For a certain optical medium the speed of light varies from a low value of 1.80 × 10 8 m/s for violet light to a high value of 1.92 × 10 8 m/s for red light. Calculate the range of the index of refraction n of the material for visible light.

DNA is a molecule that stores and transmits genetic information, and its four nucleotide bases are adenine, thymine, guanine, and cytosine. It plays a crucial role in the growth, development, and functioning of all living organisms, and its understanding has revolutionized various fields of science and medicine.

DNA (Deoxyribonucleic acid) is a molecule that contains genetic information that determines the characteristics and traits of all living organisms. It is a double-stranded helix structure made up of nucleotides, which are composed of a sugar, a phosphate group, and a nitrogenous base.

The role of DNA is to store and transmit genetic information from one generation to the next. It contains the instructions that are necessary for the growth, development, and functioning of all living organisms. DNA is responsible for determining the physical and behavioral characteristics of an individual, including eye color, hair color, height, and susceptibility to diseases.

The four nucleotide bases in DNA are adenine (A), thymine (T), guanine (G), and cytosine (C). These bases pair with each other to form the rungs of the DNA ladder, with A always pairing with T and G always pairing with C.

The sequence of nucleotide bases in DNA is the code that determines the genetic information of an individual. This code is read and translated by cells to synthesize proteins, which are the building blocks of life. DNA is essential for the survival and reproduction of all living organisms.

Therefore, Adenine, thymine, guanine, and cytosine are the four nucleotide bases in DNA, a molecule that stores and transmits genetic information. All living things' growth, development, and functioning depend on it, and our growing understanding of it has revolutionized many disciplines in science and medicine.

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Answer:

HEAT FLOWS INTO THE SURROUNDING FROM THE SYSTEM AND THE SIGN OF THE SYSTEM WILL BE POSITIVE.

Explanation:

This is so of an endothermic reaction and the contraction of the gas is an endothermic one because the total heat content of the products formed is more than that of the reactants and the gas therefore absorbs heat from the surrounding during the contraction. An endothermic reaction is one in which heat is absorbed from the surroundings and the change in heat content ΔH of is positive.

The reaction for the contraction of the gas produces + 10 J of heat.

Answer:

(e) The direction of the net force is negative Z direction.

Explanation:

Let the direction of the particle (positive y) represents direction of current

Let direction of the net torque (negative x) represents direction of the field

Let direction of net force be direction of force

Apply Fleming's right hand rule, which states that when the thumb, the middleave finger and forefinger are held mutually at right angle to each other, with the thumb pointing in the direction of current (particle, positive y ), and the forefinger pointing in the direction of the field (torque, negative x), then the middle finger will be pointing in the direction of the force.

To find direction of the net force:

Y direction is already represented, negative X direction is also represented and finally force will be in Z - direction

Since we know that torque = force x position of particle

Torque is negative, thus force will be negative.

Therefore, the direction of the net force is negative Z direction

Answer:the wave is an electromagnetic wave.

Explanation:

Answer:

A: The wave is a mechanical wave.

C: The wave moves energy through matter.

F: The wave transfers energy perpendicular to the motion of the wave.

Hope it works!

Explanation:

Answer:

 W = 1.49 10⁻¹¹ kg

Explanation:

For this problem, let's use Newton's equation of equilibrium

           F - W = 0

            F = W              (1)

Strength can be found from the definition of pressure

        P = F / A

        F = P A

The radiation pressure for a reflective surface is

            P = 2 I / c)

We substitute in equation 1

         2 I / c  A = W

The intensity is defined by the ratio of the power between the area

          I = P / A

          P = I A

We substitute

          2 P / c = W

           W = 2  2.24 10-3 / 3 108

           W = 1.49 10⁻¹¹ kg

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

Answer:

Explanation:

solution is found below

Answer:

laser's wavelength λ = 597.4 nm

Explanation:

Given:

Slit spacing, d = 1.17mm

Tenth bright fringe y = 4.57cm

Distance from slits, D = 8.95m

n = 10

λ = (d * y) / (D * n)

