Twice the amplitude of each wave
Answer:
L = 41.09 Kg m2 / s The angular momentum does not depend on the time
Explanation:
The definition of angular momentum is
L = r x p
Where blacks indicate vectors
Let's apply this definition our case. Linear momentum
p = m v
Let's replace
L = m r x v
The given function is
x = 6.00 i ^ + 4.15 t j
^
We look for speed
v = dx / dt
v = 0 + 4.15 j ^
To evaluate the angular momentum one of the best ways is to use determinants
![L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]](https://tex.z-dn.net/?f=L%20%3D%20m%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C6%264.15t%260%5C%5C0%264.15%260%5Cend%7Barray%7D%5Cright%5D)
L = m 6 4.15 k ^
The other products give zero
Let's calculate
L = 1.65 6 4.15 k ^
L = 41.09 Kg m2 / s
The angular momentum does not depend on the time
Answer:

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

Explanation:
For this case we know the mass of the water given :

And we know that the initial temperature for this water is
.
We want to cool this water to the human body temperature 
Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Where
represent the specific heat for the water and this value from tables we know that
for the water.
So then we have everything in order to replace into the formula of sensible heat and we got:

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
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