A current of 3.2 Amps flows through a 7.0 Ω resistor in a circuit. Calculate the voltage.

You must show your work.

Answer:

39 m

Explanation:

We are given that

Wavelength=[tex]\lambda=400 nm=400\times 10^{-9} m[/tex]

1 nm=[tex]10^{-9} m[/tex]

y=1 km=[tex]1000 m[/tex]

1 km=1000 m

Earth mars distance =x=80 million Km=[tex]80\times 10^9 m[/tex]

1million km=[tex]10^9 m[/tex]

[tex]sin\theta=\frac{1.22\lambda}{d}[/tex]

[tex]sin\theta=\frac{y}{x}[/tex]

[tex]\frac{y}{x}=\frac{1.22\lambda}{d}[/tex]

[tex]\frac{1000}{80\times 10^9}=\frac{1.22\times 400\times 10^{-9}}{d}[/tex]

[tex]d=\frac{1.22\times 400\times 10^{-9}\times 80\times 10^9}{1000}[/tex]

[tex]d=39.04 m\approx 39 m[/tex]

Diameter of mirror =39 m

To achieve a 1.0-km resolution on Mars with light of a 400 nm wavelength, the size of the space telescope's mirror would need to be approximately 6 mm. This calculation is based on diffraction limit theory and may not be precisely accurate in practical application due to other potential limiting factors.

This question is related to the resolution of a telescope which is primarily influenced by the aperture size and wavelength of light being observed. The bigger the telescope aperture (the diameter of the primary mirror), the greater the resolution as more light can be collected. The formula for the resolution limit due to diffraction (the smallest distinguishable detail) can be given by R = 1.22λ/D where λ is the wavelength of light and D is the diameter of the telescope's main mirror.

Here the wavelength given is 400 nm or 400 x 10^-9 m, and we need to calculate the mirror diameter required for a resolution of 1.0-km on Mars from Earth. Let's convert the resolution limit from kilometers to meters (which gives us 1000m), and then to radians (using the Earth-Mars distance), which results in an angle of 1.25x10^-11 radians approximately.

Substituting these values in the formula, we can solve for D and find that the telescope mirror size needed would be approximately 0.006 m or 6 mm for this resolution. Do note, that this is a theoretical value, in reality, the size might need to be larger due to factors like the diffraction limit, non-uniformities in mirrors, or aberrations in lenses.

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Answer:

Explanation:

The magnitude of electric field = 8 x 10⁴ V /m

there is a potential difference of 8 x 10⁴ V on a separation of 1 m

so on a separation of .5 m , potential drop or change in potential will be equal to

.5 x 8 x 10⁴

= 4 x 10⁴ V .

The increase in kinetic energy of proton = V X Q

V is potential drop x Q is charge on proton

= 4 x 10⁴ x 1.6 x 10⁻¹⁹

= 6.4 x 10⁻¹⁵ J

potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of

-  6.4 x 10⁻¹⁵ J .

The change in electric potential is 4.0 x 10^4 volts. The change in potential energy of the proton-field system, given the charge of a proton, is calculated to be 6.4 x 10^-15 joules.

The change in electric potential (or voltage) is calculated by multiplying the electric field strength by the displacement, so in this case it is (8.0 x 10^4 V/m) * (0.50 m) = 4.0 x 10^4 volts.

The change in potential energy of the proton-field system can be found by multiplying the change in electric potential by the charge of a proton. The charge of a proton is 1.6 x 10^-19 C, so the change in potential energy is (4.0 x 10^4 volts) * (1.6 x 10^-19 C), which equals to 6.4 x 10^-15 joules.

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that is... A. B

................c. D

The Punnett square for the cross between a parent with straight ears and a parent with droopy ears would result in offspring with the genotypes EE, EE, Ee, and Ee.

