What is the acceleration of a car that slows down from 25.0 m/s to 10.0 m/s in 30 seconds? The car is traveling south.

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

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Assume: Moving to the right is positive. A(n) 7.7 g object moving to the right at 22 cm/s makes an elastic head-on collision wit

krek1111 [17]

Answer:

Explanation:

We shall apply the formula for velocity in case of elastic collision which is given below

v₁ = (m₁ - m₂)u₁ / (m₁ + m₂) + 2m₂u₂ / (m₁ + m₂)

m₁ and u₁ is mass and velocity of first object , m₂ and u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.

Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm

v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 ) + 2 x 18 x - 14 / ( 7.7 + 18 )

= - 8.817 - 19.6

= - 28.4 cm / s

5 0

3 years ago

What does AS>0 mean?|<br> I

tigry1 [53]

Answer:

Significant Other

Explanation:

that comes to mind when I see that

5 0

3 years ago

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For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 78kg adult receive?

kari74 [83]

RDA stands for Recommended Daily Allowance. To determine the amount needed of a certain adult per day, we simply multiply the mass of the adult to the value of RDA. For this case, we do as follows:

Daily requirement = 12 mg/kg ( 78 kg ) = 936 mg of lysine = 936000 g

6 0

3 years ago

Hi!

dangina [55]

Answer:

the moe weight you have in the marble, the higher the speed on the way down

Explanation:

4 0

2 years ago