Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by LaTeX: \Delta L_{thick}ΔLthickand LaTeX: \Delta L_{thin}ΔLthin, respectively.

Answers

Answer 1

Answer:

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Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by $\Delta L_(thick)$ and $\Delta L_(thin)$ , respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e $(1)/(2) <1$ option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

$E = (\sigma)/(\epsilon)$

Where $\sigma$ is the stress = $(Force)/(Area)$

$\epsilon$ is the strain = $(\Delta L)/(L)$

Making Strain the subject

$\epsilon = (\sigma)/(E)$

now in this question we are that the same tension was applied to both wires so

$(\sigma)/(E)$ would be constant

Hence

$(\Delta L)/(L) = constant$

for the two wire we have that

$(\Delta L_1)/(L_1) = (\Delta L_2)/(L_2)$

Looking at young modulus formula

$E = ((F)/(A) )/((\Delta L)/(L) )$

$E * (\Delta L )/(L) = (F)/(A)$

$A * (\Delta L)/(L) = (F)/(E)$

Now we are told that a comprehensive force is applied to the wire so for this question

$(F)/(E)$ is constant

And given that the length are the same

$A_1 (\Delta L_(thin))/(L_(thin)) = A_2 (\Delta L_(thick))/(L_(thick))$

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

Assuming that

$A_2 = 2 A_1$

$A_1 (\Delta L_(thin))/(L_(thin)) = 2 A_1 (\Delta L_(thick))/(L_(thick))$

$(\Delta L_(thin))/(L_(thin)) = 2 (\Delta L_(thick))/(L_(thick))$

From the question the length are equal

$\Delta L_(thin) =2 \Delta L_(thick)$

$(\Delta L_(thick))/(\Delta L_(thin) ) = (1)/(2)$

Hence the ratio is less than 1

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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

The 1 here represented the central bright spot

$n_(max) = 2 * 2500 + 1$

$n_(max) =5001$

Plzzz helppppp!!! I need answers A, B, C & D

Answers

Answer: the answer i for c is yes 0& 10

Explanation:

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Answers

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm What is the distance between the two second-order minima?

Answers

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=(y_1)/(2)\n\Rightarrow \beta_1=(1.4)/(2)\n\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=(m_1\lambda D)/(d)\n\Rightarrow d=(m_1\lambda D)/(\beta_1)$

For second order

$\beta_2=(m_2\lambda D)/(d)\n\Rightarrow \beta_2=(m_2\lambda D)/((m_1\lambda D)/(\beta_1))\n\Rightarrow \beta_2=(m_2)/(m_1)\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\n\Rightarrow y_2=2(m_2)/(m_1)\beta_1\n\Rightarrow y_2=2(2)/(1)* 0.7\n\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

What is the wavelength of the electromagnetic radiation needed to eject electrons from a metal?

Answers

Answer:

λ = hc/(eV + h $f_(0)$ )

Explanation:

Let the work function of the metal = ∅

the kinetic energy with which the electrons are ejected = E

the energy of the incident electromagnetic wave = hf

Then, we know that the kinetic energy of the emitted electron will be

E = hf - ∅

because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.

This means that the energy of the incident e-m wave can be written as

hf = E + ∅

also, we know that the kinetic energy of the emitted electron E = eV

and the work function ∅ = h $f_(0)$

we can they combine all equations to give

hf = eV + h $f_(0)$

we know that f = c/λ

substituting, we have

hc/λ = eV + h $f_(0)$

λ = hc/(eV + h $f_(0)$ ) This is the wavelength of the e-m radiation needed to eject electrons from a metal.

where

λ is the wavelength of the e-m radiation

h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s

c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s

e is the charge on an electron

V is the voltage potential on the electron

$f_(0)$ is the threshold frequency of the metal

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