Guest #4326

2 months ago

This problem uses the relationships among current **I**, current density **J**, and drift speed **vd**. We are given the total of electrons that pass through the wire in **t = 3s **and the area **A**, so we use the following equation to to find **vd**, from **J** and the known electron density **n**,** **so:

The current** I** is any motion of charge from one region to another, so this is given by:

The magnitude of the current density is:

Being:

Finally, for the drift velocity magnitude vd, we find:

**Notice: The current I is very high for this wire. The given values of the variables are a little bit odd**

The current

The magnitude of the current density is:

Being:

Finally, for the drift velocity magnitude vd, we find:

Question

A 4.0 ω resistor, an 8.0 ω resistor, and a 13.0 ω resistor are connected in parallel across a 24.0 v battery. what is the equivalent resistance of the circuit? answer in units of ω.

Solution 1

104/47 ω

____________

____________

Question

A printer is connected to a 1.0 m cable. if the magnetic force is 9.1 × x10-5 n, and the magnetic field is 1.3 × 10-4 t, what is the current in the cable

Solution 1

The magnetic force on a current-carrying wire due to a magnetic field is given by

where

I is the current

L the wire length

B the magnetic field strength

In our problem, L=1.0 m, and , so we can re-arrange the formula to find the current in the wire:

where

I is the current

L the wire length

B the magnetic field strength

In our problem, L=1.0 m, and , so we can re-arrange the formula to find the current in the wire:

Question

An oscillating block - spring system has a mechanical energy of 1.10 j, and amplitude of 11.0 cm, and a maximum speed of 1.7 m/s. find the spring constant.

Solution 1

The total mechanical energy of the block-spring system is given by the sum of the potential energy and the kinetic energy of the block:

where

k is the spring constant

x is the elongation/compression of the spring

m is the mass of the block

v is the speed of the block

At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:

(1)

where we used x=A, the amplitude (which is the maximum displacement of the spring).

Since we know

A = 11.0 cm= 0.11 m

E = 1.10 J

We can re-arrange (1) to find the spring constant:

where

k is the spring constant

x is the elongation/compression of the spring

m is the mass of the block

v is the speed of the block

At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:

(1)

where we used x=A, the amplitude (which is the maximum displacement of the spring).

Since we know

A = 11.0 cm= 0.11 m

E = 1.10 J

We can re-arrange (1) to find the spring constant:

Question

An elephant can hear sound with a frequency of 15 hz. what is the wavelength of this wave if the speed of sound in air is 343 m/s?

Solution 1

The relationship between frequency, wavelength and speed of a wave is:

where

is the wavelength

v is the speed of the wave

f is its frequency

For the sound wave in our problem, f=15 Hz and v=343 m/s, therefore the wavelength is

where

is the wavelength

v is the speed of the wave

f is its frequency

For the sound wave in our problem, f=15 Hz and v=343 m/s, therefore the wavelength is

Question

A light source emits a beam of photons, each of which has a momentum of 2.7 × 10-29 kg·m/s. (a) what is the frequency of the photons? (b) to what region of the electromagnetic spectrum do the photons belong?

Solution 1

The **frequency** of the **photons** is equal to 1.22 ×10¹³ Hz and lies in the **infrared region **of the electromagnetic spectrum.

The **frequency** of the **photons** or light can be described as the number of **oscillations** in one second. The frequency possesses S.I. units per second or **Hertz**.

The **relationship** between **momentum** (p), **frequency** (ν), and speed of light (c) is:

p = hν/c

ν = pc/h

Given, the **momentum** of the photons, p = 2.7 ×10⁻²⁹ Kg.m/s

The **speed** of light, c = 3×10⁸ m/s

The** plank's constant**, h = 6.626 ×10⁻³⁴ Js

The **frequency** of the photons can determine from the above-mentioned relationship:

ν = (2.7 × 10⁻²⁹).( 3 × 10⁸)/ 6.626 × 10⁻³⁴

ν = 1.22 × 10¹³ Hz

Therefore, the **frequency** of the photons is 1.22 × 10¹³ Hz and lies in the **infrared region** of the spectrum.

Learn more about **frequency**, here:

#SPJ2

Solution 2

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Question

A gravitational field vector points toward the earth; an electric field vector points toward an electron. why do electric field vectors point away from protons? 1. protons have much larger mass. 2. unlike electrons, protons will produce an electric field of their own. 3. protons have more net charge than electrons. 4. protons are positively charged. 5. none of these

Solution 1

The **electric **field vectors point away from **protons **because **protons **are **positively **charged. Option 4 is the **correct **option.

The **electric **field is the **field**, which is surrounded by the** **electric charged. The **electric field **is the electric force per unit **charge**.

When the two **charges **or the charged bodies **interact **each other, the force of attraction or repulsion acts **between **them. This force of **attraction** or **repulsion **can be find out using the coulombs law.

This force of **attraction **or repulsion is identified by the sign of the particular **charge**. According to the property of **interacting **charges-

**Like**charges-For the like**charges**, when two charges**interact**to each other, then the charges repel each other.**Unlike**charges-For the unlike**charges**, when two**charges**interact to each other, then the charges attract each other.