λ = (1.17x10⁻³ * 4.57x10⁻²) / (8.95 x 10)

λ = 5.3469‬x10⁻⁵ / 8.95x10¹

λ = 0.5974 x 10⁻⁵⁻¹

λ = 0.5974 x 10⁻⁶ m

λ = 597.4 x 10⁻⁹ m

λ = 597.4 nm

Answer:

a) = 0.106 V/m

b) = 354.52 pT

c) = 6.5 W

Explanation:

Given,

Intensity, I = 15 μW/m²

Distance, d = 8.3 km

E = √(2Iμc)

where

c = speed of light = 2.99*10^8

μ = permeability of free space = 4π*10^-7

E = √(2 * 15*10^-6 * 4*3.142*10^-7 * 2.99*10^8)

E = √(30*10^-6 * 1.257*10^-6 * 2.99*10^8)

E = √1.13*10^-2

E = 0.106 V/m

B = E/c

B = 0.106 / 2.99*10^8

B = 354.52 pT

P = IA

Where A = 1/2 * 4πr²

A = 0.5 * 4 * 3.142 * (8.3*10^3)²

A = 4.33*10^8

P = I A

P = 15*10^-6 * 4.33*10^-8

P = 6495 W

P = 6.5 kW

Therefore

a) = 0.106 V/m

b) = 354.52 pT

c) = 6.5 W

Answer:

a) Amplitude of the electric field, E = 0.1063 V/m

b) Magnetic field, B = = 3.545 x 10⁻¹⁰ Tesla

c) Transmission power, P = 6492.73 W

Explanation:

Intensity, I = 15 μW/m²

Distance, R = 8.3 km = 8300 m

a) The electric field amplitude  

Relationship between the amplitude and intensity of a signal is given by:

[tex]I = \frac{E^{2} }{2c \mu}[/tex]..........(1)

c = 3 * 10⁸ m/s

μ = 4 π * 10⁻⁷

Substitute these values into equation (1)

[tex]15 * 10^{-6} = \frac{E^{2} }{2* 3 * 10^{8} * 4\pi * 10^{-7} }\\15 * 10^{-6} * 240\pi = E^{2}\\ E^{2} = 0.0113097\\E = 0.1063 V/m[/tex]

b)Amplitude of the magnetic field component at the airplane

 B = E / c

B = = 0.1063 / (3 x 10⁸ )

B = = 3.545 x 10⁻¹⁰ Tesla

3) The transmission power:

Intensity, [tex]I = \frac{Power}{Area}[/tex]

 P = I A

Area of the hemisphere, A = 2πR²

A = 2π * 8300²

A = 432848635.8116 m²

Power, P = 15 * 10⁻⁶ * 432848635.8116

P = 6492.73 W

Answer:

 v =√ G m M / r

Explanation:

For this exercise we will use Newton's second law

        F = m a

where the acceleration is centripetal

        a = v² / r

 force is the universal force of attraction

       F = G m M / r²

we substitute

       G m M / r² = m v² / r

        v² = G m M / r

       v =√ G m M / r

Final answer:

The orbital speed of a satellite in a circular orbit of radius R around a planet of mass M is given by v = sqrt(G * M / R).

Explanation:

When a satellite is in a circular orbit around a planet, it experiences a centripetal acceleration directed towards the center of the planet. This centripetal acceleration is provided by the gravitational force of the planet. Newton's second law states that the force acting on an object is equal to its mass multiplied by its acceleration. Therefore, we can equate the gravitational force to the mass of the satellite multiplied by its centripetal acceleration to find the relationship between the orbital speed and the radius of the orbit.

The gravitational force acting on the satellite is given by the formula F = G * (m * M) / R^2, where G is the universal gravitational constant, m is the mass of the satellite, M is the mass of the planet, and R is the radius of the orbit. The centripetal acceleration is given by the formula a = v^2 / R, where v is the orbital speed. Equating the force and the mass multiplied by acceleration, we can solve for v:

F = m * a => G * (m * M) / R^2 = m * v^2 / R => v^2 = G * M / R => v = sqrt(G * M / R)

Therefore, the orbital speed of a satellite in a circular orbit of radius R around a planet of mass M is given by v = sqrt(G * M / R).

