Answer:
G = 0.424
Explanation:
Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))
Where Ds = stopping sight distance = 415miles = 126.5m
G = absolute grade road
V = velocity of vehicle = 52miles/hr
f = friction = 0 because the road is wet
tr = standard perception / reaction time = 2.5s
So therefore:
Substituting to get G
We have
2479.4G = 705.6G + 751.72
1773.8G = 751.72
G = 751.72/1773.8
G = 0.424
Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
<em></em>
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound =
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds =
kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Answer:
When the expenditure increased, then the consumer's expenditure is increased to Rs.150 and when the price falls of the good it becomes Rs.5. Then, Good X will be Rs.10.
last question:If the Good X falls by 20%, then, it will be Rs.2, and according to his demand 100 units will be equal to Rs.200.
cause if one unit=rs.2, then 100units=100×2=200.
Explanation:
Air fuel ratio:
Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.
When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.
When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.
Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²

x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K

K= 58.66 lb/ft
The damper coefficient for critically damped system


Cc= 12.7 lb.sec/ft