Nope.... Weight of object changes due to change in gravity but mass remains constant...
Answer:
The new position is 0.1865 m
Explanation:
As the context of the data is not available, thus following data is utilized from the question as attached above
x_relax=0.32 m
x_stiff=0.13 m
spring stiffness k=9 N/m
mass of block =0.073 kg
t=0.07 s
Velocity of the block is to be estimated thus
Force due to compression in spring is given as
F_s=k Δx
F_s=9(0.32-0.13)
F_s=1.71 N
Force on the block is given as
F_m=mg
F_m=0.073 x 9.8
F_m=0.71 N
Net Force
F=F_s-F_m
F=1.71-0.71 N
F=1 N
As Ft=Δp
So
Δp=1x0.07=0.07 kgm/s
Δp=p_final-p_initial
0.07=p_final-0
p_final=0.07 kgm/s
p_final=m*v_f
v_f=(p_final)/(m)
v_f=0.07/0.073
v_f=0.95 m/s
So now the velocity of the block is 0.95 m/s
time is 0.07 s
y_new=y_initial+y_travel
y_new=0.12+(0.95 x 0.07)
y_new=0.12+0.065
y_new=0.1865 m
So the new position is 0.1865 m
I love this one. All the effects of gravity are vertical. There is no horizontal force on the crate, and we are neglecting air resistance, so the horizontal speed of the crate doesn't change. When it gets to the ground, its horizontal speed is still 100 meters per second.
Correct option: A
An object remains at rest until a force acts on it.
As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.