The Punnett square is a tool used in genetics to predict the possible genotypes of offspring based on the genotypes of the parents. In this case, the Punnett square for the cross between a parent with straight ears (genotype EE) and a parent with droopy ears (genotype ee) would look like:

EEeEEEEeEeEe

The genotypes of each square are as follows:
A: EE
B: EE
C: Ee
D: Ee

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Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

[tex]p_f=p_i\\mv_1+Mv_2=(m+M)v[/tex]

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

[tex](0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}[/tex]

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

[tex]E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2[/tex]

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

[tex](0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m[/tex]

hence, the heigth is 68cm

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           [tex]Em_{f}[/tex] = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m [tex]v_{cm }^{2}[/tex] + ½ [tex]I_{cm}[/tex] w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

Answer:

A) The thickness of the glass is 868 nm

B) The wavelength is 3472 nm

Explanation:

We are given;

Refractive index = 1.30

Wavelength = 496 nm

Next wavelength = 386 nm

Now, we need to calculate the thickness of the glass

Using formula for constructive interference which is given as;

2nt = (m + ½)λ

Where;

m is the order of the interference

λ is wavelength

t is thickness

Now, for the first wavelength, we have;

2nt = (m + ½)496 - - - - eq1

for the second wavelength, we have;

2nt = (m + 1 + ½)386

2nt = (m + 3/2)386 - - - - eq2

Thus, combining eq1 and eq2, we have;

(m + ½)496 = (m + 3/2)386

496m + 248 = 386m + 579

496m - 386m = 579 - 248

110m = 331

m = 331/110

m = 3

Put 3 for m in eq 1;

2nt = (3 + ½)496

2nt = 1736

t = 1736/(2 x 1)

t = 868 nm

B) now we need to calculate the longest wavelength.

From earlier, we saw that ;

2nt = (m + ½)λ

Making wavelength λ the subject, we have;

λ = 2nt/(m + ½)

The longest wavelength will be at m = 0

Thus,

λ = (2 x 1 x 868)/(0 + ½)

λ = 3472 nm

Answer:

Spiral galaxy

Explanation:

If a galaxy forms from a protogalactic cloud with a lot of angular momentum, we would expect this galaxy to be a spiral galaxy. This is due to the conservation of angular momentum, where greater angular momentum usually results in a flatter, disk-like shape seen in spiral galaxies. Other factors also affect galaxy formation.

In the subject of astrophysics, the angular momentum of the protogalactic cloud has significant implications for the type of galaxy formed. Specifically, if a galaxy forms from a protogalactic cloud with a high angular momentum, we'd expect this galaxy to be a spiral galaxy.

Angular momentum is a property of rotating bodies, determined by the mass, shape, and spin rate of the object. In the context of galaxies, a protogalactic cloud with higher angular momentum is more likely to flatten out into a spinning disk shape, leading to the formation of a spiral galaxy. This is due to the conservation of angular momentum. In contrast, lower angular momentum could lead to the formation of elliptical galaxies, which are less flat and more randomly arranged.

In the realm of galaxy formation and evolution, these are widely accepted ideas, though it's important to note that other factors, like the gas content of the protogalactic cloud and interactions with other galaxies, can also play a major role.

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The energy differences between the levels of the quantum objects and the photons in the visible spectrum determine the dark spectral lines in an absorption spectrum.

In order to determine the energies of the dark spectral lines in an absorption spectrum, we need to consider the energy differences between the energy levels of the quantum objects and the energy levels of the photons in the visible spectrum. The energy levels of the quantum objects are -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. The range of photon energies for visible light is 1.63 to 3.26 eV. Only photons with energies matching the energy differences between the levels of the quantum objects will be absorbed, resulting in dark lines in the spectrum.

Based on the given information, the dark spectral lines within the visible spectrum would be:

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Explanation:

Mass of baseball, m = 0.148 kg

Initial speed of the ball, u = 14.5 m/s

Final speed of the ball, v = 11.5 m/s

After crashing through the pane of a second-floor window, the ball shatters the glass as it passes through, and leaves the window at 11.5 m/s with no change of direction. So, the direction of the impulse that the glass imparts to the baseball is in opposite direction to the direction of the balls path.

The change in momentum of the ball is called impulse. It is given by :

[tex]J=m(v-u)\\\\J=0.148\times (11.5-14.5) \\\\J=-0.444\ kg-m/s\\\\|J|=0.444\ kg-m/s[/tex]

Hence, this is the required solution.

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = [tex]W_{y}[/tex] / W

             W_{y} = W sin 46

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         [tex]Em_{f}[/tex] = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

Answer:

(D) 3

Explanation:

The angular momentum is given by:

[tex]\vec{L}=\vec{r}\ X \ \vec{p}[/tex]

Thus, the magnitude of the angular momenta of both solar systems are given by:

[tex]L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}[/tex]

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

[tex]L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1[/tex]

but L1=L2=L:

[tex]L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3[/tex]

Hence, the answer is (D) 3

HOPE THIS HELPS!!

Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a  , a is separation of slits

Puting the values given

.51 x 10⁻³ =  502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

Answer:

a) [tex]\mu_{k} = 0.704[/tex], b) [tex]R = 0.312\,m[/tex]

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s[/tex]

[tex]\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}[/tex]

[tex]\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}[/tex]

[tex]\mu_{k} = 0.704[/tex]

b) The speed of the block is determined by using the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]v = x\cdot \sqrt{\frac{k}{m} }[/tex]

[tex]v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }[/tex]

[tex]v \approx 9.829\,\frac{m}{s}[/tex]

The radius of the circular loop is:

[tex]\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}[/tex]

[tex]\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}[/tex]

[tex]R = 0.312\,m[/tex]

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

λ = 3.2 x 10⁻⁷ m = 320 nm

So, the radiation of 320 nm wavelength is absorbed by Ozone.

Answer:

0.03052 nm

Explanation:

From c = f¥

Where c = speed of electromagnetic wave in vacuum = 3x10^8 m/s

f = frequency of the wave

¥ = wavelenght of the wave.

¥ = c/f = (3x10^8)/(9.38×10^14)

= 3.052x10^-7 m

= 0.03052 nm

Answer:

Check attachment

The question have two distance

I decided to use the one in the question "a" in attachment and I will use the other one here

Explanation:

Given that,

Speed of particle relative to the earth is

V = 0.99543c

Where c is speed of light

c = 3 × 10^8 m/s

Particle height as detected by scientist is 43km

The initial length is 43km

Lo = 43km

Lo = 43,000m

A. Time taken for the particle to reach the earth surface?

Speed = distance / Time

Time = distance / speed

t = L / V

t = 43,000 / 0.99543c

t = 43,000 / (0.99543 × 3 × 10^8)

t = 1.4399 × 10^-4 seconds

b. Initial Lenght is given as Lo = 43km

Using length contraction formula

L = Lo√(1 — u² / c²)

L = 43√[1 — (0.99543c)² / c²]

L = 43√[1 — 0.990881c² / c²]

L = 43√[1 — 0.990881]

L = 43 × √(9.1191 × 10^-3)

L = 43 × 0.095494

L = 4.1062 km

c. Using time dilation formula

∆to = ∆t√(1 — u² / c²)

∆t is gotten from question a

∆t = 1.4399 × 10^-4 seconds

∆to = ∆t√[1 — (0.99543c)² / c²]

∆to = ∆t√[1 — 0.990881c² / c²]

∆to = ∆t√[1 — 0.990881]

∆to = ∆t × √(9.1191 × 10^-3)

∆to = 1.4399 × 10^-4 × 0.095494

∆to = 1.375 × 10^-5 seconds

To check if the time dilation agree

t = L / V

t = 4.1062 × 1000 / 0.99543c

t = 4.1062 × 1000 / 0.99543 × 3 × 10^8

t = 1.375 × 10^-5 seconds

The time dilation agreed

Answer:

The answer is

D. 1230j

Explanation:

When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force

given data

Mass of gun M=10kg

Mass of bullet m=20g----kg=20/1000 =0.02kg

Propulsive speed of bullet = 350m/s

Hence the moment of the bullet will be equal and opposite to that of the gun

mv=MV

where V is the recoil velocity which we are solving for

V=mv/M

V=0.02*350/10

V=7/10

V=0.7m/s

The energy contained in the bullet can be gotten using

KE=1/2m(v-V)²

KE=1/2*0.02(350-0.7)²

KE=1/2*0.02(349.3)²

KE=1/2*0.02*122010.49

KE=1/2*2440.20

KE=1220.1J

roughly the energy is 1230J

Answer

D, 2t

Explanation:

See attached file

"2t" will be the path length difference between Wave 1 and Wave 2. A complete solution is below.

According to the question,

The path difference between Wave 1 and 2 will be:

When [tex]\Theta = 0^{\circ}[/tex],

→ [tex]P.d= \Delta x[/tex]

         [tex]= AB+BC[/tex]

hence,

Path difference,

= [tex]t+t[/tex]

= [tex]2t[/tex]

Thus the above answer i.e., "option d" is appropriate.