A **gravitational **field vector points toward the **earth**; an electric field vector **points **toward an **electron**.

It is known that the charge of **electron **is negative. The **electric **filed vector is attracted **towards **the electron, this ,means the **charge **of it should be **positive **(unlike charge to electron to attract it).

Now this **electric** field vectors point away from **protons**, as this both become same or like **charges **(positive).

Thus, the **electric **field vectors point away from **protons **because **protons **are **positively **charged. Option 4 is the **correct **option.

Learn more about **electric field **here;

Solution 2

The correct answer is

4. protons are positively charged.

In fact, the direction of the electric field of a charged particle depends on the sign of the charge of the particle. While a negatively charged particle (such as the electron) produces an electric field that points toward the charge, a positively charged particle (like a proton) produces an electric field that points away from the charge. This can be proofed by using a positive test charge: if we put a positive test charge near a proton, this test charge will move away from the proton (because it is repelled from it, since they have same charge), and the direction of its motion gives the direction of the electric field generated by the proton, so away from it.

4. protons are positively charged.

In fact, the direction of the electric field of a charged particle depends on the sign of the charge of the particle. While a negatively charged particle (such as the electron) produces an electric field that points toward the charge, a positively charged particle (like a proton) produces an electric field that points away from the charge. This can be proofed by using a positive test charge: if we put a positive test charge near a proton, this test charge will move away from the proton (because it is repelled from it, since they have same charge), and the direction of its motion gives the direction of the electric field generated by the proton, so away from it.

Question

A formatted printout (or screen display) of the contents of one or more tables or queries is a form. _________________________ a. True
b. False

Solution 1

No its false now I'm just going to type because I need 20 or more characters to answer this question....

Question

A crying baby emits sound with an intensity of 8.0 × 10-8 w/m2. calculate a reasonable estimate for the intensity level from a set of quintuplets (five babies), all crying simultaneously at the same place? the lowest detectable intensity is 1.0 × 10-12 w/m2.

Solution 1

Sound intensity of 1 baby, I = 8*10^-8 W/m^2

The sound heard should be higher by:

10*log (n) where for 5 babies, n = 5. Then

10*log (n) = 10*log (5) ≈ 7 dB

Also give is the reference sound, Io = 1.0*10^-12 W/m^2

Therefore,

Sound intensity, L1 = 10*log (I/I1) = 10*log [(8*10^-8)/(1*10^-12)] ≈ 49 dB

Therefore, total intensity for the five babies is:

Total intensity = 49+7 = 56 dB

The sound heard should be higher by:

10*log (n) where for 5 babies, n = 5. Then

10*log (n) = 10*log (5) ≈ 7 dB

Also give is the reference sound, Io = 1.0*10^-12 W/m^2

Therefore,

Sound intensity, L1 = 10*log (I/I1) = 10*log [(8*10^-8)/(1*10^-12)] ≈ 49 dB

Therefore, total intensity for the five babies is:

Total intensity = 49+7 = 56 dB

Solution 2

**The intensity level from a set of quintuplets (five babies) : ****56 dB**

**Wave intensity** is the power of a wave that is moved through a plane of one unit that is perpendicular to the direction of the wave

Can be formulated

I = intensity, W m⁻²

P = power, watt

A = area, m²

The farther the distance from the sound source, the smaller the intensity

So the intensity is inversely proportional to the square of the distance from the source

Intensity level (LI) can be formulated

Io = 10⁻¹²

For the level of intensity of several sound sources as many as n pieces can be formulated:

The intensity level of 1 baby is

LI = 10 log 8.10⁴

LI = 49

The intensity level of 5 babies :

LI5 = LI + 10 log n

LI5 = 49 + 10 log 5

LI5 = 49 + 7

**LI5 = 56 **

**The intensity of a laser beam **

**electric field **

**magnetism **

Question

Make a rule: how would you find the resistance of a parallel circuit with n identical resistors?

Solution 1

When n resistors are connected in parallel, it means they are connected to the same potential difference V:

(2)

It also means that the total current in the circuit is given by the sum of the currents flowing through each branch (each resistor) of the circuit:

(1)

By using Ohm's law:

we can rewrite (1) as

However, we said that the potential difference across each resistor is equal (eq.(2)), so we can rewrite the last formula as

From which we find an expression for the equivalent resistance of n resistors in parallel:

(2)

It also means that the total current in the circuit is given by the sum of the currents flowing through each branch (each resistor) of the circuit:

(1)

By using Ohm's law:

we can rewrite (1) as

However, we said that the potential difference across each resistor is equal (eq.(2)), so we can rewrite the last formula as

From which we find an expression for the equivalent resistance of n resistors in parallel:

Question

A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units of m/s.

Solution 1

For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,

y = 2L = 2*0.75 =1.5 m

Additionally,

y = v/f Where v = wave speed, and f = ferquncy

Then,

v = y*f = 1.5*220 = 330 m/s

1/2y = L, where L = length of the string, y = wavelength.

Therefore,

y = 2L = 2*0.75 =1.5 m

Additionally,

y = v/f Where v = wave speed, and f = ferquncy

Then,

v = y*f = 1.5*220 = 330 m/s

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