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3 years ago

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A bus traveling at 25.2mi/h speeds up at a constant rate of 9.2x10^-3 m/s^2. What velocity does it reach 25min later?

1 answer:

Softa [21]3 years ago

6 0

Answer: actually to solve this u must know what 3.2m/s2 is..do u know it ??...wel let me tell...3.2m/s2 means that in every second the speed of the body increases by 3.2m/s..so if it acts for 6.8 seconds..increase in speed will be 3.2*6.8 however note that this is the increase in speed.. have to change this in km/hr & add this to the initial speed(27km/hr) to get the answer....hope this has cleared ur doubts....

Explanation:

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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

Stella is driving down a steep hill. she should keep her car __________ to help _________.

Softa [21]

Answer:

Stella is driving down a steep hill. She should keep her car in a lower gear to help slow her vehicle.

Explanation:

8 0

3 years ago

Greeley [361]

Explanation:

q= n e

6 × 10^-11 = n (1.6 × 10^-19)

n = 6×10^-11 / 1.6 × 10^-19

n= 3.75 × 10⁸ electrons

4 0

3 years ago

How much force is needed to accelerate a 1.5 kg physics book to an acceleration of 6<br> m/s^2?

aleksley [76]

Answer:

Force = 8.0 k g m / s

Explanation:

Force = mass x acceleration

Mass = 4.0 k g Acceleration = 2.0 m / s 2

Hence,force = ( 4.0 x 2.0 ) k g m / s 2 = 8.0 k g m / s 2

3 0

3 years ago

A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop

liq [111]

Answer:

692.31 N

Explanation:

Applying,

F = ma............... Equation 1

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 0

3 years ago