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3 days ago

6

Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy

starts jumping on the trampoline too. Suddenly, Max goes twice as high even though he did not put any extra effort into jumping. What type of wave interaction is this?

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A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

kicyunya [3294]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

$U_1=k\dfrac{q_1q_2}{x_1}$

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Consequently, the work done by the electrostatic force on the moving charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Therefore, this concludes the solution.

3 0

2 months ago

The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes

Sav [3153]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass: m speed: v

2. mass: 4 m speed: v

3. mass: 2 m speed: ¼ v

4. mass: 4 m speed: v

5. mass: 4 m speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v² = mv²/2

KE of Body 2(mass₂) = ½*4m*v² = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)² = mv²/16

KE of Body 4(mass₄) = ½*4m*v ² = 2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)² = mv²/2

</psubstituting>

6 0

2 months ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

serg [3582]

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is $\sqrt{725}$ ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Now, let's calculate the derivative of distance in relation to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

1 month ago

A suspension system is being designed for a 1300 kg vehicle. When the vehicle is empty, its static deflection is measured as 2.5

Keith_Richards [3271]

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

3 0

2 months ago