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Answer:

The wavelength is  [tex]\lambda = 706nm[/tex]

Explanation:

From the question we are told that

    The refractive index of the glass is [tex]n__{glass}} = 1.52[/tex]

    The thickness of film is [tex]D = 127nm = 127*10^{-9}m[/tex]

     The refractive index of film [tex]n__{film}} = 1.39[/tex]

     The refractive index of air is [tex]n__{air}} = 1.00[/tex]

Generally the thickness of the film can be obtained mathematically from this expression

               [tex]D = \frac{\lambda}{4 * n__{film}} }[/tex]

Where [tex]\lambda[/tex] is the wavelength

           Making the wavelength the subject of the formula

                      [tex]\lambda = 4 * n__{film}} * D[/tex]

Substituting values

                     [tex]\lambda = 4 *1.39 * 127 *10^{-9}[/tex]

                       [tex]\lambda =7.06 *10^{-7}m = 706nm[/tex]

Answer:

[tex]\lambda = 706.12 nm[/tex]

Explanation:

The optical path length for the reflection of light = [tex]2 n_{film} t[/tex]

For destructive interference,  [tex]2 n_{film} t = \frac{\lambda}{2}[/tex]

The thickness of the anti-reflective film = 127 nm

The largest wavelength of the reflected light, [tex]\lambda = 4n_{film} t[/tex]

[tex]\lambda = 4 * 1.39 *127 * 10^{-9}[/tex]

[tex]\lambda = 706.12 * 10^{-9} m\\\lambda = 706.12 nm[/tex]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

The required time is: 0.395 s. Weight and wavelengths are 0.013 and 1.5. The equation for waves traveling down the string is: y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

a) To find the time it takes for a pulse to travel the full length of the string, we need to determine the velocity of the wave. The velocity of a wave on a string is given by the equation [tex]v = \sqrt{(T/\mu)[/tex], where T is the tension and μ is the linear mass density of the string.  

Given that the weight is 0.0125 N and the length is 1.50 m.

Here, mass = w/g = 0.0125/9.8 = 1.275 x 10⁻³ kg

we can calculate μ as follows: μ = m/L = (1.275 x 10⁻³)/(1.50 m) = 0.00085 kg/m.

Substituting the values into the equation:

we get v = [tex]\sqrt{W/0.00085}[/tex]. As it is given that the tension of the string is constant and equal to W.

Thus, v = [tex]\sqrt{14.7}[/tex] = 3.8 m/s

The time it takes for a pulse to travel the full length of the string can be calculated using the formula time = distance/velocity = 1.50/3.8. So, time = 0.395 s.

b) In case of string, V² = F/μ

μ x V² = F

This implies:

F = 0.00085 x [tex](3.8)^2[/tex] = 0.0123

C)  We know, f (frequency) = 1/t = 1/0.395 = 2.53

Also, v = λf. This implies:

λ = v/f = 3.8/2.53 = 1.5

D) Finally, to find the equation for waves traveling down the string, we need to adjust the sign inside the cosine function to account for the wave traveling in the opposite direction. It is given that:

y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

So, the equation becomes y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

Complete Question:

Question

A 1.50-m string of weight 0.0125 N is tied to the ceil-ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.

y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

(a) How much time does it take a pulse to travel the full length of the string?

(b) What is the weight W?

(c) How many wavelengths are on the string at any instant of time?

(d) What is the equation for waves traveling ?down? the string?

Answer:

Explanation:

Given that, .

Frequency

f = 60Hz

Number of turns

N = 125turns

Surface area of coil

A = 3 × 10^-2 m²

Magnetic field

B = 0.12T

Voltage peak to peak? I.e the EMF

EMF is given as

ε = —dΦ/dt

Where Φ is magnetic flux and it is given as

Φ = NBA Cosθ

Where N is number of turns

B is magnetic field

A is the cross sectional area

And θ is the resulting angle from the dot product of area and magnetic field

Where θ =ωt and ω = 2πf

Then, θ = 2πft

So, your magnetic flux becomes

Φ = NBA Cos(2πft)

Now, dΦ / dt = —NBA•2πf Sin(2πft)

dΦ / dt = —2πf • NBA Sin(2πft)

So, ε = —dΦ/dt

Then,

ε = 2πf • NBA Sin(2πft)

So, the maximum peak to peak emf will occur when the sine function is 1

I.e Sin(2πft) = 1

So, the required peak to peak emf is

ε = 2πf • NBA

Substituting all the given parameters

ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2

ε = 169.65 Volts

The peak to peak voltage is 169.65 V

The required value of peak output voltage is 169.65 Volts.

