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3 years ago

Without motorcycle riders are at risk of severe injury in a crash ?

Engineering

2 answers:

astra-53 [7]3 years ago

6 0

The answer is A
Proper protection

nekit [7.7K]3 years ago

5 0

Answer:

The correct option is;

A. proper protection

Explanation:

Motorcycle riders ride the motorcycle while at some level of speed while having the entire body exposed to be a major part of any collision.

Injuries sustained from motorcycle accidents are several times more severe than injuries sustained by occupants of a car that is fully protected by the metallic panel in the same and even more serious accident scenarios

Hence, motorcycle riders require adequate protection by putting on available motorcyclist safety gear

Therefore, to reduce the risk of severe injury n a crash, motorcycle riders require proper protection.

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A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of

andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ $SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}$

By putting the value, we get

⇒ $=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}$

⇒ $=3.7258 \ l$

hence,

The heat loss will be:

⇒ $Q=SF\times K(T_2-T_1)$

$=3.7258\times 1\times 1.8\times (500-300)$

$=3.7258\times 1.8\times (200)$

$=1341.288 \ W/m$

3 0

3 years ago

g An analog voice signal, sampled at the rate of 8 kHz (8000 samples/second), is to be transmitted by using binary frequency shi

slamgirl [31]

Answer:

The module is why it’s goin to work

Explanation:

4 0

3 years ago

A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the

Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2)$

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

$C_p = 520 \:\: J/kg.K$

So, the inlet temperature of argon is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$

$T_{in} = \frac{1}{2} \times 119 + 300$

$T_{in} = 360K$

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

$C_p = 5193 \:\: J/kg.K$

So, the inlet temperature of helium is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$

$T_{in} = \frac{1}{2} \times 12 + 300$

$T_{in} = 306K$

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

$C_p = 1039 \:\: J/kg.K$

So, the inlet temperature of nitrogen is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$

$T_{in} = \frac{1}{2} \times 60 + 300$

$T_{in} = 330K$

Note: Answers are rounded to the nearest whole numbers.

5 0

3 years ago

A clean machine is a _______________ machine.

solniwko [45]

A clean machine is a clean machine :-)

4 0

3 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago