(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.
(b) The velocity of the rock after 2 seconds is 7.56 m/s.
(c) The time for the block to hit the surface is 4.03.
(d) The velocity of the block at the maximum height is 0.
<h3>
Velocity of the rock</h3>
The velocity of the rock is determined as shown below;
Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m
v² = u² - 2gh
where;
- g is acceleration due to gravity in mars = 3.72 m/s²
v² = (15)² - 2(3.72)(13.14)
v² = 127.23
v = √127.23
v = 11.28 m/s
<h3>Velocity of the rock when t = 2 second</h3>
v = dh/dt
v = 15 - 3.72t
v(2) = 15 - 3.72(2)
v(2) = 7.56 m/s
<h3>Time for the rock to reach maximum height</h3>
dh/dt = 0
15 - 3.72t = 0
t = 4.03 s
<h3>Velocity of the rock when it hits the surface</h3>
v = u - gt
v = 15 - 3.72(4.03)
v = 0
Learn more about velocity at maximum height here: brainly.com/question/14638187
Answer: 117 kPa
Explanation:
For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa
For the liquid at depth 9m, the gauge pressure is equal to= P₂
Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.
So, For finding gauge pressure we have formula P= ρ * g * h
Also gravity also remains same for both liquids
So taking ratio of their respective pressures we have
= 
So
= 
Or P₂= 39 * 3 = 117 kPa
Answer:
100 ÷ 9.58 = 10.44 (approximate answer)
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
<u>W = 12.96 J</u>