Answer:

You are swimming faster

Explanation:

Since you are covering the same amount of distance in less time, you must be going faster. If you were going slower, you would cover the same amount of distance in a longer amount of time.

If you swim the same distance in less time, you are going faster.

Speed can be defined as the distance covered by an object per unit time and it is usually measured in meter per seconds (m/s).

Mathematically, speed is given by the formula;

[tex]Speed = \frac{distance}{time}[/tex]

From the above formula, we can deduce that speed is inversely proportional to time and directly proportional to distance. Thus, the faster the speed of an object, the lesser the time to cover a particular distance and vice-versa.

In this context, you are going faster when you swim the same distance in less time.

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For the NACA 44XX series airfoils, using thin airfoil theory, the angle of zero lift is determined to be -1.6 degrees. The design angle of attack, which eliminates the suction peak, coincides with this angle. The design lift coefficient is 0.04, equivalent to the maximum camber of the airfoil.

The question provided revolves around the aerodynamic properties of the NACA 44XX series airfoils, specifically in terms of their performance at cruise without the need for leading edge flaps. Given the NACA numbering system, where the second digit represents the location of the maximum camber (p) as a fraction of the chord, we can determine several characteristics for the 44XX airfoils.

(a) Angle of zero lift: 0 degrees. (b) Design angle: ~1.376°. (c) Design lift coefficient: ~0.047. Utilized thin airfoil theory.

To address these questions, we can utilize thin airfoil theory, which provides approximate solutions for the aerodynamic characteristics of airfoils at small angles of attack. Let's go through each question:

(a) Determine the angle of zero lift for the NACA 44XX airfoils:

The angle of zero lift, also known as the zero-lift angle of attack [tex](\(\alpha_{L=0}\))[/tex], can be calculated using thin airfoil theory. For symmetrical airfoils like the NACA 44XX series, the zero-lift angle of attack is simply 0 degrees.

(b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack:

The suction peak is eliminated when the airfoil is operating at its design lift coefficient [tex](\(C_{L_{\text{design}}}\))[/tex]. In thin airfoil theory, the lift coefficient is directly proportional to the angle of attack [tex](\(\alpha\))[/tex]. For cambered airfoils, this design lift coefficient is achieved at an angle of attack where the camber is aligned with the oncoming flow, resulting in no suction peak.

For NACA 44XX airfoils, the maximum camber is located at [tex]\(p = \frac{4}{10} = 0.4\)[/tex]. The angle of attack at which the camber aligns with the flow is called the camber line slope angle [tex](\(\alpha_{camber}\))[/tex]. For NACA 44XX series airfoils, the camber line slope angle can be approximated by:

[tex]\[\alpha_{camber} \approx 0.06 \times p = 0.06 \times 0.4 = 0.024 \text{ radians}\][/tex]

To convert this to degrees:

[tex]\[\alpha_{camber} \approx 0.024 \times \frac{180}{\pi} \approx 1.376^\circ\][/tex]

So, the design angle of attack where the suction peak is eliminated is approximately [tex]\(1.376^\circ\).[/tex]

(c) What is the design lift coefficient for the NACA 44XX airfoils?

At the design angle of attack, the lift coefficient [tex](\(C_L\))[/tex] can be calculated using thin airfoil theory. For symmetrical airfoils, [tex]\(C_L\)[/tex] is given by:

[tex]\[C_L = 2\pi \alpha\][/tex]

Substituting the design angle of attack [tex]\(\alpha_{camber}\)[/tex], we get:

[tex]\[C_{L_{\text{design}}} = 2\pi \times 1.376^\circ = 0.047\][/tex]

So, the design lift coefficient for the NACA 44XX airfoils is approximately [tex]\(0.047\).[/tex]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]

where :

[tex]n_1[/tex] is the index of refraction of the 1st medium

[tex]n_2[/tex] is the index of refraction of the 2nd medium

[tex]\theta_1[/tex] is the angle of incidence (the angle between the incident ray and the normal to the boundary)

[tex]\theta_2[/tex] is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

[tex]n_1=1.00[/tex] (index of refraction of air)

[tex]n_2=1.50[/tex] approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

[tex]sin \theta_2 =\frac{n_1}{n_2}sin \theta_1[/tex]

And since [tex]n_2>n_1[/tex], we have

[tex]\frac{n_1}{n_2}<1[/tex]

And so

[tex]\theta_2<\theta_1[/tex]

Which means that

The angle of incidence is greater than the angle of refraction

Answer:

This demonstrates Newton's Third Law because for every action there is an equal and opposite force.

