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2 years ago

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In a race over a distance s, runner A starts from rest and accelerates at a1 for the first distance x and then runs at constant

speed. Runner B starts from rest and accelerates at a2 for the first distance x and then runs at constant speed. Runner A begins running as soon as the race begins but B first takes a nap to rest up.What is the longest nap B can take and still not lose the race?

1 answer:

prohojiy [21]2 years ago

8 0

Answer:

dt = (x+s)/sqrt(2x)*(1/sqrt(a1) - 1/sqrt(a2))

Explanation:

- Equation of motion for constant acceleration are:

v = a*t

z =( a*t^2)/2

v^2 = 2*z*s

- Compute time t1:

x = (a*t^2)/2

t_1 = sqrt(2*x/a)

- Velocity @ time t is: v = a*t = a*sqrt(2*x/a)

- Remaining distance (s-x) is run at above calculated speed, so the time t_2 is:

t_2 = (s-x)/v = (s-x)/(a*sqrt(2*x/a))

- Compute total T:

T = t_1 + t_2 = sqrt(2*x/a) + (s-x)/(a*sqrt(2*x/a))

- Simplify: T = (x+s)/sqrt(2*a*x)

- Time taken by each runner:

T_A =(x+s)/sqrt(2*a1*x)

T_B = (x+s)/sqrt(2*a2*x)+ dt

Equating the total times (T_A = T_B):

(x+s)/sqrt(2*a2*x)+ dt = (x+s)/sqrt(2*a1*x)

Hence,

dt = (x+s)/sqrt(2*a1*x) - (x+s)/sqrt(2*a2*x))

dt = (x+s)/sqrt(2x)*(1/sqrt(a1) - 1/sqrt(a2))

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Explanation:

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System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

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For linear motion in horizontal direction it is:

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τ = RF --------(4)

Put (4) and (1) in (2)

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from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

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From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

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$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

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Time period is related to angular frequency as:

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$T = 2\pi \sqrt{\frac{3M}{2k}$

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2 years ago

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Sunny_sXe [5.5K]

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0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

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Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

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Answer:

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Explanation:

Hello!

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(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature. )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

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2 years ago