<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16
Traveling against the current:
(18 - c)*t = 8
Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16
Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16
(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0
Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c
So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3
(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3
And it's verified.</span>
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since

, and time is always positive, velocity will be positive in one direction and negative in the other direction.
Answer:
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Explanation:
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Answer:
Maximum permitted surface crack length is 1.29 mm
Explanation:
As per the question:
Tensile stress, 
Surface energy of magnesium oxide, SE = 
Modulus of elasticity of the material, E = 225 GPa = 
Now,
To calculate the maximum allowable surface crack length:

