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const2013 [10]
3 years ago
9

A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

Physics
1 answer:
QveST [7]3 years ago
3 0

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

W = mg  = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2

b) at either end

P = \frac{F}{A} = P_o + \rho gh

dF = PdA

dF = \rho g w \int h dh

F = \rho g w \frac{h^2}{2}

F = \rho g A \frac{h}{2}           [A = wh]

F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}

F = 294000 N

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Sound travels at about 360 meters/sec. What is the wavelength of sound with a
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0.24 m

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Given:

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v = λ / t [ f = 1 / t ]

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1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant
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a=0   v = v₀ + a t

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in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

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What is the angle of reflection?
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3 years ago
Read 2 more answers
A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball
lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.  

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

M_{1} u_{1} +M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

Since initial velocity for M1 ball is zero, then

M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

and

M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

So, on solving all the above equation, we get an equation for velocity and that is

\frac{2M_{2}u_{2} }{(M_{1}+M_{2}  }=final velocity of ball with mass 2.5 kg

v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is \frac{1}{2}*2.5*(4.67)^{2}  =27.09 J

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0
3 years ago
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