A 60-kg woman stands on the very end of a uniform board, which is supported one-quarter of the way from one end and is balanced. What is the mass of the board?

a. 60 kg
b. 30 kg
c. 20 kg
d. 15 kg
e. 120 kg

In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.

Malus's law warns that

[tex]I = I_0 cos^2\theta[/tex]

Where,

[tex]\theta=[/tex] Angle between the analyzer axis and the polarization axis

[tex]I_0 =[/tex]Intensity of the light before passing through the polarizer

The intensity of the beam from the first polarizer is equal to the half of the initial intensity

[tex]I = \frac{I_0}{2}[/tex]

Replacing with our the numerical values we get

[tex]I = \frac{46}{2}[/tex]

[tex]I = 23W/m^2[/tex]

Therefore the  intensity of the light that emerges from the filter is [tex]23W/m^2[/tex]

Answers:

a) [tex]8820 m/s[/tex]

b) [tex]189500 Pa[/tex]

Explanation:

We have the following data:

[tex]t=30 min \frac{60 s}{1 min}=1800 s[/tex] is the time

[tex]h=11 m[/tex] is the height the water reaches vertically

[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity

[tex]P_{air}=101.3 kPa=101.3(10)^{3} Pa[/tex] is the pressure of air

[tex]\rho_{water}=1000 kg/m^{3}[/tex] is the density of water

Knowing this, let's begin:

Here we will use the following equation:

[tex]h=h_{o}+V_{o}t-\frac{g}{2}t^{2}[/tex] (1)

Where:

[tex]h_{o}=0 m[/tex] is the initial height of water

[tex]V_{o}[/tex] is the initial speed of water

Isolating [tex]V_{o}[/tex]:

[tex]V_{o}=\frac{1}{t}(h+\frac{g}{2}t^{2})[/tex] (2)

[tex]V_{o}=\frac{1}{1800 s}(11 m+\frac{9.8 m/s^{2}}{2}(1800 s)^{2})[/tex]

[tex]V_{o}=8820.006 m/s \approx 8820 m/s[/tex] (3)

In this part we will use the following equation:

[tex]P=\rho_{water} g d + P_{air}[/tex] (4)

Where:

[tex]P[/tex] is the absolute pressure in the chamber

[tex]d=9 m[/tex] is the depth

[tex]P=(1000 kg/m^{3})(9.8 m/s^{2})(9 m) + 101.3(10)^{3} Pa[/tex]

[tex]P=189500 Pa[/tex] (5)

Answers:

a) [tex]8.009(10)^{-7} N[/tex]

b) [tex]1.8(10)^{-8} [/tex]

Explanation:

a) Accoding to the Universal Law of Gravitation we have:

[tex]F_{g}=G\frac{Mm}{d^2}[/tex] (1)

Where:

[tex]F_{g}[/tex] is the gravitational force between the eagle and the throng

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Universal Gravitational constant

[tex]M=4.5 kg[/tex] is the mass of the eagle

[tex]m=(80 kg)(3(10)^{6} people/kg)=240(10)^{6} kg[/tex] is the mass of the throng

[tex]d=300 m[/tex] is the distance between the throng and the eagle

[tex]F_{g}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} \frac{(4.5 kg)(240(10)^{6} kg)}{(300 m)^{2}}[/tex] (2)

[tex]F_{g}=8.009 (10)^{-7} N[/tex] (3) As we can see the gravitational force between the eagle and the throng is quite small.

b) The attraction force between the eagle and Earth is the weight [tex]W[/tex] of the eagle, which is given by:

[tex]W=Mg[/tex] (4)

Where [tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity on Earth

[tex]W=(4.5 kg)(9.8 m/s^{2})[/tex] (5)

[tex]W=44.1 N[/tex] (6)

Now we can find the ratio between [tex]F_{g}[/tex] and [tex]W[/tex]:

[tex]\frac{F_{g}}{W}=\frac{8.009 (10)^{-7} N}{44.1N}[/tex]

[tex]\frac{F_{g}}{W}=1.8(^{-8})[/tex] As we can see this ratio is also quite small

a. The angle between the reflected part of the beam and the surface of the glass is [tex]\(43.5^\circ\).[/tex]

b. The angle between the refracted beam and the surface of the glass is [tex]\(64.5^\circ\).[/tex]

To solve this problem, we need to apply the laws of reflection and refraction. Let's address each part separately.

