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3 years ago

What is the exact meaning of croos section and cross sectional area of a conductor?

1 answer:

4 0

The cross section is the little tiny circle you see when you cut a wire
and look at the flat, cut end.

The cross-sectional area of the wire is the area of that little circle.
It's equal to

Area = (pi) x (1/4) x (Diameter of the wire)²

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A force of 500 N acting on an object of 2 kg displaces it by 3 m. Find the work done by the force and the object's final speed a

Salsk061 [2.6K]

Answer:

buscalo en gugul nby buenas nochis

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3 years ago

Pilings are driven into the ground at a building site by dropping a 2250 kg object onto them. What change in gravitational poten

den301095 [7]

Answer: 324.135 kJ

Explanation:

Given

Mass of dropping is $m=2250\ kg$

The initial height of dropping is $h_1=16\ m$

The final height of dropping $h_2=1.3\ m$

Gravitational potential energy is the function of height i.e.

$\Rightarrow \text{G.E.}=mgh$

Change in Gravitational Energy is

$\Rightarrow \Delta \text{G.E.}=mg(h_1-h_2)=2250\times 9.8\times (16-1.3)\\\\\Rightarrow \Delta \text{G.E.}=3,24,135\ J\approx 324.135\ kJ$

5 0

3 years ago

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles

alexandr402 [8]

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4 x 10⁻¹⁵ m

8 0

4 years ago

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

RideAnS [48]

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{3}=0.333Hz$

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to $T=\frac{1}{f}=\frac{1}{0.25}=4sec$

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = $=\frac{20}{10}=2sec$

So time period T = 2 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{2}=0.5Hz$

6 0

3 years ago

Read 2 more answers