Answer:
The level of service of of compound grade freeway is LOSB.
Explanation:
Find the provided attachments for explanation
Answer:
the crown is false densty= 12556kg/m^3[/tex]
Explanation:
Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.
W=mg
W=weight(N)=31.4N
M=Mass
g=gravity=9.81m/S^2
solving for M
m=W/g

The second step is find the volume of crown remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water

Where
F=weight in water=28.9N
m=mass of crown=3.2kg
g=gravity=9.81m/S^2
α=density of water=1000kg/m^3
V= crown´s volume
solving for V

finally, we remember that the density is equal to the index between mass and volume

To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.
1.weigh the crown in the air and find the mass
2. put water in a cylindrical bucket and measure its height with a ruler.
3. Put the crown in the bucket and measure the new water level with a ruler.
4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.
5. find density by dividing mass by volume
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by


78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer: a) The technology that deals with the generation, control and transmission of power using pressurized fluids
Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.
Answer:
Fuel efficiency for highway = 114.08 miles/gallon
Fuel efficiency for city = 98.79 miles/gallon
Explanation:
1 gallon = 3.7854 litres
1 mile = 1.6093 km
Let's first convert the efficiency to km/gallon:
48.5 km/litre = (48.5 * 3.7854) km/gallon
48.5 km/litre = 183.5919 km/gallon (highway)
42.0 km/litre = (42.0 * 3.7854) km/gallon
42.0 km/litre = 158.9868 km/gallon (city)
Next, we convert these to miles/gallon:
183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon
183.5919 km/gallon = 114.08 miles/gallon (highway)
158.9868 km/gallon = (158.9868 /1.6093) miles/gallon
158.9868 km/gallon = 98.79 miles/gallon (city)