Take the derivative to find the velocity of the object:
The object stops when :
so the answer is E.
50 pounds is equal to 22,700 grams. How many pounds does this backpack full of gold weigh? Would you be able to carry out that b
1 answer:
The question is incomplete. Here is the complete question.
7) A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold (density = 19.3 g/cm³) while treasure hunting in the wilderness. How much mass would your backpack hold if you filled it with gold?
8) 50 pounds is equal to 22,700 grams. How many pounds does this backpack full of gold weigh? Would you be able to carry out that backpack filled with gold?
Answer: 7) m = 694,800g
8) It wieghts 1530.4 lbf
Explanation:
7) Volume of the backpack will be
V = 30 x 30 x 40
V = 36000 cm³
According to density of gold, each cm³ corresponds to 19.3g, then:
m = 19.3 * 36000cm³
m = 694800g
If I filled it with gold, the backpack will hold 694800g or 694.8kg.
8) 50 pounds is equal to 22700g. Then:
50 pound = 22700g
W = 694800g
W =
W = 1530.4 lbf
The backpack, in pounds, will weigh 1530.4lbf, which is very heavy to be supported by a human body.
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Answer:
a) F_net = 6.48 10⁻¹⁸ ( ), b) x = 0.15 m
Explanation:
a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.
The electric force is given by Coulomb's law
F =
Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.
F_net= ∑ F = F₁ + F₂
let's locate a reference system in the load that is on the left side, the distances are
left side - electron r₁ = x
right side -electron r₂ = d-x
let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q
we substitute
F_net = k q Q ( )
F _net = kqQ ( )
let's substitute the values
F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( )
F_net = 6.48 10⁻¹⁸ ( )
now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given
x (m) F (N)
0.05 27.0 10-16
0.10 8.10 10-16
0.15 5.76 10-16
0.20 8.10 10-16
0.25 27.0 10-16
b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m
