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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:

where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,

<u>v₂ = 30 m/s</u>
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b)

where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,

<u>m₁ = 12600 kg</u>
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c)

where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,

<u>m₂ = 12600 kg</u>
11 Years
Usually sunspot and solar flare cycles vary. But the average duration occurs every 11 years, sometimes 9-14 years as well.
The rod transferred negatively charged particle to the metal sphere. The rod becomes less negatively charged than it was originally (originally it was charged to -21 nc, now it has -8, and -8>-21), so it must have moved 13 nc of negative charge to the sphere.