The work done on the filled bucket in raising out of the hole is 2, 925 Joules
<h3>How to determine the work done</h3>
Using the formula:
Work done = force * distance
Note that force = mass * acceleration
F = mg + ma
F = 4. 5 * 10 + 28 * 10
F = 45 + 280
F = 325 Newton
Distance = 9m
Substitute into formula
Work done = 325 * 9
Work done = 2, 925 Joules
Therefore, the work done is 2, 925 Joules
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Answer:
The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.
Explanation:
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Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
Their "airspeeds" (speed through the air) are equal, but the one traveling in the
same direction as the jet-stream appears to move along the ground faster.
