Answer:
- 1.07 ft
Explanation:
V1 = (-5, 7, 2)
V2 = (3, 1, 2)
Projection of v1 along v2, we use the following formula
=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}
So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4
The magnitude of vector V2 is given by
=
So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft
Thus, the projection of V1 along V2 is - 1.07 ft.
so we need to find the direction of v2
Explanation:
36-4/4= 9 m/squared. meter per squared.
acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)
Answer:
At point A, the cart has high potential energy. At point b, the cart is pulled down by gravity. At point c, the cart gains its highest kinetic energy. At point d, the cart returns back to the same state but with lower potential energy.
Answer:
The speed of water must be expelled at 6.06 m/s
Explanation:
Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:
with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:
with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:
Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so :
We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for :
The minus sign indicates the velocity of the water is opposite the velocity of the octopus.