Given data:

The frequency of ac generator is, f = 60 Hz.

The number of turns of coil is, n = 125 turns.

The area of each coil is, [tex]A =3.0 \times 10^{-2} \;\rm m^{2}[/tex].

The strength of magnetic field is, B = 0.12 T.

To start with this problem, we need to find the peak emf first. The peak output voltage is nothing but the value of this peak emf only. The expression for the  peak EMF is given as,

ε = —dΦ/dt

here,

Φ is magnetic flux and it is given as

Φ = NBA Cosθ

Here,

θ is the resulting angle from the dot product of area and magnetic field

Where θ =ωt and ω = 2πf

Then, θ = 2πft

So, the expression for the magnetic flux becomes,

Φ = NBA Cos(2πft)

Now, dΦ / dt = —NBA•2πf Sin(2πft)

dΦ / dt = —2πf • NBA Sin(2πft)

So, ε = —dΦ/dt

Then,

ε = 2πf • NBA Sin(2πft)

So, the maximum peak to peak emf will occur when the sine function is 1

 Sin(2πft) = 1

So, the required peak to peak emf is

ε = 2πf • NBA

Substituting all the given parameters

ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2

ε = 169.65 Volts

Thus, we can conclude that the required value of peak output voltage is 169.65 Volts.

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The object strikes the surface of the Earth with a speed of approximately 11.2 km/s.

Ignoring atmospheric friction, the object's total mechanical energy (the sum of its gravitational potential energy and kinetic energy) is conserved throughout its fall. We can use this principle to calculate the object's speed at the Earth's surface.

Here's how:

Calculate the gravitational potential energy (PE) at the initial altitude:

PE_i = -G * M_earth * m / (R_earth + h)

where:

G is the gravitational constant (6.67 × 10^-11 N m²/kg²)

M_earth is the mass of the Earth (5.97 × 10^24 kg)

m is the mass of the object (910 kg)

R_earth is the Earth's polar radius (6357 km)

h is the initial altitude (1200 km)

Set the final PE (PE_f) to 0:

Since all the PE will be converted to kinetic energy when the object reaches the surface, PE_f = 0.

Apply the conservation of mechanical energy:

E_i = PE_i + KE_i = PE_f + KE_f = KE_f (as PE_i = PE_f = 0)

where:

KE_i is the initial kinetic energy (0 J, as the object is at rest)

KE_f is the final kinetic energy

Solve for the final velocity (v_f):

KE_f = 1/2 * m * v_f^2

v_f = sqrt(2 * E_i / m)

Plug in the values and calculate v_f:

v_f = sqrt(2 * (-G * M_earth * m / (R_earth + h)) / m)

v_f ≈ 11,180 m/s

Convert m/s to km/s:

v_f ≈ 11.2 km/s

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  [tex]v_A= 4m/s[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]L = 1m[/tex]

     The initial speed of block A is [tex]u_A[/tex]

     The final speed of block A is  [tex]v_A = \frac{1}{2}u_A[/tex]

      The initial speed of block B is [tex]u_B = 0[/tex]

      The mass of block A  is  [tex]m_A = 7kg[/tex]  gh

      The mass of block B is  [tex]m_B = 2 kg[/tex]

According to the principle of conservation of momentum

       [tex]m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}[/tex]

Since block B at initial is at rest

       [tex]m_A u_A = m_Bv_B + m_A \frac{u_A}{2}[/tex]

      [tex]m_A u_A - m_A \frac{u_A}{2} = m_Bv_B[/tex]

          [tex]m_A \frac{u_A}{2} = m_Bv_B[/tex]

  making [tex]v_B[/tex] the subject of the formula

             [tex]v_B =m_A \frac{u_A}{2 m_B}[/tex]

Substituting values

               [tex]v_B =\frac{7 u_A}{4}[/tex]  

