Answer:
0.78m (rounded to nearest hundredth of a meter)
explanation:
time taken for going up=time taken for drop down after reaching the highest point. at the highest point, the velocity becomes 0.
now all thats left is dropping an object from a height (h) and seeing how long it takes to reach the ground. then find out the flight’s total time divided by 2 (0.8/2=0.4)
lets say the velocity is v and the height she jumped to is h. we can make a kinematic expression:
s=vt+½gt²
once we put it all together you should get this:
h=0×0.4+½(9.81) 0.4²
.
∴
Time taken for downward drop
=
0.8
2
=
0.4
s
Suppose that she jumped with initial velocity
=
u
Also suppose that she jumped to a height
h
Using following kinematic expression
s
=
u
t
+
1
2
g
t
2
and inserting various quantities we get
h
=
0
×
0.4
+
1
2
(
9.81
)
0.4
2
h
=
0.78
m
rounded to nearest hundredth of a meter.
Answer:
c. Water
Explanation:
The solvent is the substance in which a solute is dissolved to make the solution.
In this case, the water is the solvent, the flavored powder is the solute, and the solution is the final drink.
The air drag is a force that depends on the speed of an object relative to the wind. Under certain conditions, it can be modeled as:

Where b is a constant.
As a falling object reaches a speed so that its weight is cancelled out by the air drag, the object will reach a maximum velocity.
In a speed vs time gaph, the speed would approach the maximum speed like an asymptote.
On the other hand, since the object falls from rest, the initial speed on the graph must be zero.
Taking these considerations into account, the correct graph for the movement of an object that falls from rest if air drag is not ignored, is option B.
Given,
The momentum of the object A before the collision, p₁ =80 Ns
The momentum of the object B before the collision, p₂=-30 Ns
Given that the objects stick together after the collision.
From the law of conservation of momentum, the total momentum of a system should always remain the same. Thus the total momentum of the objects before the collision must be equal to the total momentum of the objects after the collision.
Thus,

Where p is the total momentum of the system at any instant of time.
On substituting the known values,

Therefore the total momentum of the system is 50 Ns
Thus the momentum of the object AB after the collision is 50 Ns