Answer:
Explanation:
The forces exerted by each mass is best understood in terms of their momentum.
Momentum is a sort of compelling force or impulse. It is given as:
Momentum = mass x velocity
Let us consider the momentum of the balls;
Substance C;
Mass = 1kg
Velocity = 5m/s
Momentum of C = 1 x 5 = 5kgm/s
Substance D:
Mass = 100kg
Velocity = 5m/s
Momentum of D = 100kg x 5m/s = 500kgm/s
Body D has a higher momentum compared to Body C. This suggests that body D will exert a higher force than C when they collide.
The higher the momentum, the more the force of impact it has.
The answer is A: can change
Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a




h=5.15 m
The depth is 5.15 m.
Answer:
ΔK = 24 joules.
Explanation:
Δ
Work done on the object
Work is equal to the dot product of force supplied and the displacement of the object.
* Δ
Δ
can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δ
= (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).
This gives us
*
=
=
J