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Find the force on a 5 pС charge in a place where the electric field is 400 N/C.

Answers

Answer 1

Answer:

Answer:

Electric force, $F=2* 10^(-9)\ N$

Explanation:

It is given that,

Charge on the particle, $q=5\ pC=5* 10^(-12)\ C$

Electric field, $E=400\ N/C$

Let F is the electric force acting on the charged particle. The electric force per unit electric charge is called electric field. Mathematically, it is given by :

$F=qE$

$F=5* 10^(-12)\ C* 400\ N/C$

$F=2* 10^(-9)\ N$

So, the force acting on the charged particle is $2* 10^(-9)\ N$ . Hence, this is the required solution.

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Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....? Listed following are locations and times atwhich different phases of the Moon are visible from Earth’sNorthern Hemisphere. Match these to the appropriate moon phase.

The three given moon phases are:

Waxing Crescent Moon

Waning Crescent Moon

Full Moon

The things we need to match to the above three topicsare:

*visible near western horizon an hour after sunset
*rises about the same time the sun sets
*visible near eastern horizon just before sun rises
*occurs about 3 days before new moon (i know this is waningcrescent)
*visible due south at midnite
*occurs 14 days after new moon (i know this is full moon)

Answers

Answer:

Explanation:

*visible near western horizon an hour aftersunset WAXING CRESCENT

*rises about the same time the sun sets FULLMOON

*visible near eastern horizon just before sun rises WANING CRESCENT

*occurs about 3 days before new moon WANING CRESCENT

*visible due south at midnite FULL MOON

*occurs 14 days after new moon FULL MOON

From her bedroom window a girl drops a water-filled balloon to the ground, 4.77 m below. If the balloon is released from rest, how long is it in the air?

Answers

We need to use the equation x = vt + (1/2)at^2. We know x = 4.77, v = 0, and a = 9.81m/s^2. Plug in the values. 4.77 = (0)t + (1/2)(9.81)t^2 Solve for t. 4.77 = (4.905)t^2 0.972 = t^2 t = (sq.rt)_/0.972 t = 0.985 So it's in the air 0.985 seconds.

Since the density of air decreases with an increase in temperature, but the bulk modulus B is nearly independent of temperature, how would you expect the speed of sound waves in air to vary with temperature?

Answers

To develop this problem it is necessary to apply the concept related to the speed of sound waves in fluids.

By definition we know that the speed would be given by

$v=\sqrt{(\beta)/(\rho)}$

$\beta =$ Bulk modulus

$\rho =$ Density of air

From the expression shown above we can realize that the speed of sound is inversely proportional to the fluid in which it is found, in this case the air. When the density increases, the speed of sound decreases and vice versa.

According to the statement then, if the density of the air decreases due to an increase in temperature, we can conclude that the speed of sound increases when the temperature increases. They are directly proportional.

If 3.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.59 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t; ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5 = I x 2.59 x 3600

1469.5 = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Answers

Answer:

$v_(su) = 19.44 m/s$

Explanation:

$m_(su)=5.68x10^(29)kg\nm_(sa)=5.68x10^(26)kg$

$T=9.29x10^8\nr_(o)=1.43x10^(12)$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_(su)*r_(o)=(m_(sa)+m_(su))*r_(1)$

solve to r1

$r_1=(m_(sa)*r_(o))/(m_(sa)+m_(su))=(5.68x10^(26)*1.43x10^(12))/(5.68x10^(26)+5.68x10^(26))$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = (m_(sa)*2*pi*r_1^2)/(T)$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_(sun)=(m_(sa)*2*\pi *( 2r_(o)*r_1 -r_1^2))/(T)$

Since

$L_(su)= m_(su)*v_(su)*r_1$

then

$v_(su)=(m_(sa)*2*pi*(2r_(o)*r_(1)-r_(1)^2))/(T*m_(sa)*r_1)$

$v_(su) = 19.44 m/s$

Final answer:

In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.

Explanation:

The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.

However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.

Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.

So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.

This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.

Learn more about Orbital Physics here:

brainly.com/question/12805730

#SPJ3

What are 4 ways individuals can influence the government?

Answers

Voting, running, speaking in front of government

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