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Jobisdone [24]
4 years ago
5

A rectangular trough, 1.5 m long, 0.50 m wide, and 0.40 m deep, is completely full of water. One end of the trough has a small d

rain plug right at the bottom edge. Part APart complete When you pull the plug, at what speed does water emerge from the hole
Physics
1 answer:
Vlad1618 [11]4 years ago
6 0

Answer:

V = 2.801 m/s

Explanation:

Given.

- The dimension of the tank = 1.5 x 0.5 x 0.4 m

- A plug at the bottom of the tank

Find:

When you pull the plug, at what speed does water emerge from the hole

Solution:

- We can compute the solution either by an energy balance or Bernoulli's equation as follows:

         P_1 + 0.5*p*V_1^2 + p*g*z_1 = P_2 + 0.5*p*V_2^2 + p*g*z_2

Where,

P: Pressure of the fluid (gauge)

V: Velocity of the fluid

p: Density of the fluid

z: The elevation of fluid from datum(free surface)

g: The gravitational acceleration constant

- We will consider the two states by setting a datum at the bottom of the surface.

State 1 (Free surface of the fluid at level):

            P_1 = 0 (gauge - atm)

            V_1 = 0 (Free surface)

            z_1 = 0.40 m

State 2 (Plug level @ bottom):

            P_2 = 0 (gauge - atm)

            V_2 = unknown (need to calculate)

            z_1 = 0  ... (datum)

- Lets plug the values at each state in the equation above:

               0 + 0 + p*g*z_1 = 0 + 0.5*p*V_2^2 + 0

                              V^2 = 2*g*z_1

                               V = sqrt ( 2*g*z_1)

                               V = sqrt ( 2*9.81*0.4)

                              V = 2.801 m/s

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20.7 s

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In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum
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Answer:

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By method of dimension show that the following equation are homogenous.
il63 [147K]

Answer:

Proof in explanataion

Explanation:

The basic dimensions are as follows:

MASS = M

LENGTH = L

TIME = T

i)

Given equation is:

H = \frac{u^2Sin^2\phi}{2g}

where,

H = height (meters)

u = speed (m/s)

g = acceleration due to gravity (m/s²)

Sin Ф = constant (no unit)

So there dimensions will be:

H = [L]

u = [LT⁻¹]

g = [LT⁻²]

Sin Ф = no dimension

Therefore,

[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]

<u>[L] = [L]</u>

Hence, the equation is proven to be homogenous.

ii)

F = \frac{Gm_1m_2}{r^2}\\\\

where,

F = Force = Newton = kg.m/s² = [MLT⁻²]

G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²

G = [M⁻¹L³T⁻²]

m₁ = m₂ = mass = kg = [M]

r = distance = m = [L]

Therefore,

[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

<u>[MLT⁻²] = [MLT⁻²]</u>

Hence, the equation is proven to be homogenous.

8 0
3 years ago
A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What
kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let p_{1} = momentum of the 1st ball

p_{2} = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}

or

(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}

Since we are dealing with identical balls, all the m terms cancel out so we are left with

(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} +  (v_{2})_{f}\cos \theta_{2}

Putting in the numbers, we get

1.33 = (0.750) \cos(33.30)  + (v_{2})_{f} \cos \theta_{2}

=  > (v_{2})_{f} \cos \theta_{2} = 0.703

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}

or

=  > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30)  = 0.412

Taking the ratio of the sine equation to the cosine equation, we get

\frac{ \sin \theta _{2}}{ \cos \theta_{2} }  =  \tan \theta_{2}  =  \frac{0.412}{0.703}  = 0.586

or

\theta_{2}  =  { \tan}^{ - 1} (0.586) = 30.4

Solving now for (v_{2})_{f},

(v_{2})_{f}  =  \frac{0.412}{ \sin(30.4) }  = 0.815 \:  \frac{m}{s}

3 0
3 years ago
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