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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.

You might be interested in

What's a disadvantage of highest MERV-rated filters?

Arte-miy333 [17]

Answer:

3) the pressure drop across high MERV filters is significant.

Explanation:

MERV (Minimum-Efficiency Reporting Value) is used to measure the efficiency of filter to remove particles. A filter of high MERV can filter smaller particles but this causes an increase in reduced air flow that is an increase in pressure drop. High MERV filters capture more particles causing them to get congested faster and thereby increasing pressure drop.

Excessive pressure drop can cause overheating and lead to damage of the filter. The pressure drop can be reduced by increasing the surface area of the filter.

3 0

3 years ago

Read 2 more answers

I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal

tigry1 [53]

Answer:

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

where, Nₐ₀-Nₐ = is the amount in moles of A used up

Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

from conservation of mass law, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if 3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

5 0

3 years ago

Superheated water vapor at a pressure of 20 MPa, a temperature of 500oC, and a flow rate of 10 kg/s is to be brought to a satura

katrin2010 [14]

Answer:

1.96 kg/s.

Explanation:

So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;

=> Superheated water vapor at a pressure = 20 MPa,

=> temperature = 500°C,

=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."

=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."

K1 = 3241.18, k2 = 93.28 and 2725.47.

Therefore, m1 + m2= m3.

10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.

=> 1.96 kg/s.

7 0

3 years ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat to the surroundings.

8 0

3 years ago

Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra

VashaNatasha [74]

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

int *p;

}arr[1000];

2) struct array{

int *p;

}arr[1000];

6 0

3 years ago