On a world map, which are the temperate zones

Readme [11.4K]

Answer:

The dimension is L

Explanation:

Dimension analysis is a method of representing quantities majorly with respect to some fundamental quantities of mass (M), length (L), time (T).

A sphere has a definite volume which relates to its radius by:

V = $\frac{4}{3}$ $\pi$ $r^{3}$

In this equation $\pi$ is a dimensionless quantity, and the unit of v is $m^{3}$ .

But, metre is a measure of length, thus it has a dimension of L.

So that,

$m^{3}$ ≅ $L^{3}$

Then,

$L^{3}$ = $r^{3}$

Find the cube root of both sides to have,

r = L

Therefore, the dimension of the radius of a sphere is L.

5 0

3 years ago

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago

Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

kaheart [24]

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

$f'=f_{1}-f$ ...(I)

$f'=f_{1}+f$ ...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

$f'=462-5=457\ Hz$

$f'=462+5=467\ Hz$

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

4 0

3 years ago

The mass of a meteoroid was found to be 1.24 × 107 kilograms. Write the mass in standard notation.

Musya8 [376]

Answer:

The value we are given in the question is 1.24 * 10^7. This form of writing number is called scientific notation. The standard notation is the normal, regular way of writing numbers. Scientific notation and standard notations are interchangeable.

1.24 * 10^7 written in standard notation will be = 1.24 * 10000000 = 12400000.

Thus, the mass of the meteoroid was 12400000 kg.

Explanation:

8 0

3 years ago

Does the mass of a pendulum affects the period of oscillation

pshichka [43]

Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.

8 0

3 years ago