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3 years ago

Please help me and Can you show your work

Physics

1 answer:

RSB [31]3 years ago

4 0

Answer:

7. 1ml = 20 drops

325ml = 6500

( 325 x 20 = 6500)

there are 6500 drops in a pepsi can

8. not sure sorry ; (

9. 1 mile per hour = 88 feet per minute

45 miles per hour = 3960 feet per minute

( 88 x 45 = 3960)

there are 3960 feet per minute in 45 miles per hour

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harge of uniform density (100 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical sur

Elden [556K]

Answer:

The magnitude of the electric field is 8.47 N/C

Explanation:

Given;

uniform charge density, λ = 100 nC/m³

inner radii of the cylinder, r = 1.0 mm and 3.0 mm

distance from the symmetry axis, R = 2.0 mm

$Volume =\pi (R^2 -r^2)l\\\\Volume =\pi ((2*10^{-3})^2 -(1*10^{-3})^2)l\\\\Volume =\pi (4*10^{-6} - 1*10^{-6})l\\\\Volume = 3*10^{-6} \pi l \ m^3$

Area = 2πrl

Area =2π(2 x 10⁻)l

Volume = A x d

d = Volume / Area

$d = \frac{V}{A} = \frac{3*10^{-6}*\pi*l}{4\pi *10^{-3} l} = 75 *10^{-5} \ m$

the magnitude of the electric field

$E = \frac{\lambda *d}{\epsilon_o} = \frac{100*10^{-9} *75*10^{-5}}{8.854*10^{-12}} \\\\E = 8.47 \ N/C$

Therefore, the magnitude of the electric field is 8.47 N/C

6 0

3 years ago

High pressure center of dry air

Virty [35]

Anticyclone is the high pressure center of dry air

5 0

3 years ago

Read 2 more answers

give me much information about Hubble space telescopes as possible you can. I have final research thing and I need ti write abou

zubka84 [21]

The Hubble space telescope is a space telescope that was launched to lower the orbit in 1990 and remains in operation. it was not the first space telescope, but it is one of the largest and most versatile, well known as both as a vital research tool and as a public relations boon for astronomy. The Hubble telescope is named after astronomer Edwin Hubble, and is one of NASA’s great observatories, along with Compton GammaRay observatory, the Chandra x-ray observatory, and the Spitzer Space Telescope.

Hubble features a 2.4 meter mirror, and it’s four main instruments observed in the ultraviolet, visible, and near in-fraired regions of the electromagnetic spectrum.
Hobbles orbit outside the distortion of earths atmosphere allows it to capture extremely high resolution images with the substantially lower background light than ground base telescopes. it has recorded some of the most detailed visible light images, allowing a deep view into space. Many Hubble observations have led to the breakthroughs in astrophysics, such as determining the right of expansion of the universe.

The Hubble telescope was built by the United States space agency NASA with the contributions from that European space agency. The space telescope science Institute select hobbles targets and processes the resulting data while Goddard space flight Center controls the spacecraft. Space telescope for repost as early as 1923. Hubble was founded in 1970s with the proposed launch in 1983, but the project was reset by technical delays, budget problems, and the 1986 challenger disaster. It was finally launched by space shuttle discovery in 1990, but its main mirror had been ground incorrectly, resulting in spherical aberration that can promise to telescopes capabilities. The optics were corrected to their intent to quality by a servicing mission in 1993. How old is only telescope designed to be maintained in space by astronauts. Five space shuttle missions have repaired, upgraded, and replace systems on the telescope, including all five of the main instruments. The first mission was initially canceled on safety grounds following the Columbia disaster which was in 2003, but nice administrator Michael D Griffin approved the fifth servicing mission which was completed in 2009. The telescope was opening as of April 24, 2020 is 30th anniversary, and could last until 2030-2040. One successor to the Hubble telescope is the James Webb space telescope which is scheduled to be launched in late 2021.

3 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

7 0

3 years ago

Briefly describe the relationship between an equipotential surface and an electric field, and use this to explain why we will pl

baherus [9]

Answer:

E = - dV/dx

Explanation:

Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.

El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.

De los antes expuesto las dos magnitudes están relacionadas

E = - dV/dx

por lo cual el potenical es el gradiente del potencial eléctrico.

Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial

7 0

3 years ago