A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o
f 3.24 km. The bombardier releases one bomb. (a) How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance.
1 answer:
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
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Answer : The cell potential for this cell 0.434 V
Solution :
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First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the concentration of cell potential for this cell.
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
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Now put all the given values in the above equation, we get:
Therefore, the cell potential for this cell 0.434 V