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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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Answer:

Nuclear fission is almost 8,000 times more efficient than traditional fossil fuels at producing energy. That's a lot of energy packed into a small space. Nuclear energy is more efficient, which means it uses less fuel to power the plant and produces less waste.


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A motor car shaft consists of a steel tube 30 mm internal diameter and 4 mm thick. The engine develops 10 kW at 2000 r.p.m. Find
tresset_1 [31]

The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.

<h3>What is power?</h3>

Power is the energy transferred per unit time.

Torque is find out by

P = 2πNT/60

10000 = 2π x 2000 x T / 60

T =47.74 N.m

The gear ratio Ne / Ns =4/1

Ns =2000/4 = 500

Ts =Ps x 60/(2π x 500)

Ts =190.96 N.m

Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))

τ max =T/J x D/2
where d₁ = 30mm = 0.03 m

           d₀ = 30 +(2x 4) = 38mm =0.038 m

Substitute the values into the equation, we get

τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)

τ max = 28.98 MPa.

Thus, the maximum shear stress in the tube is 28.98 MPa.

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brainly.com/question/13385520

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7 0
2 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

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