This [tex]v__B[/tex] is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of [tex]v__B'[/tex] at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   [tex]a= \frac{v^2_{B}'}{L}[/tex]

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         [tex]T + mg = m \frac{v^2_{B}'}{L}[/tex]

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 [tex]\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L[/tex]

From above  

                [tex](T + mg) L = m v^2_{B}'[/tex]

Substitute this into above equation

             [tex]\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L[/tex]  

             [tex]\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L[/tex]

          [tex]\frac{49 mv_A^2}{16} = T + 5mgL[/tex]

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      [tex]\frac{49 mv_A^2}{16} = 5mgL[/tex]

making  [tex]v_A[/tex] the subject

            [tex]v_A = \sqrt{\frac{80mgL}{49m} }[/tex]

substituting values

          [tex]v_A = \sqrt{\frac{80* 9.8 *1}{49} }[/tex]

              [tex]v_A= 4m/s[/tex]

Answer:

a) [tex]v = 2.886\,\frac{m}{s}[/tex], b) [tex]\mu_{k} = 0.014[/tex]

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

[tex](62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v[/tex]

[tex]v = 2.886\,\frac{m}{s}[/tex]

b) The deceleration experimented by the system person-sled is:

[tex]a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}[/tex]

[tex]a = -0.139\,\frac{m}{s^{2}}[/tex]

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

[tex]-\mu_{k}\cdot m\cdot g = m\cdot a[/tex]

[tex]\mu_{k} = -\frac{a}{g}[/tex]

[tex]\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\mu_{k} = 0.014[/tex]

Answer:

See explanation

Explanation:

Solution:-

Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.

Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.

( See attachments )

- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:

                                  [tex]B = \frac{uo*I}{2\pi *r}[/tex]

Where,

I : The current,

r : The shortest distance to the wire,

uo : The permeability of free space. = 4π * 10^-7  T. m/A

-  Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.

- The magnetic field of a long straight wire has more implications than one might first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. The Biot-Savart law is written in its complete form as:

                             [tex]B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}[/tex]      

Where the integral sums over,

 1) The wire length where vector dl = direction of current (in or out of plane)

 2) r is the distance between the location of dl and the location at which the magnetic field is being calculated

 3)  r^ is a unit vector in the direction of r.

Answer:

B

Explanation:

Answer:

The answer is 0.001 m

Explanation:

Solution

Recall that,

A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at = 1200°C

The diffusion coefficient for nitrogen in steel at this temperature is =6 * 10-11 m2/s

The diffusion flux is = 1.2 *10^-7 kg/m2s

Concentration of nitrogen in the steel at the high-pressure surface is= 4 kg/m3.

The high-pressure side will the concentration is estimated to be = 2.0 kg/m3

Now,

The flux = -D dC/dx

1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx

∫ˣ₀ dx = -5x^10-4 ∫²₄ dC

so,

x = (2-4) kg/m3 (-5x10-4 m4/kg)

where x = .001 m

Therefore x = 0.001 m

The distance from this high-pressure side will the concentration is  0.001 m

Since the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, the diffusion flux is found to be 1.2x 10-7 kg/m2 -s.

We know that

The flux =[tex]-D\ dC\div dx[/tex]

So,

1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx

dC/dx = -2000

Now

-2000 = 4 - 2/0-x_B

x_B = 2/2000

x_B = 1*10^-3m

x_B = 0.001m

Hence, The distance from this high-pressure side will the concentration is  0.001 m

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Final answer:

The volume of the gas at 408 K and 24 N/cm² will be 1.76 liters, calculated using the combined gas law that integrates Boyle's law and Charles's law for the given changes in pressure and temperature.

Explanation:

The volume of a gas is dependent on both its pressure and temperature, following Boyle's law and Charles's law, respectively. The relationship is such that volume varies inversely with pressure and directly with temperature (in kelvins) when the amount of gas is constant. To find the new volume when both the pressure and temperature change, we use the combined gas law, which integrates both Boyle's and Charles's laws:

V₁/T₁ * P₁ = V₂/T₂ * P₂

Given that at P₁ = 16 N/cm², T₁ = 340 K, and V₁= 2.2 liters, we can find V₂ when P₂ = 24 N/cm² and T₂ = 408 K using the formula:

V₂ = (V₁ * T₂/T₁) * (P₁/P₂)

Plugging in the values:

V₂ = (2.2 liters * 408 K / 340 K) * (16 N/cm² / 24 N/cm²)

V₂ = (2.2 liters * 1.2) * (2 / 3)

V₂ = 2.64 liters * (2 / 3)

V₂ = 1.76 liters

So, the volume of the gas when the temperature is 408 K and the pressure is 24 N/cm² would be 1.76 liters.

