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Pavel
10 days ago
6

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

eir original values). How is the resistance of the wire affected?
Physics
1 answer:
Softa [2K]10 days ago
4 0

Response:

The new resistance is half of the original resistance.

Explanation:

Resistance in a wire is represented by:

R=\dfrac{\rho L}{A}

\rho = resistivity of the material

L and A are the physical dimensions

If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r

The updated resistance of the wire can be calculated as follows:

R'=\dfrac{\rho L'}{A'}

R'=\dfrac{\rho (2L)}{\pi (2r)^2}

R'=\dfrac{1}{2}\dfrac{\rho L}{A}

R'=\dfrac{1}{2}R

The new resistance equals half of the original resistance. Thus, this provides the solution needed.

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A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
Yuliya22 [2420]

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

<pHence, the magnetic flux through a 2.0 m² area of the same plane is 10 Wb.

Option "b"

3 0
25 days ago
A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of
ValentinkaMS [2425]
<span>3.834 m/s. To solve this problem, we must ensure that the centripetal force equals or exceeds the gravitational force acting on the object. The formula for centripetal force is F = mv^2/r while the equation for gravitational force is F = ma. Since the mass (m) cancels out in both equations, we can equate them, leading to a = v^2/r. Now, inserting the given values (where the radius is half the diameter) allows us to find v: 9.8 m/s^2 <= v^2/1.5 m, which simplifies to 14.7 m^2/s^2 <= v^2. Therefore, we find that the minimum velocity required is 3.834057903 m/s <= v. Thus, the necessary speed is 3.834 m/s.</span>
4 0
16 days ago
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
serg [2593]

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.

6 0
27 days ago
Read 2 more answers
A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
Softa [2029]

The overall force acting on the vehicle is zero

Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

  • The weight component acting perpendicular to the slope, mgcos \theta, directed into the slope
  • The normal force N, directed outward from the slope

Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

  • The weight component aligned with the slope, mgsin \theta, directed down the slope
  • The frictional force F_f, acting up the slope

The car moves at a constant speed in this direction, indicating that its acceleration is zero.

a=0

Thus, according to Newton's second law,

F=ma

implying the net force is zero:

F=0

Learn more about slopes and friction:

5 0
21 day ago
Question 1
Keith_Richards [2256]

Answer:

1)  g = 4π² / m, 3) on the x-axis we have the pendulum lengths, while the y-axis shows the squared periods.

Explanation:

a) learners can model this system as a simple pendulum, where the angular velocity is given by

         w = √ g / l

Here, angular velocity, frequency, and period are interconnected:

         w = 2π f = 2π / T

Substituting yields:

         T = 2π√ l / g

Using this formula, students can calculate the gravitational acceleration by measuring the period for several pendulum lengths and plotting:

        T² = 4π²  l / g

We plot T² against l.

This represents a linear equation where T² is on the y-axis and l is on the x-axis:

        y = (4π² / g) l

The slope is given by:

         m = 4π² / g

Solving for g gives:

         g = 4π² / m

The slope is determined from the line's values rather than experimental data.

2) To perform the experiment, the string is secured to the sphere, then the pendulum length from the pivot to the sphere's center is measured using a tape measure. A slight angle (less than 10 degrees) is released, allowing the first swing to occur. Generally, the time for several oscillations, usually 10 or 20, is tracked to find the period:

    T = t / n

Next, a table is created comparing T² to the length, plotted with length on the x-axis to find the slope, from which the gravitational acceleration is derived.

3) The independent variable, which is the length of the pendulums, is plotted on the x-axis, while the dependent variable, the squared period, is on the y-axis.

4) Referring to the line equation:

            m = 4π² / g

             resulting in:

            g = 4π² / m

5) Once the spring is cut, the sphere continues to be influenced by gravitational acceleration. The harmonic motion ceases, and the sphere moves vertically.

5 0
1 month ago
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