Part (a): Angle Between the Reflected Beam and the Surface of the Glass

The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is given as 43.5° with respect to the surface of the glass. However, angles in optics are typically measured with respect to the normal (a line perpendicular to the surface).

So, the angle of incidence with respect to the normal (which we'll call [tex]\(\theta_i\)[/tex] ) is:

[tex]\[ \theta_i = 90^\circ - 43.5^\circ = 46.5^\circ \][/tex]

Since the angle of incidence equals the angle of reflection:

[tex]\[ \theta_r = \theta_i = 46.5^\circ \][/tex]

Therefore, the angle between the reflected part of the beam and the surface of the glass is:

[tex]\[ 90^\circ - \theta_r = 90^\circ - 46.5^\circ = 43.5^\circ \][/tex]

So, the angle between the reflected beam and the surface of the glass is:

[tex]\[ 43.5^\circ \][/tex]

Part (b): Angle Between the Refracted Beam and the Surface of the Glass

For the refracted beam, we need to apply Snell's Law, which is:

[tex]\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \][/tex]

Where:

- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air), [tex]\( n_1 = 1.00 \)[/tex],

- [tex]\( \theta_i \)[/tex] is the angle of incidence with respect to the normal, [tex]\( \theta_i = 46.5^\circ \),[/tex]

- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass), [tex]\( n_2 = 1.68 \)[/tex],

- [tex]\( \theta_t \)[/tex] is the angle of refraction with respect to the normal.

Using Snell's Law, we can solve for [tex]\(\theta_t\):[/tex]

[tex]\[ 1.00 \sin(46.5^\circ) = 1.68 \sin(\theta_t) \][/tex]

[tex]\[ \sin(\theta_t) = \frac{\sin(46.5^\circ)}{1.68} \][/tex]

Calculating [tex]\(\sin(46.5^\circ)\):[/tex]

[tex]\[ \sin(46.5^\circ) \approx 0.723 \][/tex]

So,

[tex]\[ \sin(\theta_t) = \frac{0.723}{1.68} \approx 0.430 \][/tex]

Now we find [tex]\(\theta_t\):[/tex]

[tex]\[ \theta_t = \sin^{-1}(0.430) \approx 25.5^\circ \][/tex]

The angle between the refracted beam and the surface of the glass is:

[tex]\[ 90^\circ - \theta_t = 90^\circ - 25.5^\circ = 64.5^\circ \][/tex]

So, the angle between the refracted beam and the surface of the glass is:

[tex]\[ 64.5^\circ \][/tex]

Final answer:

To find the highest available gain for an op-amp circuit with a required bandwidth, we use the Gain-Bandwidth Product (GBP), which is a constant. For a GBP of 768 kHz and a bandwidth of 32 kHz, the maximum available gain is 24 V/V.

Explanation:

The question concerns determining the highest available gain for an op-amp circuit given a required bandwidth.

The Gain-Bandwidth Product (GBP) is a constant for an op-amp and is found by multiplying the current gain with its corresponding 3-dB frequency.

With an initial gain of 96 V/V and a 3-dB frequency of 8 kHz, the GBP can be calculated as 96 V/V * 8 kHz = 768 kHz.

To meet the requirement of a 32 kHz bandwidth with the same GBP (because GBP is constant), we can calculate the maximum gain as follows:

GBP = Gain * Bandwidth, which gives us

Gain = GBP / Bandwidth.

Plugging in the numbers, we get

Gain = 768 kHz / 32 kHz, resulting in a maximum gain of 24 V/V under the conditions of a 32 kHz bandwidth.

Final answer:

The rate at which the concrete loses thermal energy by conduction through the air layer can be calculated using Fourier's Law of Heat Conduction. The formula involves the thermal conductivity, area, temperature difference, and thickness of the air layer. However, without the thermal conductivity value for air, the calculation cannot be completed.

Explanation:

To calculate the rate at which the concrete slab loses thermal energy by conduction through the surrounding air layer at sunset, we can apply Fourier's Law of Heat Conduction. This law states that the heat transfer rate (Q) through a material is directly proportional to the temperature difference across the material (ΔT), the area through which heat is being transferred (A), and the thermal conductivity (k), and inversely proportional to the thickness of the material (L).

The formula to calculate the rate of heat loss is given by Q = k*A*(ΔT/L), where ΔT is the temperature difference between the two sides of the material, A is the contact area, k is the thermal conductivity of the material, and L is the thickness of the material.

Unfortunately, without the thermal conductivity value for air in the provided data, we cannot calculate the exact rate of heat loss for this specific scenario. Thermal conductivity is required to solve this problem, and it's typically obtained from tables in textbooks or scientific references.

To solve the problem it is necessary to apply the concepts related to the flow rate of a fluid.

The flow rate is defined as

[tex]Q = Av[/tex]

Where,

[tex]Q = Discharge (m^3/s)[/tex]

[tex]A = Area (m^2)[/tex]

v = Average speed (m / s)

And also as

[tex]Q = \frac{V}{t}[/tex]

Where,

V = Volume

t = time

Let's start by finding the total volume according to the given dimensions, that is to say

[tex]V = 5*11*2.4[/tex]

[tex]V = 132m^3[/tex]

The entire cycle must be completed in 50 min = 3000s

In this way we know that the [tex]132m ^ 3[/tex] must be filled in 3000s, that is to say that there should be a flow of

[tex]Q = \frac{V}{t}[/tex]

[tex]Q = \frac{132}{3000}[/tex]

[tex]Q = 0.044m^3/s[/tex]

Using the relationship to find the speed we have to

[tex]Q = Av[/tex]

[tex]v = \frac{Q}{A}[/tex]

Replacing with our values,

[tex]v = \frac{0.044}{900*10^{-4}m^2}[/tex]

[tex]v = 0.488m/s[/tex]

Therefore the air speed in the duct must be 4.88m/s

Answer:

449.37412 N

Explanation:

G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²

m = Mass of satellite = 438 kg

M = Mass of planet

T = Time period of the satellite = 24 h

r = Radius of planet = [tex]1.94\times 10^8\ m[/tex]

The time period of the satellite is given by

[tex]T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg[/tex]

The gravitational force is given by

[tex]F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N[/tex]

The force acting on this satellite is 449.37412 N

Gravity, is often known as gravitation. The gravitational force between the planet and the satellite can be written as 0.4332 N.

Gravity, often known as gravitation, is the universal force of attraction that acts between all matter in mechanics. It is the weakest known force in nature, and so has no bearing on the interior properties of ordinary matter.

[tex]F =G\dfrac{m_1m_2}{r^2}[/tex]

As it is given that the value of the gravitational constant is G = 6.67259 × 10−11 N m² /kg², while the radius of the planet is 1.94×10⁵. And the time taken by the satellite to revolve around the planet is 24 hours. therefore, the Mass of the planet can be written as,

The Time period of the satellite is given as:

[tex]T = 2\pi \sqrt{\dfrac{r^3}{GM}}[/tex]

Substitute the values in the formula,

[tex]24 = 2\pi \sqrt{\dfrac{(1.94 \times 10^5)^3}{6.67259 \times 10^{-11}M}}\\\\M = 5.78686 \times 10^{17}\rm\ kg[/tex]

Thus, the mass of the planet can be written as 5.5786×10¹⁷ kg.

Now, the gravitational force can be written as,

[tex]F = G\dfrac{Mm}{r^2}\\\\F = 6.67259\times 10^{-11} \times \dfrac{5.5786 \times 10^{17} \times 438}{1.94 \times 10^5}\\\\ F = 0.4332\rm\ N[/tex]

Hence, the gravitational force between the planet and the satellite can be written as 0.4332 N.

Learn more about Gravitational Force:

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Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R  to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

The correct answer is option A. The statement is true because in celestial mechanics, angular momentum and total energy are conserved under gravity, regardless of whether an object moves from a smaller to a larger radius or vice versa.

Let's explain these concepts:

1. Conservation of Angular Momentum (L):

The angular momentum of an object with respect to the center of the planet is given by the product of its linear momentum (p) and the radius (r) of its orbit:

[tex]\[ L = p \times r = mvr \][/tex]

where m is the mass of the object, v is its velocity, and r is the radius of the orbit. For a central force like gravity, the torque on the object is zero, which means that the angular momentum is conserved.

2. Conservation of Total Mechanical Energy (E):

The total mechanical energy of an object in orbit is the sum of its kinetic energy (K) and potential energy (U):

[tex]\[ E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r} \][/tex]

where G is the gravitational constant, M is the mass of the planet, and m is the mass of the object.

The complete question is:

The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth).

A. True

B. False

Answer:

[tex]D_{s}[/tex] ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

     I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere               [tex]Is_{cm}[/tex] = 2/5 M R²

Spherical shell   [tex]Ic_{cm}[/tex] = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

    I = [tex]I_{cm}[/tex] + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

     D = 2R

    Ic = [tex]Ic_{cm}[/tex] + M D²

    Ic = 2/3 MR² + M (2R)²

    Ic = M R² (2/3 + 4)

    Ic = 14/3 M R²

The sphere

    Is =[tex]Is_{cm}[/tex] + M [[tex]D_{s}[/tex]²

    Is = Ic

    2/5 MR² + M [tex]D_{s}[/tex]² = 14/3 MR²

    [tex]D_{s}[/tex]² = R² (14/3 - 2/5)

    [tex]D_{s}[/tex] = √ (R² (64/15)

    [tex]D_{s}[/tex] = 2,066 R

Answer:

[tex]v_s=27.8m/s[/tex]

Explanation:

If the person hearing the sound is at rest, then the equation for the frequency heard [tex]f[/tex] given the emitted frequency [tex]f_0[/tex], the speed of the truck [tex]v_s[/tex] and the speed of sound [tex]c[/tex] will be:

[tex]f=f_0\frac{c}{c+v_s}[/tex]

Where [tex]v_s[/tex] will be positive if the truck is moving away from the person, and negative otherwise. We then do:

[tex]\frac{f}{f_0}=\frac{c}{c+v_s}[/tex]

[tex]\frac{f_0}{f}=\frac{c+v_s}{c}=1+\frac{v_s}{c}[/tex]

[tex]v_s=c(\frac{f_0}{f}-1)=(340m/s)(\frac{238Hz}{220Hz}-1)=27.8m/s[/tex]

Answer:

The gauge pressure is 1511.11 psi.

Explanation:

Given that,

Flow rate = 94 ft³/min

Diameter d₁=3.3 inch

Diameter d₂ = 5.2 inch

Pressure P₁= 15 psi

We need to calculate the pressure on other side

Using Bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2[/tex]

We know that,

[tex]V=Av[/tex]

[tex]v=\dfrac{V}{A}[/tex]

Where, V = volume

v = velocity

A = area

Put the value of v into the formula

[tex]P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2[/tex]

Put the value into the formula

[tex]15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]

[tex]P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]

[tex]P_{2}=1525.8\ psi[/tex]

We need to calculate the gauge pressure

Using formula of gauge pressure

[tex]P_{g}=P_{ab}-P_{atm}[/tex]

Put the value into the formula

[tex]P_{g}=1525.8-14.69[/tex]

[tex]P_{g}=1511.11\ psi[/tex]

Hence, The gauge pressure is 1511.11 psi.

When the separation between the plates of a parallel plate capacitor is increased, the amount of charge on the plates decreases due to the decrease in capacitance (option e), with the voltage remaining constant.

When parallel plate capacitor plates are pulled away from each other while connected to a battery maintaining a constant potential difference, the capacitance decreases. This is because the capacitance is inversely proportional to the distance between the plates. As the capacitance decreases, the charge on the plates also decreases since the voltage (V) remains constant, and the relation between charge (Q), capacitance (C), and voltage (V) is given by Q = CV. Therefore, the amount of charge on the plates decreases because the capacitance decreases (option e).

Answer:

[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g  = 10 × 9.8 = 98 N  

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied  

10 kg block will slide on 40 kg slab and net force on it  

= 100 N - kinetic friction  

[tex]=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)[/tex]

= 100 - 39.2

= 60.8 N

[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}[/tex]

[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}[/tex]

[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}[/tex]

[tex]\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }[/tex]

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4  

Frictional force on 40 kg slab by 10 kg block = 39.2 N  

[tex]40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}[/tex]

[tex]40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}[/tex]

40 kg slab will move with = [tex]0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]

[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]

Answer:

[tex]27.57713\ MPa\sqrt{m}[/tex]

Explanation:

Y = Fracture parameter

a = Crack length

[tex]\sigma[/tex] = Stress in part

Plane strain fracture toughness is given by

[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow Y=\frac{K_I}{\sigma\sqrt{\pi a}}\\\Rightarrow Y=\frac{39}{270\times \sqrt{\pi 0.00282}}\\\Rightarrow Y=1.53462[/tex]

When a = 1.41 mm

[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow K_i=1.53462\times 270\sqrt{\pi 0.00141}\\\Rightarrow K_I=27.57713\ MPa\sqrt{m}[/tex]

The value of plane strain fracture toughness is [tex]27.57713\ MPa\sqrt{m}[/tex]

Answer:

W=972.83 J

Explanation:

Given,

angle of inclination of ramp = θ = 24.3°

Force exerted on the block = F= 289 N

Crate is raised to height of = 2.39 m

coefficient of friction = 0.33

where g is the acceleration due to gravity = g = 9.8 m/s²

Work done = ?

let m be the mass of the crate

Gravity force along ramp = m g sinθ

Friction force =  μm g cosθ

now writing all the force

F = m g sinθ  + μm g cosθ

By putting the values

F = m g (sinθ  + μcosθ)              

289 = 9.8 x m  ( sin 24.3°+ 0.33 cos 24.3°)                 ( take g =10 m/s²)

289 = 9.8 m(0.71)

m = 41.53 kg

Work done will be equal to

W= m g h

W= 41.53 x 9.8 x 2.39 J

W=972.83 J

Answer:

0.35701 T

Explanation:

[tex]B_i[/tex] = Initial magnetic field

[tex]B_f[/tex] = Final magnetic field

[tex]\phi[/tex] = Magnetic flux

t = Time taken = 4.17 ms

N = Number of turns = 510

[tex]\epsilon[/tex] = Induced emf = 10000 V

r = Radius = 0.27 m

A = Area = [tex]\pi r^2[/tex]

Induced emf is given by

[tex]\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T[/tex]

The magnetic field strength needed is 0.35701 T

Answer:

a) 1.13 10-8 T.  b) +y direction

Explanation:

a)

For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.

At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:

E(z,t) = Emax cos (kz-ωt-φ) +x

B(z,t) = Bmax cos (kz-ωt-φ) +y

In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:

Bmax = Emax/c

So, solving for Bmax:

Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.

b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.

Answer:v=3.08 m/s

Explanation:

Given

mass of student [tex]m=21 kg[/tex]

distance moved [tex]d=10 m[/tex]

Force applied [tex]F=10 N[/tex]

acceleration of system during application of force is a

[tex]a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2[/tex]

using [tex]v^2-u^2=2 as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex]v^2-0=2\times 0.476\times 10[/tex]

[tex]v=\sqrt{9.52}[/tex]

[tex]v=3.08 m/s[/tex]

Answer:

option D.

Explanation:

The correct answer is option D.

When an object is in equilibrium torque calculated at any point will be equal to zero.

An object is said to be in equilibrium net moment acting on the body should be equal to zero.

If the net moment on the object is not equal to zero then the object will rotate it will not be stable.

In a system in equilibrium, the choice of axis about which torques are calculated can be anywhere because for such a system, the sum of torques about any point is zero.

When applying the torque equation, ∑τ = 0, to an object in equilibrium, the chosen axis about which torques are calculated can be located anywhere on or off the object. This principle is a result of the fact that in a system in equilibrium, the sum of torques about any point is zero, not just specific points like the object's pivot, edge, or center of gravity. For example, if you have a seesaw in balance, you could calculate the torques from the center, one of the seats, or even a point suspended above it in the air. As long as the object is in equilibrium and not moving, the torques calculated from any point will sum to zero because they counteract each other in direction and magnitude to maintain the state of balance.

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Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

r_2= larger radius curve

r_2= 2r_1..............i

let u be the speed of larger radius curve

now, [tex]\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}[/tex]................ii

form i and ii we can write

[tex]v^2= \frac{1}{2} u^2[/tex]

⇒u= √2 v

therefore, safe speed for the larger radius track u= √2 v

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]

Where

[tex]\delta =[/tex] Horizontal distance between two points

[tex]\lambda =[/tex] Wavelength

From our values we have,

[tex]\lambda = 500nm = 5*10^{-6}m[/tex]

[tex]\theta = 1\°[/tex]

The horizontal distance between this two points would be given for

[tex]\delta = dsin\theta[/tex]

Therefore using the equation we have

[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]

[tex]\Phi = \frac{2\pi(dsin\theta)}{\lambda}[/tex]

[tex]\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}[/tex]

[tex]\Phi= 1.096 rad \approx = 1.1 rad[/tex]

Therefore the correct answer is C.

The phase difference for waves from the top and bottom of the slit can be calculated by using the formula for calculating phase difference. With the provided values for wavelength, angle and slit width, the calculated phase difference is 0.55 rad.

The phase difference of the waves is directly related to the path difference between them and can be calculated by using the formula:

Φ = 2 π × (d / λ) × sin(θ).

Where φ is the phase difference, π is Pi, d is the slit width, λ is the wavelength and θ is the incident angle.

Let's plug the provided numbers into our formula:

Φ = 2 π × (5.0x10-6 m / 500x10-9 m) × sin(1.0°).

Φ = 2 π × 10 × sin(1.0°)= 0.55 rad.

So, the correct answer is (B) 0.55 rad.

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Answer:

a) v = 7,207 m / s

, b) a = 42.3 m / s²

Explanation:

We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression

Initial

    Em₀ = K = ½ m v²

Final  

    [tex]Em_{f}[/tex] = 2 Ke = ½ k x²

The two is placed because each barred has two springs and each does not exert the same force

    Emo = [tex]Em_{f}[/tex]

     ½ m v² = 2 ½ k x²

     v = √(2k/m) x

     v = √ (2 3,134 10⁵/4550) 0.614

     v = 7,207 m / s

Let's take this speed to km / h

     v = 5,096 m / s (1km / 1000m) (3600s / 1h)

     v = 25.9 km / h

This speed is common in school zones

Let's use kinematics to calculate the average acceleration

      vf² = v₀² - 2 a x

       0 = v₀² - 2 a x

       a = v₀² / 2 x

       a = 7,207²/2 0.614

       a = 42.3 m / s²

We buy this acceleration with the acceleration of gravity

       a / g = 42.3 / 9.8

       a / g = 4.3

This acceleration is well below the maximum allowed

To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.

From Snell's law we know that

[tex]n_1sin\theta_1 = n_2 sin\theta_2[/tex]

Where,

n_i = Refractive indices of each material

[tex]\theta_1[/tex] = Angle of incidence

[tex]\theta_2[/tex] = Refraction angle

Our values are given as,

[tex]\theta_1 = 38\°[/tex]

[tex]n_1 = 1[/tex]

[tex]n_2 = 1.4[/tex]

Replacing

[tex]1*sin38 = 1.4*sin\theta_2[/tex]

Re-arrange to find [tex]\theta_2[/tex]

[tex]\theta_2 = sin^{-1} \frac{sin38}{1.4}[/tex]

[tex]\theta_2 = 26.088°[/tex]

Therefore the  angle will the beam make with the normal in the glass is 26°

Answer:

Angular velocity will be 2.843 rad/sec

Explanation:

We have given the radius r = 5.17 m

Centripetal acceleration [tex]a_c=1.5g=1.5\times 9.8=14.7m/sec^2[/tex]

We know that centripetal acceleration is given by

[tex]a_c=\frac{v^2}{r}[/tex]

And linear velocity is given by [tex]v=\omega r[/tex]

[tex]a_c=\frac{(\omega r)^2}{r}=\omega ^2r[/tex]

[tex]14.7=\omega ^2\times 5.17[/tex]

[tex]\omega =2.843rad/sec[/tex]

Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

To take place the process of nuclear fusion basically seeks to reach heavy nuclei through light nuclei. Reaching this process implies a release of energy that is what makes this process attractive because it is possible to obtain significant volumes of energy. The procedure to arrive at this process also implies a high cost concerning high temperatures and exorbitant pressures as it is necessary to be able to overcome the barrier of electrostatic repulsion.

