1,000 watts = 1 kilowatt
2,000 watts = 2 kilowatts
3,000 watts = 3 kilowatts
4,000 watts = 4 kilowatts
<em>5,000 watts = 5 kilowatts</em>
The work done by the applied force on the object is (2ab²i + 3b²j) J.
<h3>Magnitude of the force on the object</h3>
The magnitude of the force on the object is calculated as follows;
f = (2xyi + 3yj)
when;
x = a, and y = b
f = (2abi + 3bj)
<h3>Work done by the force</h3>
The work done the applied force is the product of force and displacement of the object.
W = fΔs
where;
- Δs is displacement of the object
Δx = a - a = 0
Δy = 0 - b = -b
Δs = √(Δx² + Δy²)
Δs = √(-b)²
Δs = b
W = (2abi + 3bj) x b
W = (2ab²i + 3b²j) J
Thus, the work done by the applied force on the object is (2ab²i + 3b²j) J.
The complete question is below;
An object moving in the xy-plane is subjected to the force f = (2xyi + 3yj), where x and y are in m. The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis. How much work does the force do?
Learn more about work done here: brainly.com/question/8119756
Considering the answers.
1. Halve the voltage across it
2. Quarter the voltage across it
3. Double the voltage across it
4. Quadruple the voltage across
The appropriate answer would be 3. Double the voltage. According to ohms law the current through a conductor between two points is directly proportional to the voltage across the two points at a constant resistance. Therefore a increase in voltage causes a corresponding increase in the current.
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × 
slug/ft³
so here P1 + 
= P2 + 
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + = P2 +
0 + 
solve it we get
P2 = 87.9 lb/ft³
Alkali metal
Explanation:
In general, an alkali metal will be more reactive than an alkaline earth metal in the same period.
What determines reactivity of metals?
The electropositivity of metals determines how reactive they are.
- Electropositivity or metallicity is a measure of the tendency of atoms of an element to lose electrons.
- It is closely related to ionization energy and the electronegativity of an element.
- The lower the ionization energy of an element, the more electropositive or metallic it is
- Also, the more reactive it will be because, it can lose electrons more readily.
- Across a period from left to right, electropositivity decreases and from top to down a group, it increases.
- Since alkali metals are more electropositive than alkali earth metals, they are more reactive and readily lose their only electron.
- This is why the most reactive metal is francium found in the lower left corner on the periodic table.
- For non-metals, electronegativity is the most important factor.
Learn more:
Alkali metals brainly.com/question/6324347
Electronegativity brainly.com/question/11932624
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