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3 years ago

8

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

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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Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12

Nina [5.8K]

Answer:

W = - 184.8 kW

Explanation:

Given data:

$P_1 = 1.05 bar$

$T_1 = 300K$

$\dot V_1 = 84 m^3/min$

$P_2 = 12 bar$

$T_2 = 400 K$

We know that work is done as

$W = - [ Q + \dor m[h_2 - h_1]]$

for $P_1 = 1.05 bar, T_1 = 300K$

density of air is 1.22 kg/m^3 and $h_1 = 300 kJ/kg$

for $P_2 = 12 bar, T_2 = 400 K$

$h_2 = 400 kj/kg$

$\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s$

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

8 0

3 years ago

Select the correct answer. Felix aspires to be an engineer working for the government. What credentials will Felix require to ap

Aneli [31]

Answer:

OB

Explanation:

5 0

3 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

<u> Take Re=100,000:</u>

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

<u> Take Re=500,000:</u>

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago

A car C is traveling up a spiral ramp at a constant and traveling as fast as it can without losing traction. The grade of the ra

arsen [322]

Answer:

26.1 ft/s²

Explanation:

See attached pictures for detailed explanation.

4 0

3 years ago

A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr

Hoochie [10]

Answer:

$\Delta m = 102.25\,kg$

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

$m_{o} = \rho_{air}\cdot V_{tank}$

$m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{o} = 31.25\,kg$

The final mass inside the rigid tank is:

$m_{f} = \rho \cdot V_{tank}$

$m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{f}= 133.5\,kg$

The supplied air mass is:

$\Delta m = m_{f}-m_{o}$

$\Delta m = 133.5\,kg-31.25\,kg$

$\Delta m = 102.25\,kg$

4 0

3 years ago