Answer:
The maximum range 
m
Explanation:
Given,
The initial velocity of the car, u = 30 m/s
The height of the cliff, h = 50 m
Let the car drives off the cliff with a horizontal velocity of 30 m/s.
The formula for a projectile that is projected from a height h from the ground is given by the relation
m
Where,
g - acceleration due to gravity
Substituting the values in the above equation
= 132.72 m
Hence, the car lands at a distance, m
To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well
Where,
m = mass
h = height
g = Gravitational constant
When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore
Replacing in our equation we have to
The correct answer is 4.
Answer:
in the direction of the applied force
Explanation:
The volume of the object must be no larger than 
.
Explanation:
In order for an object to be able to float in water, its density must be equal or smaller than the water density.
The density of water is:
This means that the density of the object must be no larger than this value.
We also know that the density of an object is given by
where
m is the mass of the object
V is its volume
For the object in this problem, the mass is
Therefore, we can re-arrange the equation to find its volume:
So, the volume of the object must be no larger than .
Learn more about density:
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Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
