In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

Response:

(b) 10 Wb

Clarification:

Given;

angle of the magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ = 5.0 Wb

The equation for magnetic flux is;

Φ = BACosθ

where;

B denotes the magnetic field strength

A represents the area of the plane

θ is the inclination angle

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Next, calculate the magnetic flux through a 2.0 m² section of the same plane:

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

Option "b"

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

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.

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.