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coldgirl [10]
4 years ago
6

Find the speed of a transverse wave on a 75-cm length of a cord when the tension in the cord is 320 N. The mass of the cord is 1

20 g.
Engineering
1 answer:
tino4ka555 [31]4 years ago
8 0

Answer:

The speed of transverse wave will be 28.2842 m/sec  

Explanation:

We have given length of the card = 75 cm = 0.75 m

Tension on the card = 320 N

Mass of the card = 120 gram = 0.12 kg

So linear density =\frac{mass}{length}=\frac{0.12}{0.75}=0.4kg/m

We have to find the speed of the transverse wave

Speed is given by v=\sqrt{\frac{T}{linear\ density}}

v=\sqrt{\frac{320}{0.4}}=28.2842m/sec

So the speed of transverse wave will be 28.2842 m/sec

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A cantilever beam is 4000 mm long span and has a u.d.l. of 0.30 kN/m. The flexural stiffness is 60 MNm². Calculate: 1. Slope 2.
Viefleur [7K]

Answer:

1. Slope = 53.3 x 10⁻⁶

2. Deflection = -0.00016m

Explanation:

given:

let L = 4 m (span of cantilever beam)

let w = 300 N/m (distributed load)

let EI =60 MNm² (flexural stiffness)

                 dy      w * L³        300 x 4³

1. slope = ------- = --------- =  ------------------- =  53.3 x 10⁻⁶

                  dx        6EI           6 x 60x10⁶

                                  wL⁴               300 x 4⁴

2. Deflection = y = - ----------- =  - ------------------ =   -0.00016m

                                    8EI                8 x 60x10⁶

therefore the deflection is 0.16mm downwards.

3 0
3 years ago
Can anyone pls help.Will mark as branliest
poizon [28]

Answer:

Explanation:

Hope this helped you!

5 0
3 years ago
The water level in a tank z1, is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end
Alika [10]

Answer:

the maximum height h to which the the water steam could rise is 40.65m

Explanation:

Given;

Z_{1} = 20m,

P_{1gage} = 2atm ≈ 20265 N/m

Density of water ρ = 1000 kg/m^{2}

Note: we take point 1 at the free surface of the water in the tank and point 2 at the top of the water trajectory. we also take reference level at the bottom of the tank.

fluid velocity at the free surface of tank is very low (V_{1} ≅ 0) and at the top of the water trajectory V_{2} = 0

Step 1: Applying Bernoulli equation between poin 1 and point 2

P_{1}/ρg + V_{1} ^{2}/2g + Z_{1} = P_{2}/ρg + V_{2} ^{2}/2g + Z_{2}

P_{1}/ρg + Z_{1} = P_{atm}/ρg + Z_{2}

Z_{2} = (P_{1} - P_{atm})/ρg + Z_{1}

Substituting values into Z_{2} we have,

Z_{2} = \frac{2atm}{(1000 kg/m^{3} )(9.81 m/s^{2} )} (\frac{101325N/m^{2} }{1atm} )(\frac{1 kg.m/s^{2} }{1N} ) + 20 = 40.65 m

6 0
3 years ago
What is the built-in pollution control system in an incinerator called
Kobotan [32]

Explanation:

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7 0
3 years ago
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype
ella [17]

Answer:

Explanation:

Given

scale i.e. L_r=1:15

Using Reynolds number similarity

(Re)_m=(Re)_p

(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p

Properties of air

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Properties of sea water

\nu _{sea}=1.26\times 10^{-5} ft/s

\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )

V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )

V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}

V_p=0.963\ ft/s

8 0
3 years ago
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