Answer:
14.1mins
Explanation:
We know that
Time =Distance/Speed
So
= 1.7 au/ (3 E8 m /s )
= 1.7 x 1.496 x 10^11 m /(3 x 10^8 m /s
Converting Au to meters (1.7 x 1.496 x 10^11 m)
= 847.7 s
=14.1 mins
All continental coastlines match up with continental drifts. As a matter of fact, you could literally cut out the continents from a map and put them together like a puzzle and they would fit. Easiest to notice is the west of Africa and the east of South America.
Answer:
(a) 1.054 m/s²
(b) 1.404 m/s²
Explanation:
0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ)) - μk·m·g·(1 - sin(θ)) = m·a
Which gives;
0.5·g·cos(θ) - μ·g·(1 - sin(θ) = a
Where:
m = Mass of the of the block
μ = Coefficient of friction
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the block
θ = Angle of elevation of the block = 20°
Therefore;
0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°) - μk×9.81×(1 - sin(20°) = a
(a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;
0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°) - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²
(b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;
0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°) - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².
Answer:
[1, 6, -2]
Explanation:
Given the following :
Initial Position of spaceship : [3 2 4] km
Velocity of spaceship : [-1 2 - 3] km/hr
Location of ship after two hours have passed :
Distance moved by spaceship :
Velocity × time
[-1 2 -3] × 2 = [-2 4 -6]
Location of ship after two hours :
Initial position + distance moved
[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]
= [3-2, 2+4, 4-6] = [1, 6, -2]
