Hi there!
I. They are the same.
Due to the Conservation of Momentum, an explosion as such means that the TOTAL momentum of the system (all of the pieces) is conserved.
Answer:
As collision is elastic,thus we can use conservation of momentum equation
mA=0.2 kg
(vB)1=0 m/s.......................as it is on rest before collision
(vA)1=4 m/s
(vA)2=-1 m/s
(vB)2=2 m/s
using equation
(mA*vA+mB*vB)1= (mA*vA+mB*vB)2
Where 1 and 2 represents before and after collision
(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)
0.8=-0.2+(2mB)
mass of object B=mB=0.3 Kg
B is the right answer glad I could help !!
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C) 
= 2.3 10³⁹
Explanation:
A) It is asked to find the force of attraction due to the masses of the particles
Let's use the law of universal attraction
F = 
let's calculate
F = 
F_g = 4.05 10⁻⁴⁷ N
B) in this part it is asked to calculate the electric force
Let's use Coulomb's law
F = 
let's calculate
F = 
F_e = 9.2 10⁻⁸N
C) It is asked to find the relationship between these forces
= 2.3 10³⁹
therefore the electric force is much greater than the gravitational force
