A mover loads a 100 kg box into the back of a moving truck by pushing it up a ramp. The ramp is 4 m long and the back of the tru
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Refer to the diagram shown below.
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.
By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore
Here,
m = mass
g =Gravitational energy
The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore
Remember the expression for which you can determine the relationship between mass, volume and density, in which
In this case the density would be that of the object, replacing
Since the displaced volume of water is 0.429 we will have to
The density of water under normal conditions is , so
The density of the object is