Answer:
The work done by the gravel to stop the truck is 520.44 kJ
Explanation:
<u>Step 1</u>: Data given
Mass of the truck = 3047.8 kg
The ramp has an angle of 9.5 °
Velocity of the truck = 20.68 m/s
distance = 26.6 meters
<u>Step 2:</u> Calculate initial kinetic energy
sin 9.5° = 0.165
h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m
Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy
<u>Step 3: </u>Calculate potential energy
Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ
<u>Step 4:</u> What work is done by the truck on the gravel?
Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ
( 3 yr) · (186,282.397 mile/s) · (86,400 s/day) · (365 day/yr)
= (3 · 186,282.397 · 86,400 · 365) mile
= 1.762380502 x 10¹³ miles
= 1.8 x 10¹³ miles (rounded to the nearest trillion miles)
Answer:
Choice C is not equivalent to 2.50 miles.
Explanation:
The given data is now converted into feet, inches, kilometers, yards and centimeters:
mi - ft


(Choice A)
mi - in


(Choice B)
mi - km

(Different from Choice C)
mi - yd


(Choice D)
mi - cm


(Choice E)
Choice C is not equivalent to 2.50 miles.
Answer:
Explanation:
This is case of interference in thin films
for constructive interference in thin film the condition is
2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .
2 x 1.28 x t = λ/2 , if n = 0
2 x 1.28 x t = 605 /2
t = 118.16 nm .
the minimum non-zero thickness of the oil film required = 118.16 nm.
Id say d because it releases hydrogen and on the other hand a base receives it
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