Answer:
Explanation:
Give that,
Spring constant (k)=40N/m
Force applied =75N
Since the force is applied to the right, we don't know if it is compressing or stretching the spring
So let assume it compress
Using hooke's law
F=-ke
e=-F/k
Then, e=-75/40
e=-1.875m
The deformation is 1.875m.
Let assume it stretch
Using hooke's law
-F=-ke
e=F/k
Then, e=75/40
e=1.875m
The elongation is 1.875m
The charge on the moving particle
Answer:
W₂= 10000 N
Explanation:
Pascal´s Principle can be applied in the hydraulic press:
If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:
Pressure is defined as the force (F) applied per unit area (A)
P=F/A (N/m²)
P1=P2

Equation (1)
Data
W₁ = weight sits on the small piston
F₁ = W₁= 500 N
A₁ = 2.0 cm²
A₂ = 40 cm²
Calculation of the weight (W₂) can the large piston support
We replace data in the equation (1)
F₂ = 10000 N
W₂= F₂= 10000 N
Even tho one is stronger then the other... they are both alike because they are still nuclear forces.