Explanation:

The cat's paw hitting her toy string would be an action therefore when is swings to and from her it it creating and equal and opposite force. In other words, the cat hits it and when the string comes back to her it's coming back the same force just opposite of the direction she hit.

Answer:

Refractive index = 1.33

Explanation:

Refractive index is of a material explains how light passes through a medium. It is the ratio of speed of light in vacuum to the speed of light in the medium.

We know the speed of light in water is [tex] 2.25* 10^8 [/tex] and there is a change iñ speed of light in every medium it passes (except when it passes through vacuum). The change in speed of light is known to be = [tex] 2.99*10^8[/tex]

Given speed of light = 2.25*10^8 m/s

Formula for refractive index:

[tex]n = \frac{c}{v}[/tex]

Where,

change in speed of light, c= [tex] 2.99*10^8[/tex]

speed of light, v = 2.25*10^8 m/s

[tex] Refractive index, n = \frac{2.99*10^8}{2.55*10^8} [/tex]

= 1.328

= 1.33

The liquid is water, and the index of refraction of water is 1.33

As we know,

The refractive index will be:

→ [tex]n = \frac{c}{v}[/tex]

By substituting the values,

      [tex]= \frac{2.99\times 10^8}{2.55\times 10^8}[/tex]

      [tex]= 1.33[/tex]

Thus the approach above is right.

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Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = [tex]3.1[/tex] m

Width of the window = [tex]2.1[/tex] m

Area of the window = [tex](3.1\times 2.1) = 6.51\ m^2[/tex]

Difference in air pressure = Inside pressure - Outside pressure

                                           = [tex](1.0-0.954)[/tex] atm = [tex]0.046[/tex] atm

Conversion of the pressure in its SI unit.

⇒  [tex]1[/tex] atm = [tex]101325[/tex] Pa

⇒ [tex]0.046[/tex] atm = [tex]0.046\times 101325 =4660.95[/tex] Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ [tex]Pressure=\frac{Force }{Area}[/tex]

⇒ [tex]Force =Pressure\times Area[/tex]

⇒ Plugging the values.

⇒ [tex]Force =4660.95\times 6.51[/tex]

⇒ [tex]Force=30342.78[/tex] Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

Explanation:

Let l is the length of the wire. such that, side of square is l/4. Area of square loop is, [tex]A_s=\dfrac{l^2}{16}[/tex]

Radius of the circular loop,

[tex]l=2\pi r\\\\r=\dfrac{l}{2\pi}[/tex]

Area of the circular loop,

[tex]A_a=\pi r^2\\\\A_a=\pi \times (\dfrac{l}{2\pi})^2\\\\A=\dfrac{l^2}{4\pi}[/tex]

Torque in magnetic field is given by :

[tex]\tau=NIAB\sin\theta[/tex]

It is clear that, [tex]\tau\propto A[/tex]

So,

[tex]\dfrac{\tau_s}{\tau_a}=\dfrac{A_s}{A_a}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\dfrac{l^2}{16}}{\dfrac{l^2}{4\pi}}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\pi}{4}[/tex]

So, the ratio of maximum torque on the square loop to the maximum torque on the circular loop is [tex]\pi :4[/tex].

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

Thanks.

4.9 Earth minutes have elapsed when the clock face on the moon reads 12:12.

To determine how many Earth minutes have elapsed when the grandfather clock on the moon shows 12:12, follow these steps:

Step 1: Calculate the period of the pendulum on the moon

The period (T) of a pendulum is given by the formula:

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

where (L) is the length of the pendulum (1.0 m) and (g) is the acceleration due to gravity on the moon (1.62 m/s²).

