Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough surface. The exercise involves

Response:

Magnitude of the electrostatic force acting on the +32 µC charge,

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

The resultant electrostatic force on the 32 µC charge is

Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

We are tasked with finding the pressure here

Applying the pressure formula

Here,

Where, v refers to velocity

Insert the values into the equation

Therefore, the pressure at this moment is 0.875 mPa

Answer:

Explanation:

Within a duration of 60 seconds, six waves are observed.

With a total of 6 waves,

this equates to 3 wavelengths.

As a result,

the period for each wavelength is calculated as 60 divided by 3.

Thus, period = 20 seconds.

According to the frequency-period relationship,

f = 1 / T

f = 1 / 20

f = 0.05 Hz

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

where:

represents the pressure difference across the two ends

denotes the viscosity of the fluid

L signifies the length of the pipe

indicates the volumetric flow rate, where is the cross-sectional area of the tube and refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

is the dynamic viscosity of water at

L=4.0 cm=0.04 m

and r=1 mm=0.001 m

Substituting these values into the formula yields:

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.