Which of the device has the highest resistance?

A ring of charge with uniform charge density is completely enclosed in a hollow donut shape. An exact copy of the ring is comple

Naily [24]

Answer:

Explanation:

According to Gauss's theorem , total flux coming out of a closed surface area is equal to 1/ε₀ times charge enclosed

This flux coming out of enclosed charge does not depend upon the shape of enclosing surface area . It only depends upon the charge enclosed .

It the present context , in both the case ie in case of donut or sphere , since amount of charge enclosed is same , flux coming out too will be same in both the case

Hence the required ratio = 1 : 1

3 0

3 years ago

A driver in a 1000 kg car traveling at 20 m/s slams on the brakes and skids to a stop. If the coefficient of friction between th

KATRIN_1 [288]

Answer:

The car's skid marks will be 25.4842 m long.

Explanation:

According to Newton's second law:

Force = Mass × acceleration due to car

Also, The formula for frictional force,

Frictional force = μ × Normal Force

Also, Normal force = mass × acceleration due to gravitation(g)

So,

Frictional force = μ × mass × g

The two forces acting horizontally on the tire in opposite directions. So,

Mass × acceleration due to car = μ × mass × g

Solving,

Acceleration due to car = μ × mass × g

Given,

μ = 0.80

Also, 9.81 ms⁻²

So,

Acceleration due to car = 7.848 ms⁻²

Considering the Equation of motion as:

v² = u² - 2.a.s

Brakes are applied ad the car stops. The final velocity of the car (v) = 0 ms⁻¹

Given: Initial velocity of car (u) = 20 ms⁻¹

Acceleration, above calculated = 7.848 ms⁻²

Applying in the equation to calculate the distance as:

(0)² = (20)² - 2×(7.848)×s

So, Distance:

$s=\frac{400}{2\times 7.848}$

<u>The skid marks are 25.4842 m long.</u>

3 0

3 years ago

Vector a s is 2.80 cm long and is 60.0° above the x-axis in the first quadrant. vector b s is 1.90 cm long and is 60.0° below th

zysi [14]

Refer to the diagram shown below.

Vector a is 2.8 cm long and 60° above the x-axis in the first quadrant. Therefore it s represented by
$\vec{a} = 2.8 (cos60^{o} \vec{i}+sin60^{o}\vec{j})$

Vector b is 1.9 cm long and 60° below the x-axis in the fourth quadrant. Therefore it is represented by
$\vec{b}=1.9(cos60^{o}\vec{i}-sin60^{o}\vec{j})$

The vectors may be written as
a = 2.8*(cos60°, sin60°) = (1.445, 2.5028)
b = 1.9*(cos60°, -sin60°) = (0.950, -1.6454)

Part (a)
a + b = (1.445+0.950, 2.5028-1.6454)
= (2.395, 0.8574)
The magnitude of this vector is
√[2.395² + 0.8574² ] = 2.5438 cm
Its direction is
tan⁻¹ (0.8574/2.395) = 19.7° above the x-axis, in the first quadrant.
The resultant vector is shown in the figure below.

Part (b)
Similarly obtain
a - b = (0.495, 4.1485)
The magnitude is 4.1777.
The direction is 83° above the x-axis, in the 1st quadrant.
The resultant vector is shown in the figure below.

Part (c)
Similarly, obtain
b - a = (-0.495, -4.1483)
The magnitude is 4.1777.
The direction is 83° below the negative x-axis, in the 3rd quadrant.
The resultant vector is shown in the figure below.