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3 years ago

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A small rock is thrown vertically upward with a speed of 21.0 m/sm/s from the edge of the roof of a 21.0-mm-tall building. The r

ock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance.A. What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
B. How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.

1 answer:

FromTheMoon [43]3 years ago

4 0

Answer:

Explanation:

A.

Given:

Vo = 21 m/s

Vf = 0 m/s

Using equation of Motion,

Vf^2 = Vo^2 - 2aS

S = (21^2)/2 × 9.8

= 22.5 m.

B.

Given:

S = 22.5 + 21 mm

= 22.521 m

Vo = 0 m/s

Using the equation of motion,

S = Vo × t + 1/2 × a × t^2

22.521 = 0 + 1/2 × 9.8 × t^2

t^2 = (2 × 22.521)/9.8

= 4.6

t = 2.14 s

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Answer:

47 m/s

Explanation:

golf club mass, mc = 180 g

golf ball mass, mb = 46 g

initial golf club speed, vc1 = 47 m/s

final golf club speed, vc2 = 35 m/s

initial golf ball speed, vb1 = 0 m/s

final golf ball speed, vb2 = ? m/s

The total momentum is conserved, then:

mc*vc1 + mb*vb1 = mc*vc2 + mb*vb2

Replacing with data and solving (dimension are omitted):

180*47 + 46*0 = 180*35 + 46*vb2

vb2 = (180*47 - 180*35)/46

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3 years ago

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State and prove bessel inequality

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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

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

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2 years ago

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Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,

Leviafan [203]

Answer:

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

Explanation:

As we know that the charge per unit length of the long cylinder is given as

$\lambda = \frac{q}{L}$

here we know that the electric field between two cylinders is given by

$E = \frac{2k\lambda}{r}$

now we know that electric potential and electric field is related to each other as

$\Delta V = - \int E.dr$

$\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr$

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2 years ago

A ski lift carries a 75.0-kg skier at 3.00 m/s for 1.50 min along a cable that is inclined at an angle of 40.0° above the horizo

Nookie1986 [14]

Answer

given,

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time = 1.50 min = 90 s

angle of inclination, θ = 40°

distance = s x t

= 3 x 90 = 270 m

a) W = F. d cos θ

W = mg. d cos θ

W = 75 x 9.8 x 270 x cos 40°

W = 152021.52 J

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b) Power extended by the ski

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2 years ago

What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a

viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

$COP = 8.49$

b

$COP_1 = 9.49$

Explanation:

From the question we are told that

The lower operation temperature of refrigerator is $T_1 = -8.00^oC = 265 \ K$

The upper operation temperature of the refrigerator is $T_2 = 23.2 ^oC = 296.2 \ K$

Generally the refrigerators coefficient of performance is mathematically represented as

$COP = \frac{T_1}{T_2 - T_1 }$

=> $COP = \frac{265}{296.2 - 265 }$

=> $COP = 8.49$

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

$COP_1 = \frac{T_2}{ T_2 - T_1}$

=> $COP_1 = \frac{296.2}{ 296.2 - 265 }$

=> $COP_1 = 9.49$

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3 years ago