Plugging in the values:

[tex]\[ T_{\text{moon}} = 2\pi \sqrt{\frac{1.0}{1.62}} \][/tex]

[tex]\[ T_{\text{moon}} = 2\pi \sqrt{0.617} \][/tex]

[tex]\[ T_{\text{moon}} \approx 2\pi \times 0.786 \][/tex]

[tex]\[ T_{\text{moon}} \approx 4.937 \text{ seconds} \][/tex]

Step 2: Calculate the period of the pendulum on Earth

For comparison, the period of the pendulum on Earth with [tex]\(g_{\text{earth}} = 9.81 \text{ m/s}^2\)[/tex] is:

[tex]\[ T_{\text{earth}} = 2\pi \sqrt{\frac{1.0}{9.81}} \][/tex]

[tex]\[ T_{\text{earth}} = 2\pi \sqrt{0.102} \][/tex]

[tex]\[ T_{\text{earth}} \approx 2\pi \times 0.319 \][/tex]

[tex]\[ T_{\text{earth}} \approx 2.006 \text{ seconds} \][/tex]

Step 3: Calculate the number of periods in 12 minutes on the moon

12 minutes on the moon clock is:

[tex]\[ 12 \text{ minutes} \times 60 \text{ seconds/minute} = 720 \text{ seconds} \][/tex]

The number of periods in 720 seconds is:

[tex]\[ \frac{720}{T_{\text{moon}}} = \frac{720}{4.937} \approx 145.9 \text{ periods} \][/tex]

Step 4: Convert the number of periods to Earth seconds

Since each period on Earth takes approximately 2.006 seconds:

[tex]\[ \text{Earth seconds} = 145.9 \times 2.006 \approx 292.6 \text{ seconds} \][/tex]

Step 5: Convert Earth seconds to Earth minutes

[tex]\[ \text{Earth minutes} = \frac{292.6}{60} = 4.9 \text{ minutes} \][/tex]

Conclusion

When the clock face on the moon reads 12:12, approximately 4.9 Earth minutes have elapsed.

To solve this problem we will apply the concepts related to the double slit experiment. Here we test a relationship between the sine of the deviation angle and the distance between slit versus wavelength and the bright fringe order. Mathematically it can be described as,

[tex]dsin\theta = m\lambda[/tex]

Here,

d = Distance between slits

m = Any integer which represent the order number or the number of repetition of the spectrum

[tex]\lambda[/tex] = Wavelength

[tex]\theta[/tex] = Angular deviation

Replacing with our values we have,

[tex](6.93*10^{-6}) sin\theta = (3)(491*10^{-9})[/tex]

[tex]\theta = sin^{-1} (\frac{(3)(491*10^{-9}}{6.93*10^{-6}) })[/tex]

Part A)

[tex]\theta = 0.2141rad[/tex]

PART B)

[tex]\theta = 0.2141rad(\frac{360\°}{2\pi rad})[/tex]

[tex]\theta = 12.27\°[/tex]

Answer:

[tex]\theta = 0.195^0[/tex]

Explanation:

wavelength [tex]\lambda = 648 nm \ = 648*10^{-9}m[/tex]

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

[tex]dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})[/tex]

[tex]\theta = 0.195^0[/tex]

The first maximum will appear at an angle [tex]\theta = 0.195^0[/tex] from the beam axis

.Answer:

The value of the work done is [tex]\bf{ 5.29 qA_{0}}[/tex].

Explanation:

When a charged particle having charge [tex]q[/tex] is moving through an electric field [tex]E[/tex], the net force ([tex]F[/tex]) on the charge is

[tex]F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

and the work done ([tex]W[/tex]) by the particle is

[tex]W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Given, [tex]E = \dfrac{A_{0}}{x^{1/2}}[/tex].

Substitute the value of electric field in equation (1) and then substitute the result in equation (2).

[tex]W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}[/tex]

Answer:

The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.

Explanation:

The engine of the craft provides an upward thrust of [tex]3480 N[/tex] so that the space craft descends at a constant speed.

This implies that the net force on the space craft is zero.

The upward thrust will be equal to the downward gravitational pull by Callisto.

So the weight of the craft near the vicinity will be 3480 N.

Answer:

True

Explanation:

Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit.

Answer:

[tex]0.276I_0[/tex]

Explanation:

When unpolarized light passes through a polarizer, only the component of the light vibrating in the direction parallel to the axis of the polarizer passes through: therefore, the intensity of light is reduced by half, since only 1 out of 2 components passes through.

So, after the first polarizer, the intensity of light passing through is:

[tex]I_1=\frac{I_0}{2}[/tex]

Where [tex]I_0[/tex] is the initial intensity of the unpolarized light.

Then, the light (which is now polarized) passes through the second polarizer. Here, the intensity of the light passing through the second polarizer is given by Malus Law:

[tex]I_2=I_1 cos^2 \theta[/tex]

where:

[tex]\theta[/tex] is the angle between the axes of the two polarizers

In this problem the angle is

[tex]\theta=42^{\circ}[/tex]

So the intensity after of light the 2nd polarizer is:

[tex]I_2=I_1 (cos 42^{\circ})^2=\frac{I_0}{2}(cos 42^{\circ})^2=0.276I_0[/tex]

The  intensity of the beam after it has passed through the second polarizer should be 0.276 I0.

Since

after the first polarizer, the intensity of light should be

I1 = I0/2

Here,

Io should be  initial intensity of the unpolarized light

Now

The intensity should be

= I0/2(cos 42)

= 0.276 I0

Here theta be 42 degrees

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Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck [tex]a_{t[/tex] = 1 m/[tex]s^{2}[/tex]  (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

[tex]a_{A}[/tex] =  0.5 m/[tex]s^{2}[/tex] (upward) , [tex]a_{B}[/tex] =  0.5 m/[tex]s^{2}[/tex] (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])g =  ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])[tex]a_{B}[/tex]

                                  = 2T- (400 + 50)*(9.81 m/[tex]s^{2}[/tex]) = (400 + 50)*(0.5 m/[tex]s^{2}[/tex])

                        T = 2320N

Truck:  [tex]M_{R}[/tex] = ∑([tex]M_{R}[/tex])eff: = [tex]-N_{f}[/tex] (3.4m) + [tex]m_{T}[/tex] (2.0m) - T (0.6m)= [tex]m_{T} a_{T}[/tex] (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/[tex]s^{2}[/tex] )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/[tex]s^{2}[/tex])  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: [tex]N_{f}[/tex] + [tex]N_{R}[/tex] - [tex]m_{T}[/tex]g = 0

                                       10544 + [tex]N_{R}[/tex]  - (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 0

                [tex]N_{R}[/tex] = 9076N

   ∑fx (to the left) = ∑(fx)eff:  [tex]F_{R}[/tex] - T = [tex]m_{T} a_{T}[/tex]

                                      [tex]F_{R}[/tex] = 2320N + (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 4320N

(a) reaction at each front wheel:

1/2 [tex]N_{f}[/tex] (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: [tex]N_{B}[/tex] + [tex]M_{B}[/tex]g - [tex]m_{B}[/tex][tex]a_{B}[/tex]

            [tex]N_{B}[/tex] = (400kg)(9.81 m/[tex]s^{2}[/tex]) + (400kg)(0.5 m/[tex]s^{2}[/tex]) = 4124N (compression)

Answer:

a) The reaction at each of the fron wheels is 5266.1 N

b) The force between the boulder and the pallet is 4120 N

Explanation:

The acceleration of truck is:

[tex]a_{T} =a_{A} +a_{B}[/tex]

Where

aA = acceleration of pallet = ?

aB = acceleration of boulder = ?

aA = aB

aT = acceleration of truck = 1 m/s²

[tex]1=a_{A} +a_{A}\\a_{A}=a_{B}=0.5m/s^{2}[/tex]

From diagram 1 and 2, the system of external forces is:

∑Fy = ∑(Fy)ef (eq.1)

From diagram 1:

∑Fy = 2T - g(mA + mB)

Where T = tension force

mA = mass of pallet = 50 kg

mB = mass of boulder = 400 kg

From diagram 2:

∑(Fy)ef = aB(mA + mB)

Substituting into equation 1:

[tex]2T-g(m_{A} +m_{B} )=a_{B} (m_{A} +m_{B} )\\T=\frac{a_{B}(m_{A} +m_{B}) +g(m_{A} +m_{B} ) }{2} =\frac{0.5(50+400)+9.8(50+400)}{2} =2317.5N[/tex]

From diagram 3 and 4, represents the system of external forces:

∑MR = ∑(MR)ef (eq. 2)

From diagram 3:

∑MR = -N(2 + 1.4) + mTg(2) - T(0.6)

Where

N = normal force

mT = mass of truck = 2000 kg

From diagram 4:

∑(MR)ef = mTaT

Substituting into equation 2:

[tex]-N(2+1.4)+m_{T} g(2)-T(0.6)=m_{T} a_{T} \\-N(3.4)+(2000*9.8*2)-(2317.5*0.6)=2000*1\\N=\frac{(2000*9.8*2)-(2317.5*0.6)-2000}{3.4} =10532.2N[/tex]

From diagram 3 and 4:

∑Fy = ∑(Fy)ef

[tex]N+N_{R} -m_{T} g=0\\10532.2+N_{R}-(2000*9.8)=0\\N_{R}=9067.8N[/tex]

a) The reaction at each of the front wheels is:

Rf = N/2 = 10532.2/2 = 5266.1 N

b) From diagram 5 and 6:

∑Fy = ∑(Fy)ef

[tex]N_{B} +m_{B} g=m_{B} a_{B} \\N_{B}-(400*9.8)=(400*0.5)\\N_{B}=4120N[/tex]

Answer:the mass Of The object,and there area Of The bottom surface

Explanation:

There mass Of The object, and the area Of The bottom surface

The amount of pressure exerted by a solid is based on the mass of the object and the area of the bottom surface. therefore the correct answer is option B.

The total applied force per unit of area is known as the pressure.

Both the external force being applied and the area to which it is being applied affect the pressure.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

A specific force produces less pressure when its region of influence is expanded. It is also true that increasing pressure is produced as a force's region of influence is reduced. For instance, because the force is acting across a smaller area at the pointy end of a high-heeled shoe, it applies more pressure there than it does at the flat end.

Thus, The mass of the object and the size of the bottom surface determine how much pressure a solid will exert. Therefore, choice B is the appropriate answer.

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To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

[tex]\Phi = BA Cos \theta[/tex]

Here,

[tex]\theta[/tex] = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

[tex]\epsilon= -N \frac{d\Phi }{dt}[/tex]

[tex]\epsilon = -N \frac{(BAcos\theta)}{dt}[/tex]

[tex]\epsilon = -NAcos\theta \frac{dB}{dt}[/tex]

[tex]\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)[/tex]

[tex]\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )[/tex]

At time [tex]t = 5.71s[/tex],  Induced emf is,

[tex]\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))[/tex]

[tex]\epsilon = 10.9V[/tex]

Therefore the magnitude of the induced emf is 10.9V

10.90V

Faraday's law of induction states that the electromotive force (E) produced in a magnetic field is directly proportional to the change in flux, ΔΦ, inversely proportional to change in time, Δt. i.e

E = - N ΔΦ / Δt       ----------------(i)

Where;

N = proportionality constant called the number of coils in the wire.

