The answer c for your question
Answer:
Explanation:
Generally, length of vector means the magnitude of the vector.
So, given a vector
R = a•i + b•j + c•k
Then, it magnitude can be caused using
|R|= √(a²+b²+c²)
So, applying this to each of the vector given.
(a) 2i + 4j + 3k
The length is
L = √(2²+4²+3²)
L = √(4+16+9)
L = √29
L = 5.385 unit
(b) 5i − 2j + k
Note that k means 1k
The length is
L = √(5²+(-2)²+1²)
Note that, -×- = +
L = √(25+4+1)
L = √30
L = 5.477 unit
(c) 2i − k
Note that, since there is no component j implies that j component is 0
L = 2i + 0j - 1k
The length is
L = √(2²+0²+(-1)²)
L = √(4+0+1)
L = √5
L = 2.236 unit
(d) 5i
Same as above no is j-component and k-component
L = 5i + 0j + 0k
The length is
L = √(5²+0²+0²)
L = √(25+0+0)
L = √25
L = 5 unit
(e) 3i − 2j − k
The length is
L = √(3²+(-2)²+(-1)²)
L = √(9+4+1)
L = √14
L = 3.742 unit
(f) i + j + k
The length is
L = √(1²+1²+1²)
L = √(1+1+1)
L = √3
L = 1.7321 unit
Answer:
Gravitational potential energy to kinetic energy
Explanation:
In this case you have a case about conservation of energy.
When the mass is released and allowed to fall, its energy is completely gravitational potential energy with a value of U = mgh. m is the mass, g is the gravitational constant and h is the height to the floor from the mass.
While the mass is falling down part of its potential energy converts to kinetic energy of value K=1/2mv^2, because the mass has been acquiring more and more velocity.
Thus, the kinetic energy is increasing while the potential energy is decreasing.
When the mass is just above the floor (the moment just before the mass hits the floor) all its potential energy has been converted to kinetic energy.
Then, you have that the kinetic energy of the mass when the mass is just above the floor, is equal to the potential energy when the mass is at height of h. That is:
This is how the law of conservation of energy is fulfilled.