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2 years ago

An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m

Physics

1 answer:

Svetllana [295]2 years ago

5 0

The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

<h3>Work done by the force experienced by the object</h3>

The magnitude of the work done by force experience by the object is calculated as follows;

W = f.d

where;

F is the applied force (2xyi + 3yj), where x and y are in meters
d is the displacement of the object = (a, b)

The work done by the force is determined from the dot product of the force and the displacement of the object.

F = (2xyi + 3yj).(a + b)

W = (2abi + 3bj).(ai + bj)

W = (2a²b + 3b²)J

Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

The complete question is below:

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

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When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

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Answer:

460.52 s

Explanation:

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dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

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V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

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V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

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3 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

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The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

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B is the bulk modulus

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$\Delta \rho$ is the variation of density

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$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

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brainly.com/question/9805263

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