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28 days ago

The kinetic energy of the pom-pom came from potential energy in the system.

Physics

1 answer:

Sav [3.1K]28 days ago

4 0

I understand that potential energy, in the realm of physics, refers to the energy stored in an object due to its position relative to other objects, internal stresses, its electric charge, or other factors. Notable types of potential energy comprise gravitational potential energy based on an object's mass and its distance from the central mass of another object, elastic potential energy from an extended spring, and electric potential energy from an electric charge within an electric field.

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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

1 month ago

Six pendulums of mass m and length L as shown are released from rest at the same angle (theta) from vertical. Rank the pendulums

Sav [3153]

Answer: 1m, 1m, 2m, 2m, 4m, 4m.

It’s important to remember that the masses attached do not influence the number of oscillations.

Explanation:

To determine the number of oscillations (complete cycles), we can apply the formula n = t / T ……equation 1

The variables that impact the period of a simple pendulum are solely its length and gravitational acceleration. The period remains unaffected by factors such as mass.

period (T)= 2 x π x √(L/g) ….equation 2

where π = 3.142, L= rope length, and g = 9.8 m/s (gravitational acceleration)

According to the question, the time (t) is 60 seconds.

By merging equations 1 and 2, we obtain

number of oscillations = time / (2 x π x √(L/g))

Case 1: for L = 4m

number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))

= 14.9 = 14 complete cycles (the problem specifies complete cycles)

Case 2: for L = 2m

number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))

= 21.4 = 21 complete cycles

Case 3: where L = 4m, results in the same as case 1, yielding 14 complete cycles

Case 4: where L = 2m, mirrors the outcome in case 2, producing 21 complete cycles

Case 5: in the instance of L = 1m

number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))

= 30.1 = 30 complete cycles

Case 6: when L = 1m, which repeats case 5, also gives 30 complete cycles

From these findings, the order of the pendulums from the highest to lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.

Remember, the number of oscillations is independent of their respective masses.

3 0

2 months ago

Read 2 more answers

A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano

serg [3582]

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

$AD=AB+BD$

Insert the values into the equation

$D= 4+4.0\cos40^{\circ}$

$D=7.06\ km$

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

$AC=\sqrt{(CD)^2+(AD)^2}$

Insert the values into the equation

$AC=\sqrt{(4.0\sin40)^2+(7.06)^2}$

$AC=7.51$

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.

7 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

An experiment based at New Mexico’s Apache Point observatory uses a laser beam to measure the distance to the Moon with millimet

Ostrovityanka [3204]

Answer:

Unfortunately, I do not possess the information

Explanation:

6 0

1 month ago