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bija089 [108]
2 years ago
7

O.

Physics
1 answer:
Tatiana [17]2 years ago
5 0

Answer:

Answer???????????????

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State and prove bessel inequality​
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
Read 2 more answers
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
(a) Calculate the buoyant force on a 2.00-L helium balloon.
iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

\rho = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N

The buoyant force is 0.0239364 N

Net vertical force is given by

F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N

The net vertical force is 0.0057879 N

6 0
3 years ago
Please help I'm stuck on this question ​
Afina-wow [57]

Answer:

increase

decrease

Explanation:

using formula

Vt=mg/6πηr

so if m increases V increases

r is the denominator so if r increases V decreases

8 0
2 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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