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)
Answer:
The answer is 2.32 m/s
Explanation:
Solution
Given that:
U a-b = ΔT
Thus
F cos 30° * hₐ - F sin 30° * hₐ + Whₐ = 1/2 m (v²b - v²ₐ)
Now,
vb =√2Fhₐ (cos 30° - sin 30°)+ mghₐ/ m + v²ₐ
Where
F = This is the force acting on the collar
m = this is the mass on the collar C
g = The acceleration due to gravity
hₐ = The height of the collar at the position A
vb = The velocity of the collar at position B
vₐ = The velocity of the collar at A
So,
We replace 5N for F, 0.2 m for hₐ, 0.5 kg for m, 9.81 m/s for g, 0 for vₐ
Now,
vb =√ 2 * (5 *0.2 * ( cos 30° - sin 30°) +0.5 * 9.81 * 0.2 / 0.5 + 0
=√2.694/0.5
=2.32 m/s
Hence, the he velocity v with which the collar strikes the end B is 2.32 m/s
Answer:
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Answer:
The minimum diameter for each cable should be 0.65 inches.
Explanation:
Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,
(1/2)(Weight/Cross Sectional Area) = Allowable Stress
Here,
Weight = 1000 lb
Cross-sectional area = πr²
where, r = minimum radius for each cable
(1/2)(1000 lb/πr²) = 1500 psi
500 lb/1500π psi = r²
r = √1.061 in²
r = 0.325 in
Now, for diameter:
Diameter = 2(radius) = 2r
Diameter = 2(0.325 in)
<u>Diameter = 0.65 in</u>