Answer:
magnitude = 3
unit vector = 
Explanation:
Given vectors:
u = 2i + 2j - k
v = -i + k = -i + 0j + k
(a) u x v is the cross product of u and v, and is given by;
![u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]](https://tex.z-dn.net/?f=u%20X%20v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%262%26-1%5C%5C-1%260%261%5Cend%7Barray%7D%5Cright%5D)
u x v = i(2+0) - j(2 - 1) + k(0 - 2)
u x v = 2i - j - 2k
Now the magnitude of u x v is calculated as follows:
| u x v | = 
| u x v | = 
| u x v | = 
| u x v | = 3
Therefore, the magnitude of u x v is 3
(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e
û = 
u x v = 2i - j - 2k [<em>calculated in (a) above</em>]
|u x v| = 3 [<em>calculated in (a) above</em>]
∴ û = 
∴ û =
The distance of tiger's leap from the base of rock is 5.58 m
It is a question of two dimensional motion
The time of motion in two dimensional motion is given by:
t= 
where y is the height and g is the acceleration due to gravity
y is given to be 7.5m and let us assume g to be 9.8 m/s^2
t = 
= 1.24s
Using time and speed,
We know that distance is the product of speed and time,
Distance= speed x time
speed is given to be 4.5 m/s
distance from the base of rock = 4.5 x 1.24
= 5.58m
Hence the distance of tiger's leap from the base of rock is 5.58 m
Disclaimer:
The acceleration due to gravity is assumed to be 9.8 m/s^2
For further reference:
brainly.com/question/11213880?referrer=searchResults
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Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it ha
Answer:
The graph line that doesn't change in amounts.
Explanation:
Meaning if its a straight line horizontally across it is in equilibrium. If you don't know what I mean, search up equilibrium graph, and it will show you what I am talking about.
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
where F is the force applied, and
is the stretch of the spring with respect to its equilibrium position. Using the data, we find
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