This process does not generate any type of radioactive waste like other processes, therefore it is not as dangerous as nuclear fission. For this reason the correct answer is A. Fusion products are generally not radioactive.

Answer:

e. All of these statements are false.

Explanation:

As we know that heat transfer take place from high temperature to low temperature.

It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.

The first law of thermodynamics is the energy conservation law.

Second law of thermodynamics  states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.

By using heat pump ,heat can transfer from cooler body to the hotter body.

Therefore all the answer is False.

The true statement among the given options is that the entropy of a system can be reduced by cooling it.

The correct answer to the student's question regarding true statements about thermodynamics is option (c) It is always possible to reduce the entropy of a system, for instance, by cooling it. This statement aligns with the principles of thermodynamics, which affirm that entropy, a measure of disorder or randomness, can decrease in a system if energy is removed from the system, such as by lowering its temperature. However, this does not violate the second law of thermodynamics because entropy may still increase in the overall process when considering the surroundings.

In contrast, options (a) and (b) are false because they wrongly imply irreversibility in scenarios where reversibility is possible. Specifically, it is possible to reverse an entropy increase by cooling a system and it is also possible to convert some amount of thermal energy back into mechanical energy, although not with 100% efficiency due to inherent thermodynamic losses.

The second law also articulates that heat transfer occurs spontaneously from a higher to a lower temperature body and not in the reverse direction without external work, implying that a spontaneous flow of heat from a colder to a warmer body is impossible, as is complete conversion of heat to work in a cyclical process.

Answer:

0.00288 J

Explanation:

We know that

W= Fds

F = −α∆s^4

α = 45 N/m^4 and ∆s = displacement

W= −α∆s^4ds

integrating both the sides from s= 0 to 0.2

W= 45/5×0.2^5= 0.00288 J

The workdone on the stone-elastic band system is mathematically given as

W= 0.00288 J

Question Parameter(s):

F = −α∆s4 , where α = 45 N m4

∆s is the displacement of the elastic band

Generally, the equation for the Workdone  is mathematically given as

W= Fds

Thereofre

F = −α *ds^4

Where

α = 45  

ds = displacement

In conclusion

W= 45/5*0.2^5

W= 0.00288 J

Read more about  Work

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Answer:

2.771208%

Explanation:

[tex]\Delta V[/tex] = Change of volume

[tex]V_0[/tex] = Initial volume

[tex]\Delta T[/tex] = Change in temperature = (0.3-1050)

[tex]\beta[/tex] = Volumetric coefficient of thermal expansion = [tex]-26.4\times 10^{-6}\ /K[/tex]

Volumetric expansion of heat is given by

[tex]\frac{\Delta V}{V_0}=\beta \Delta T\\\Rightarrow \frac{\Delta V}{V_0}=-26.4\times 10^{-6}\times (0.3-1050)\\\Rightarrow \frac{\Delta V}{V_0}=0.02771208[/tex]

Finding percentage

[tex]\frac{\Delta V}{V_0}=0.02771208\times 100=2.771208\%[/tex]

The ratio of change of volume to initial volume is 2.771208%

To calculate the ratio ΔV/V0 for zirconium tungstate, the volume change can be determined using the volumetric coefficient of thermal expansion. The formula for the ratio is ΔV/V0 = (Volume change/Initial volume) × 100.

The ratio ΔV/V0 can be calculated using the formula:

ΔV/V0 = (Volume change/Initial volume) × 100

Given that the volumetric coefficient of thermal expansion for zirconium tungstate is -26.4 × 10^(-6)/K, we can use this value to calculate the volume change. The volume change can be found by multiplying the coefficient of thermal expansion by the change in temperature:

Volume change = (-26.4 × 10^(-6)/K) × (1050 K - 0.3 K)

Using this value, we can calculate the ratio ΔV/V0:

ΔV/V0 = (Volume change/Initial volume) × 100

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