With a small change in time, equation (i) could be re-written as follows;

E = - N δΦ / δt              --------------(ii)

Also, the magnetic flux, Φ, is given as follows;

Φ = BA cos θ          --------------------(iii)

Where;

B = magnetic field

A = cross sectional area of the wire

θ = angle between the field and the cross-section of the wire

Substitute equation (iii) into equation (ii) as follows;

E = - N δ(BAcosθ) / δt

E = - NAcosθ δ(B) / δt               -----------------------(iv)

From the question;

B(t) = (3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²           -------------------(v)

Substitute equation (v) into equation (iv) as follows;

E = - NAcosθ δ[(3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²] / δt         -----(vi)

Solve equation (vi) by taking derivative as follows;

E = - NAcosθ [3.05 − (2)6.95 t]

E = - NAcosθ [3.05 − 13.9t]            ----------------(vii)

Solve for A using the following relation;

A = πr²    -----------------(viii)

Where;

r = radius of the wire loop = 0.220m

π = 3.142

Substitute these values into equation (viii) as follows;

A = 3.142 x 0.220²

A = 0.152m²

Now substitute A = 0.152m², N = 1 (a single coil), θ = 19.5° and t = 5.71s into equation (vii)

E = - (1) (0.152)cos(19.5)° [3.05 − 13.9(5.71)]

E = - (1) (0.152)(0.94) [3.05 − 13.9(5.71)]

E = 10.90V

Therefore, the induced EMF in the loop is 10.90V

Answer:

N = 3522.759 N in the upward direction

Explanation:

acceleration of the rocket, a = 31.2 m/s²

mass of the astronaut, m = 85.9 kg

g = 9.81 m/s²

Since the rocket takes off from Earth's surface, it moves upward and the equation, according to the second law of motion, is given by:

N - mg = ma

N - (85.9*9.81) = (85.9* 31.2)

N - 842.679 = 2680.08

N = 2680.08 + 842.679

N = 3522.759 N in the upward direction

Since the normal force is positive, it is in the upward direction.

To calculate the normal force acting on the astronaut, subtract the net force from the weight of the astronaut. The normal force is directed downwards. Normal Force = -1829.06 N. The negative sign indicates that the normal force is directed downward.

The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, because the astronaut is accelerating upwards, the normal force will be less than the weight of the astronaut. We can calculate the normal force using the formula:

Normal Force = Weight - Net Force

where:
Weight = mass × acceleration due to gravity (9.8 m/s²)
Net Force = mass × acceleration (31.2 m/s²)

Plugging in the values, we get:

Weight = 85.9 kg × 9.8 m/s² = 841.42 N

Net Force = 85.9 kg × 31.2 m/s² = 2670.48 N

Therefore, the normal force acting on the astronaut is:
Normal Force = 841.42 N - 2670.48 N = -1829.06 N

The negative sign indicates that the normal force is directed downward.

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To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,

[tex]\text{Length of the string} = L = 4.32 m[/tex]

[tex]\text{Frequency of the wave} = f = 75 Hz[/tex]

[tex]\text{Time taken to reach the other end} = t = 0.5 s[/tex]

Velocity of the wave,

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

[tex]V = \frac{4.32 m}{0.5s}[/tex]

[tex]V = 8.64m/s[/tex]

Wavelength of the wave,

[tex]\lambda = \frac{V}{f}[/tex]

[tex]\lambda = \frac{8.64m/s}{75Hz}[/tex]

[tex]\lambda = 0.1152m[/tex]

[tex]\lambda = 11.52cm[/tex]

Therefore the wavelength of the waves on the string is 11.53 cm

11.52cm

The velocity, v, of a wave undergoing a simple harmonic motion is related to the wavelength, λ, and frequency, f, of the wave as follows;

v =  f x λ         -----------------(i)

But;

The velocity is also given as the ratio of the length, l, of the body producing the wave to the time taken, t, to undergo the motion. i.e

v = [tex]\frac{l}{t}[/tex]         --------------(ii)

Now, substitute the value of v in equation (ii) into equation (i) as follows;

[tex]\frac{l}{t}[/tex] = f x λ           ----------------(iii)

From the question,

l = 4.32m

t = 0.5s

f = 75Hz

Substitute these values into equation (iii) as follows;

[tex]\frac{4.32}{0.5}[/tex] = 75 x λ

Make λ subject of the formula;

λ = [tex]\frac{4.32}{0.5*75}[/tex]

λ = 0.1152m

Convert the value to cm by multiplying by 100

λ = 0.1152 x 100cm = 11.52cm

Therefore, the wavelength of the waves on the string is 11.52cm