
assuming no friction or relatively small friction \
mechanical advantage can be calculated as : Length of input arm / length of output arm = L1/ L2
Answer:
18.63 N
Explanation:
Assuming that the sum of torques are equal
Στ = Iα
First wheel
Στ = 5 * 0.51 = 3 * (0.51)² * α
On making α subject of formula, we have
α = 2.55 / 0.7803
α = 3.27
If we make the α of each one equal to each other so that
5 / (3 * 0.51) = F2 / (3 * 1.9)
solve for F2 by making F2 the subject of the formula, we have
F2 = (3 * 1.9 * 5) / (3 * 0.51)
F2 = 28.5 / 1.53
F2 = 18.63 N
Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.
Answer:
72km
Explanation:
30 mins --> 30 x 60 s = 1800 s
Distance --> Speed x Time
= 40m/s x 1800s
= 72 000 m
= 72 km (1km is 1000m)
Answer:
6.6 atm
Explanation:
Using the general gas law
P₁V₁/T₁ = P₂V₂/T₂
Let P₂ be the new pressure
So, P₂ = P₁V₁T₂/V₂T₁
Since V₂ = 2V₁ , P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius
T₂ = 273 + 2t (since its Celsius temperature doubles).
Substituting these values into the equation for P₂, we have
P₂ = P₁V₁(273 + 2t)/2V₁(273 + t)
P₂ = 12(273 + 2t)/[2(273 + t)]
P₂ = 6(273 + 2t)/(273 + t)]
assume t = 30 °C on a comfortable spring day
P₂ = 6(273 + 2(30))/(273 + 30)]
P₂ = 6(273 + 60))/(273 + 30)]
P₂ = 6(333))/(303)]
P₂ = 6.6 atm