Answer:

(C) T

The tension T at equilibrium will be equal to the Buoyant force.

The Buoyant force is given by:

Fb = density x acceleration due to gravity x volume displaced

The change in height doesn't affect the Buoyant force and hence the tension.

Note: The figure of question is added in the attachment

The tension in the thread when the block is at the new depth is :

(C) T

The definition of buoyancy refers to whether something can drift in water or discuss, or the control of water or other fluids to keep water above water, or an idealistic disposition.

The tension T at equilibrium will be equal to the Buoyant force.

The Buoyant force is given by:

Fb = density x acceleration due to gravity x volume displaced

The change in height doesn't affect the Buoyant force and hence, the tension.

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Answer:

The gravity at planet X is  [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]

The value of [tex]\alpha[/tex] on the moon is  [tex]\alpha = 730.38 \ revolutions[/tex]

Explanation:

  From the question we are told that

        The clocks hour hand makes [tex]\alpha[/tex] revolution every 1 hour which is 3600 sec

this implies that the time peroid for 1 revolution would be  [tex]= \frac{3600}{\alpha }sec[/tex]

   The peroid  for a pendulum is mathematically represented as

            [tex]T = 2 \pi\sqrt{\frac{L}{g} }[/tex]

Where L is the pendulum length

            g is the acceleration due to gravity

Let assume that we have a pendulum that counts in second on earth

   This implies that its peroid would be = 2 second

   i.e one second to swimg forward and one second to swing back to its original position

  Now the length of this pendulum on earth is

          [tex]L = \frac{gT^2}{4 \pi^2}[/tex]         [Making L the subject in above equation]

 Substituting values

        [tex]L = \frac{9.8 * (2)^2}{4 * (3.142)^2}[/tex]

           [tex]= 1[/tex]

When the same pendulum is taken to planet X  the peroid would be

                 [tex]T = \frac{3600}{\alpha }[/tex]

Recall this value was obtained above for 1 revolution (from start point to end point back to start point)

      So the acceleration due to gravity on this planet would be mathematically represented as

                [tex]g = \frac{4 \pi L }{T^2}[/tex]   [making g the subject in the above equation]

     substituting values

               [tex]g = \frac{4 * 3.142^2 * 1}{[ \frac{3600}{\alpha } ]^2}[/tex]

                    [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]

On moon the acceleration due to gravity has a constant value of

     [tex]g = 1.625 m/s^2[/tex]

The period of this pendulum  on the moon can be mathematically evaluated as

      [tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]

 substituting value

     [tex]T = 2 *3.142 \sqrt{\frac{1}{1.625} }[/tex]

        [tex]= 4.929s[/tex]

 given that  

       [tex]1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s[/tex]  {Note 1 revolution takes a peroid }

Making [tex]\alpha[/tex] the subject of the formula

           [tex]\alpha =\frac{3600}{4.929}[/tex]

              [tex]\alpha = 730.38 \ revolutions[/tex]

The principal stress in the pipe at point A, which is located on the outer surface of the pipe is 132.773 psi

It is given that l = 12 in

The force F = 60 lb

The radiuses are as follows:

outer radius, R = 3.90 in/2

R = 1.95 in

Inner radius, r = 3.65 in/2

r = 1.825 in

The angle between the force applied and the distanced from the axis is 30°

So we get the torque:

T = l×Fsin 30°

T = 12 × 60 × (0.5)

T = 360 lb-in

Now, the angle of rotation or angular displacement ωt is given as

ωt = π(R⁴ - r⁴)/(2R)

ωt = π((1.95 in)⁴ - (1.825 in)⁴)/(2×1.95 in)

ωt = 2.7114 in³

Then the principal stress in the pipe at point A is:

principal stress = T/ωt

principal stress = (360 lb-in)/(2.7114 in³)

principal stress = 132.773 lb/in²

principal stress = 132.773 psi

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Answer:

35.35 is the final intensity

Explanation:

See attached file for calculation