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4 years ago

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Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.

036 N. What is the magnitude of the charge on each sphere? (a) 2x10-3 C (h) 3 x 10-4C

1 answer:

Mars2501 [29]4 years ago

8 0

Answer:

$0.5\times 10^{-6}C$

Explanation:

According to coulombs law force between two charges is given by $F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2}$ here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So $0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)}$ As $\epsilon _0$ is constant which value is $8.85\times 10^{-12}$

$Q^2=0.250\times 10^{-12}$

$Q=0.5\times 10^{-6}C$

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<h2>An artesian system is present when groundwater <u>under pressure rises above the aquifer level.</u></h2>

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3 years ago

The vertical displacement of the wave is measured from the what? And what is the term for vertical displacement?

navik [9.2K]

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3 years ago

Read 2 more answers

A certain car is capable of accelerating at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it

gogolik [260]

Solution: 85.11 m

Given:

acceleration of the car, $a= 0.85 m/s^{2}$

initial velocity, $u=83km/h=23.05 m/s$

final velocity, $v=94 km/h=26.11 m/s$

we need to find the displacement (s) of the car.

we would use the equation of motion:

$2as=v^{2}-u^{2}\\
\Rightarrow s=\frac{v^{2}-u^{2}}{2a} \\
\Rightarrow s=\frac{26.11^{2}-23.05^{2}}{2\times0.85}\\
\Rightarrow s=85.11 m$

hence, the displacement done by the car is = 85.11 m

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3 years ago

An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a

Blababa [14]

Answer:

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

Explanation:

Given:

cross sectional area of the wire, $A=5.75\times 10^{-6}\ m^2$

density of aluminium wire, $\rho=2.7\times 10^3\ kg.m^{-3}$

young's modulus of the material, $E=7\times 10^{10}\ N.m^{-2}$

wave speed, $v=118\ m.s^{-1}$

<u>We have mathematical expression for strain as:</u>

$\frac{\Delta L}{L} =\frac{\sigma}{E}$ ...............................(1)

and since, $\sigma =\frac{T}{A}$

where, T = tension force in the wire

equation (1) becomes:

$\frac{\Delta L}{L} =\frac{T}{A.E}$ ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

$v=\sqrt{\frac{T}{\mu} }$ ...................................(3)

where: $\mu=$ linear mass density of the wire

$\therefore \mu=\rho \times A$

Now, equation (3) becomes

$v=\sqrt{\frac{T}{\rho \times A} }$

$T=v^2.\rho \times A$ ............................(4)

Using eq. (2) & (4) for tension T

$v^2.\rho \times A=A.E\times \frac{\Delta L}{L}$

$\frac{\Delta L}{L} =\frac{v^2.\rho}{E}$

putting the respective values

$\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}$

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

6 0

3 years ago

A quarter-wave monopole radio antenna (also called a Marconi antenna) consists of a long conductor of one quarter the length of

sasho [114]

Answer:

a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m

b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m

Explanation:

(a) Radiowave wavelength= λ = c/f

As we know, Radiowave speed in the air = c = 3 x 10^8 m/s

f = frequency = 604 kHz = 604 x 10^3 Hz

Hence, wavelength = (3x10^8/604x10^3) m

λ = 496.69 m

So the height of the antenna BROADCASTING AT 604 kHz = λ /4 = (496.69/4) m

= 124.17 m

(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz (1kHZ = 1000 Hz)

Hence, wavelength = λ = (3 x 10^8/1710 x 10^3) m

λ= 175.44 m

So, height of the antenna = λ /4 = (175.44/4) m

= 43.86 m

5 0

4 years ago