PART D

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.

Response:

n (a sin θ) = m λ₀

Since n > 1, this indicates that the fringes separate further apart

Clarification:

In a diffraction experiment, the equation for constructive interference fringes is provided by

a sin θ = m λ₀

It is presumed that the air has been evacuated from the experiment, setting n = 1

When this experiment is conducted in water, the wavelength alters

λₙ = λ₀ / n

for achieving constructive interference

a sin θ = m λₙ

we replace

a sin θ = m λ / n

n (a sin θ) = m λ₀

Given that water's refractive index is n = 1.33, the distance between the fringes increases due to n > 1, causing the fringes to move apart

Answer:

1.32.225 N/C, moving away from the point charge

2. 8.972*10^-12 C

3. the field is oriented away from the axon

Explanation:

The calculation for the electric field is illustrated below:

E = k*|q|/r²

Where:

E = electric field; k = 8.98755*10⁹ N*m²/C²; r = distance separating the field being measured from the point charge = 0.05 m; q = point charge

For a length of 0.100 m of the axon, the value of q is calculated as:

q = (5.6*10¹¹)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Therefore:

q = (5.6*10¹¹)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Consequently:

E = (8.98755*10⁹)*(8.972*10^-12)/0.05² = 32.255 N/C

A positive point charge produces an electric field that radiates outward, while a negative point charge creates an electric field directed inward.

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F =?

According to Hooke's law:

F = Ke

= 0.7×1.5

= 1.05 N

a) 3.56 x 10^22 N. b) 3.56 x 10^22 N. The sun’s mass is M = 2 x 10^30 kg, while the Earth's mass is m = 6 x 10^24 kg, with a distance of R = 1.5 x 10^11 m separating them. Applying Newton's law for gravitational force F = G (mM / R²), where G = 6.67 × 10^-11 m^3 kg^-1 s^-2 gives us F = 3.56 x 10^22 N. A) The gravitational force by the sun on Earth equates to the force exerted by Earth on the sun, which is also 3.56 x 10^22 N.