Probably for the umbilical cord that connects babies (from their early stages in the womb to their removal) to their mothers. The cord is cut, forming the belly button. This is analogous to astronauts in space.
<span>a. The magnitude of the vector is doubled as well.
Let's say we have a 2-dimensional vector with components x and y.
It's magnitude lâ‚ is given by:
lâ‚ = âš(x² + y²)
If we double the components x and y, the new magnitude lâ‚‚ is:
lâ‚‚ = âš((2x)² + (2y²))
With a bit of algebra...
lâ‚‚ = âš(4x² + 4y²)
lâ‚‚ = âš4(x² + y²)
lâ‚‚ = 2âš(x² + y²)
We can write the new magnitude lâ‚‚ in terms of the old magnitude lâ‚.
lâ‚‚ = 2lâ‚
Therefore, the new magnitude is double the old one.
It should be clear that this relationship applies to 3D (and 1D) vectors as well.
b. The direction angle is unchanged.
The direction angle θ₠for a 2-dimensional vector is given by:
θ₠= arctan(y / x)
If we double both components, we get:
θ₂ = arctan(2y / 2x)
θ₂ = arctan(y / x)
θ₂ = θâ‚
The new direction angle is the same as the old one.</span>
Vf ( final velocity) = ?
Vo ( initial velocity) = 1.5 m/s
a (acceleration) = .65 m/s^2
t (time) = 2.6 seconds
The equation you want is Vf = Vo + (a)(t).
Plugging in the variables gives Vf = 1.5 + (.65)(2.6).
The final velocity is 1.5 + 1.69, which comes out to +3.19 m/s.
Answer:
Explanation:
Given that:
The weight of ball = 4 pounds
The spring stretch x = 1/15 feet
Using the relation of weight on an object:
W = mg
m = W/g
m = 4 / 32
m = 1/8
Now, from Hooke's law:
F = kx
4 =k(1/5)
k = 20 lb/ft
However, since the air resistance is 4 times the velocity;
Then, we can say:
C = 4
Now, for the damped vibration in the spring-mass system, we have:
Solving the differential equation:
m² + 32m + 160 = 0
Solving the equation:
m = -25.80 or m = -6.20
So, the general solution for the equation is:
y(0) = 0 ; y'(0) = 8
At y'(0) = 8
From (1), let 
, then replace the value of c_1 into equation (2)
From
∴
The required solution in terms of t is:
