D. Hydroelectric (hydro = water)
Understanding the given:
85 kg mountain climber
6.50 m long rope
gravity = 10m/s2
If we want to identify the work done on this scenario
we get f = 85kg x 10m/s2 = 850 N
w = 850N x 6.5 m = 5525 J
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Answer:
281.25 J
Explanation:
We are told that the two objects with masses m and 3m.
Also that energy stored in the spring is 375 joules.
Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3
Thus, from principle of conservation of energy, we have;
½mv² + ½(3m)(v/3)² = 375J
(m + 3m/9)½v² = 375
(4/3)m × ½v² = 375
Multiply both sides by ¾ to get;
½mv² = 375 × ¾
½mv² = 281.25 J
Therefore, energy of lighter body is 281.25 J
Answer: The ice cube would float on top of the water and the rock would sink to the bottom.
Explanation: The ice cube has a smaller density than the rock which allows the ice cube to float but makes the rock sink to the bottom of the glass of water.
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
