You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p
1 answer:
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Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
Answer:
The distance traveled will be 156.0 m
Explanation:
Please, see the attached figure 1 for a description of the problem.
The distance traveled by the prey is the magnitude of the vector "r" (see figure).
If the components of "r" are "rx" (horizontal component) and "ry" (vertical component), its magnitude will be:
We already know the vertical component of r, ry, because the prey falls from an altitude of 156 m (see figure). Then ry = 156 m
To find the value of rx we will use the function that describes the trajectory of the prey (see figure 2):
y = -39 · x² + 156
When the prey hits the ground, the height is 0 (y = 0). rx will be the value of x for which y = 0.
0 = -39 · x² + 156
-156/-39 = x²
x = 2.0 m
Then r = (2.0 m, -156.0 m)
The distance traveled will be:
