For the following reaction, 24.8 grams of diphosphorus pentoxide are allowed to react with 13.2 grams of water . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer:

C. The two cart system has a kinetic energy of 2.4 J.

Explanation:

Hi there!

The momentum of the two cart system is conserved. That means that the momentum of the system before the collision is equal to the momentum of the system after the collision. The momentum of the system is calculated by adding the momenta of the two carts:

initial momentum of the system = final momentum of the system

pA + pB =  p (A + B)

mA · vA + mB · vB = (mA + mB) · v

Where:

pA and pB = initial momentum of carts A and B respectively.

p (A +B) = momentum of the two cart system after the collision.

mA and mB = mass of carts A and B respectively.

vA and vB = initial velocity of carts A and B.

v = velocity of the two cart system.

We have the following data:

mA = 0.4 kg

mB = 0.8 kg

vA = 6 m/s

vB = 0 m/s

Solving the equation for v:

mA · vA + mB · vB = (mA + mB) · v

0.4 kg · 6 m/s + 0.8 kg · 0 m/s = (0.4 kg + 0.8 kg) · v

2.4 kg m/s = 1.2 kg · v

v = 2.4 kg m/s / 1.2 kg

v = 2 m/s

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where m is the mass of the object and v its speed.

Replacing with the data we have obtained:

KE = 1/2 · 1.2 kg · (2 m/s)²

KE = 2.4 J

The two cart system has a kinetic energy of 2.4 J.

After the collision, the two carts stick together and move with a common velocity. The kinetic energy of the two carts after the collision is 2.4 J.

First, we need to calculate the initial momentum of cart A and the final momentum of the two carts combined.

The initial momentum of cart A is given by the formula: momentum = mass * velocity.

So, momentum of cart A = 0.4 kg * 6 m/s = 2.4 kg*m/s.

After the collision, the two carts stick together and move with a common velocity.

Using the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.

Hence, 2.4 kg*m/s = (0.4 kg + 0.8 kg) * final velocity.

Solving for the final velocity, we get: final velocity = 2.4 kg*m/s / 1.2 kg = 2 m/s.

Finally, we can calculate the kinetic energy of the two carts after the collision using the formula: kinetic energy = (1/2) * mass * velocity^2.

Kinetic energy = (1/2) * (0.4 kg + 0.8 kg) * (2 m/s)^2 = 2.4 J.

Therefore, the correct answer is option C. 2.4 J.

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Explanation:

Given that,

Frequency of the AM radio station, [tex]f=1020\ kHz=1020\times 10^3\ Hz[/tex]

(a) Let [tex]\lambda[/tex] is the wavelength of this electromagnetic radiation. It can be calculated as :

[tex]\lambda=\dfrac{c}{f}[/tex]

[tex]\lambda=\dfrac{3\times 10^8\ m/s}{1020\times 10^3\ Hz}[/tex]

[tex]\lambda=294.11\ m[/tex]

Since the wavelength of one cycle is 294.12 m, then the total number of cycles over a 3 km distance is:

[tex]n=\dfrac{3000}{294.12}=10.19\ cycles[/tex]

Let the period is the duration of one cycle is given by:

[tex]T=\dfrac{1}{f}[/tex]

[tex]T=\dfrac{1}{1020\times 10^3}[/tex]

[tex]T=9.8\times 10^{-7}\ s[/tex]

So, total time required is t as :

[tex]t=n\times T[/tex]

[tex]t=10.19\times 9.8\times 10^{-7}[/tex]

[tex]t=9.98\times 10^{-6}\ s[/tex]

The wavelength of the electromagnetic radiation from the AM radio station is approximately 293.1 meters. It takes approximately 10 milliseconds for the radiation to propagate from the broadcasting antenna to a radio 3 km away.

The wavelength of electromagnetic radiation can be calculated using the formula:

wavelength = speed of light / frequency

Given that the frequency of the AM radio station is 1020 kilohertz and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula to find the wavelength:

wavelength = (3 x 10^8 m/s) / (1020 x 10^3 Hz)

Simplifying the expression gives us a wavelength of approximately 293.1 meters.

To calculate the time required for the radiation to propagate from the broadcasting antenna to a radio 3 km away, we can use the formula:

time = distance / speed

Given that the distance is 3 km and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula:

time = (3 km) / (3 x 10^8 m/s)

Converting kilometers to meters gives us a distance of 3000 meters:

time = (3000 m) / (3 x 10^8 m/s)

Simplifying the expression gives us a time of 0.01 seconds or 10 milliseconds.

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Answer:

Pressure, [tex]P=2.05\times 10^6\ Pa[/tex]

Explanation:

Given that,

The combined mass of the man and chair is 95.0 kg, m = 95 kg

The radius of the circular leg of the chair, r = 0.6 cm

Area of the 4 legs of the chair, [tex]A=4\times \pi r^2[/tex]

Let P is the pressure each leg exert on the floor. The total force acting per unit area is called pressure exerted. Its expression is given by :

[tex]P=\dfrac{F}{A}[/tex]

[tex]P=\dfrac{mg}{4\times \pi r^2}[/tex]

[tex]P=\dfrac{95\times 9.8}{4\times \pi (0.6\times 10^{-2})^2}[/tex]

[tex]P=2.05\times 10^6\ Pa[/tex]

So, the pressure exerted by each leg on the floor is [tex]2.05\times 10^6\ Pa[/tex]. Hence, this is the required solution.

To calculate the pressure each chair leg exerts on the floor, convert the combined mass of the man and chair into force using the equation F = mg, and then calculate the area of each chair leg using the formula A = πr². Finally, divide the total force by the number of legs and then that value by the area of each leg.

The key to solving this problem is recognizing that the pressure exerted by the man and chair on each chair leg is equal to the force of their combined weight divided by the area of contact the chair legs make with the floor. First, convert the combined mass of the man and chair (95.0 kg) into force in newtons, using the equation F = mg, where F is the force in Newtons, m is the mass in kilograms, and g is the acceleration due to gravity (9.8 m/s²). In this case, F = 95.0 kg * 9.8 m/s² = 931 N.

Next, calculate the area of each chair leg that is in contact with the floor. Since each leg is circular, its area, A, can be calculated using the formula A = πr², where r is the radius. Remember to convert the radius from centimeters to meters. Therefore, A = π*(0.006 m)² = 0.000113 m².

Finally, divide the total force by four (since the weight is distributed evenly over four legs) and then that value by the area of each leg to find the pressure each leg exerts: P = F/(4*A) = 931 N / 4 / 0.000113 m² = 2063363.4 Pascal. Remember to use the correct units of pressure (Pascal).

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Answer:

Speed of the water that emerge out of the pipe is 18.8 m/s

Explanation:

Since we know that water drops projected upwards to maximum height of 18 m

So here we can use kinematics equations here

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we have

[tex]v_f = 0[/tex]

[tex]d = 18 m[/tex]

[tex]a = -9.80 m/s^2[/tex]

so we will have

[tex]0 - v_i^2 = 2(-9.80)(18)[/tex]

[tex]v_i = 18.8 m/s[/tex]

To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.

By torque it is understood that

[tex]\tau = F*d[/tex]

Where,

F= Force

d = Distance

The value of the given Torque acts from the center of mass causing it to rotate clockwise.

The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.

I enclose a graph that allows us to understand the problem in a more didactic way.

The correct answer is D.

Answer:

Explanation:

Given

no of moles [tex]n=2.43[/tex]

Temperature raised [tex]\Delta T=11.9 k[/tex]

Work done by gas

[tex]W=\int_{V_1}^{V_2}PdV[/tex]

[tex]W=P\Delta V[/tex]

[tex]W=nR\Delta T[/tex]

[tex]W=2.43\times 8.314\times 11.9[/tex]

[tex]W=240.41 kJ[/tex]

(b)Energy Transferred as heat

[tex]Q=nc_p\Delta T[/tex]

[tex]c_p[/tex]=specific heat at constant Pressure

[tex]c_p[/tex] for ideal Mono atomic gas is [tex]\frac{5R}{2}[/tex]

[tex]Q=2.43\times \frac{5R}{2}\times 11.9[/tex]

[tex]Q=601.03 kJ[/tex]

(c)Change in Internal Energy

[tex]\Delta U=Q-W[/tex]

[tex]\Delta U=601.03-240.41=360.62 kJ[/tex]

(d)Change in average kinetic Energy [tex]\Delta k[/tex]

[tex]K.E._{avg}=\frac{3}{2} \times k\times T[/tex]

[tex]\Delta K.E.=\frac{3}{2} \times k\times \Delta T[/tex]  ,where k=boltzmann constant

[tex]\Delta K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 11.9[/tex]

[tex]\Delta K.E.=2.46\times 10^{-22} J[/tex]

Answer:

Answer:

Answer:

1. f = 5.94 x 10^14 Hz

2. 3.94 x 10^-19 J or  2.46 eV

Explanation:

wavelength, λ = 505 nm = 505 x 10^-9 m

speed of light,c = 3 x 10^8 m/s

1. Let the frequency of the light is f.

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

[tex]f=\frac{3 \times 10^{8}}{505 \times 10^{-9}}[/tex]

f = 5.94 x 10^14 Hz

2. Energy is given by

E = h x f

where, h is the Plank's constant.

E = 6.63 x 10^-34 x 5.94 x 10^14

E = 3.94 x 10^-19 J

Now, we know that 1 eV = 1.6 x 10^-19 J

E = 2.46 eV

Explanation:

Explanation:

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is [tex]-1.74 rad/s^2[/tex]

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

[tex]\omega_i = 2000 rpm[/tex]

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

[tex]1 rev = 2\pi rad[/tex]

[tex]1 min = 60 s[/tex]

We find:

[tex]\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s[/tex]

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

[tex]\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s[/tex]

Now we can find the angular acceleration, given by

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_i = 209.3 rad/s[/tex] is the initial angular speed

[tex]\omega_f = 104.7 rad/s[/tex] is the final angular speed

t = 60 s is the time interval

Substituting,

[tex]\alpha = \frac{104.7-209.3}{60}=-1.74  rad/s^2[/tex]

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

[tex]\omega' = \omega_i + \alpha t[/tex]

where we have

[tex]\omega_i = 209.3 rad/s[/tex]

[tex]\alpha = -1.74 rad/s^2[/tex]

And substituting t = 40 s, we find

[tex]\omega' = 209.3 + (-1.74)(40)=139.7 rad/s[/tex]

d)

The angular displacement of the wheel in a certain time interval t is given by

[tex]\theta=\omega_i t + \frac{1}{2}\alpha t^2[/tex]

where

[tex]\omega_i = 209.3 rad/s[/tex]

[tex]\alpha = -1.74 rad/s^2[/tex]

And substituting t = 60 s, we find:

[tex]\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad[/tex]

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

[tex]\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev[/tex]

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Answer:

(a) 5.91 rad/s

(b) 2.96 rad/s

(c) 2.25 rad

(d) 0.81 m

Explanation:

The torque generated by tension force from the string is:

T = FR = 14*0.36 = 5.04  Nm

This torque would then create an angular acceleration on the uniform-density wheel with moments of inertia of

[tex] I = 0.5mR^2 = 0.5*10*0.36^2 = 0.648kgm^2[/tex]

[tex]\alpha = \frac{T}{I} = \frac{5.04}{0.648}=7.78rad/s^2[/tex]

(a) The wheel turns for 0.76s, this means the final angular speed is

[tex]\omega_f = t\alpha = 0.76*7.78 = 5.91 rad/s[/tex]

(b) Since the force is constant, the torque is also constant and so is the angular acceleration. This means angular speed is rising at a constant rate. That means the average angular speed is half of the final speed

[tex]\omega_a = 0.5\omega_f = 0.5*5.91 = 2.96 rad/s[/tex]

(c) The total angle that the wheel turns is the average angular speed times time

[tex]\theta = t\omega_a = 2.96*0.76 = 2.25 rad[/tex]

(d) The string length coming off would equal to the distance swept by the wheel

[tex]d = R\theta = 0.36*2.25 = 0.81 m[/tex]

The final angular speed of this uniform-density wheel of mass is equal to 5.91 radians/s.

Given the following data:

Mass = 10 kg.

Radius = 0.36 m.

Initial velocity = 0 m/s (since it's starting from rest).

Force = 14 Newton.

Time = 0.76 seconds.

First of all, we would determine the torque produced due to the tensional force that is acting on the string by using this formula:

[tex]\tau = Fr\\\\\tau = 14 \times 0.36[/tex]

Torque = 5.04 Nm.

Also, we would determine the moment of inertia by using this formula;

[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 10 \times 0.36^2\\\\I=5 \times 0.1296[/tex]

I = 0.648 [tex]kgm^2[/tex]

Next, we would determine the angular acceleration by using this formula;

[tex]\tau=\alpha I\\\\\alpha =\frac{\tau}{I} \\\\\alpha =\frac{5.04}{0.648}\\\\\alpha = 7.78 \;rad/s^2[/tex]

Now, we can calculate the final angular speed:

[tex]\omega_f = t\alpha \\\\\omega_f = 0.76 \times 7.78\\\\\omega_f = 5.91 \;rad/s[/tex]

[tex]\omega_A = \frac{1}{2} \omega_f\\\\\omega_A = \frac{1}{2} \times 5.91\\\\\omega_A =2.96\;rad/s[/tex]

[tex]\theta = t\omega_A \\\\\theta = 0.76 \times 2.96[/tex]

Angle = 2.25 rad.

In order to calculate the length of the string that came off the wheel, we would determine the distance swept by the wheel:

[tex]d=r\theta\\\\d=0.36 \times 2.25[/tex]

d = 0.81 meter.

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Answer:

9.7 x 10¹¹ .

Explanation:

2.5 % of 110 W = 2.75 J/s

energy of one photon

= hc / λ

=[tex]\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}[/tex]

= .036 x 10⁻¹⁷ J

No of photons emitted

= 2.75 / .036 x 10⁻¹⁷

= 76.38 x 10¹⁷

Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance

photons passing per unit area of this sphere

= 76.38 x 10¹⁷  / 4π ( 2.8)²

Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing

= [tex]\frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2[/tex]

= 9.7 x 10¹¹ .

Answer:

E. radiation.

Explanation:

As we know that heat transfer due to conduction depends on thermal conductivity of the materials and heat transfer due to convection depends on the velocity of the fluid.But on the other hand heat transfer due to radiation depends on the surface properties like  emmisivity .So when bottle is silvered then it will leads to minimize the radiation heat transfer.

Therefore answer is --

E. radiation.

The answer is option C.

The interior of a thermos is silvered to minimize heat transfer by conduction, convection, and radiation. The silvering acts like a mirror, reflecting heat, and the vacuum between the thermos walls almost eliminates conduction and convection.

The interior of a thermos bottle is silvered to minimize heat transfer due to C. conduction, convection and radiation.

This is because the silvering on the inner surface of the thermos acts like a mirror, reducing the amount of heat that can be transferred by all three modes: conduction, convection, and especially radiation.

Conduction is the transfer of heat through direct contact of molecules, minimized in a thermos by the vacuum between its double walls. Convection is the transfer of heat in a fluid (like air or liquid) through the motion of the fluid itself, which is also nearly eliminated by the vacuum. Radiation is the transfer of heat through electromagnetic waves, which the silvering reflects back, greatly reducing heat loss this way.

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Answer:

 v > 133.5 m/s

Explanation:

Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.

Let's start by using the concepts of energy to find the velocity at the top of the circle

Initial. Top circle

    Em₀ = K + U = ½ m v² + m g y

If we place the reference system at the bottom of the cycle y = 2R = L

    Em₀ = ½ m v² + m g y

final. Low circle

    [tex]Em_{f}[/tex] = K = ½ m v₁²

    Emo =  [tex]Em_{f}[/tex]

    ½ m v² + m g y = 1/2 m v₁²

    v₁² = v² + (2g L)

    v₁² = v² + 2 g L

The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point

     v₁ = √2g L

We already have the speed system at the bottom we can use the moment

Starting point before crashing

    p₀ = m v₀

End point after collision at the bottom of the circle

    [tex]p_{f}[/tex] = (m + M) v₁

The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved

    p₀ = [tex]p_{f}[/tex]

    m v₀ = (m + M) v₁

   v₀ = (m + M) / m v₁

Let's replace

   v₀ = (1+ M / m) √ 2g L

Let's reduce to the SI system

   m = 40 g (kg / 1000g) = 0.040 kg

Let's calculate  

    v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)

    v₀ = 34.75 3.8417

    v₀ = 133.5 m / s

the velocity must be greater than this value

    v > 133.5 m/s

The concept used to solve this problem is the conservation of momentum and the conservation of energy.

For conservation of the moment we have the definition:

[tex]m_1v_1 = (m_1+m_2)v_f[/tex]

Where,

m = Mass

[tex]v_1[/tex] = Initial velocity for object 1

[tex]v_f[/tex] = Final velocity

Replacing the values we have to,

[tex]m_1v_1 = (m_1+m_2)v_f[/tex]

[tex]5*12=(5+5)v_f[/tex]

[tex]v_f = 6m/s[/tex]

By conservation of energy we know that the potential energy is equal to the kinetic energy then

[tex]mgh = \frac{1}{2} m(v_f^2-v_i^2)[/tex]

[tex]gh = \frac{1}{2} v_f^2[/tex]

[tex]h = \frac{1}{2} g*v_f^2[/tex]

[tex]h = \frac{1}{2} (9.8)(6)^2[/tex]

[tex]h = 1.837m[/tex]

Therefore after the collision the height when the combined chinks will go is 1.837m

The combined chunks of ice will rise about 3.67 meters above the valley floor after their collision. This is calculated using the principles of conservation of momentum and energy where the initial kinetic energy of the moving ice chunk is conserved and then converted into gravitational potential energy as the combined chunk of ice ascends.

This problem involves the concepts of conservation of momentum and energy. The two chunks of ice are initially sliding with a certain momentum and kinetic energy. When they collide and stick together, the total momentum must still be conserved because the system is isolated (i.e., no external forces are acting). Although the speed will be halved (since the total mass doubled), the total kinetic energy is conserved in the collision. This kinetic energy will then be converted to potential energy as the combined chunk of ice ascends. Using the equation for gravitational potential energy (PE = mgh), where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height, we can solve for h.

The initial kinetic energy of 5kg chunk of ice is KE = 1/2 mv^2, which equals 1/2 * 5kg * (12m/s)^2 = 360 Joules. After the collision, the kinetic energy is equally shared with the other chunk of ice. Since the initial mass is doubled but the speed is halved, the kinetic energy remains the same. Therefore, we set the gravitational potential energy equal to the initial kinetic energy when it reaches its peak: mgh = KE, 10kg * 9.8 m/s^2 * h = 360 Joules. Solving for h, we get that h is approximately 3.67 meters. Thus, the combined chunks of ice will rise about 3.67 meters above the valley floor after the collision.

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The distance [tex]\( d_1 \)[/tex] from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, maintaining static equilibrium, is approximately 46.8 cm.

1. To achieve static equilibrium, the torques on both sides of the pivot point must balance out.  

2. The torque due to the meter stick with mass ( M ) is [tex]\( M \times g \times \frac{L}{2} \)[/tex], where ( L ) is the length of the meter stick (100 cm) and \( g \) is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]).

3. The torque due to the coin on the 0-cm end is [tex]\( m \times g \times d_1 \)[/tex], where [tex]\( m \)[/tex] is the mass of the coin.

4. Since the torques balance out, we have the equation: [tex]\( M \times g \times \frac{L}{2} = m \times g \times d_1 \).[/tex]

5. Rearrange the equation to solve for [tex]\( d_1 \): \( d_1 = \frac{M \times \frac{L}{2}}{m} \).[/tex]

6. Substitute the given values: [tex]\( d_1 = \frac{95 \, \text{g} \times \frac{100}{2} \, \text{cm}}{3.1 \, \text{g}} \).[/tex]

7. Calculate [tex]\( d_1 \)[/tex]to find the distance from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, which is approximately 46.8 cm.

Answer:

low pressure

Explanation:

The streamlines of air particles are squeezed together as the air goes over the top of the bulb. Then, by the law of conservation of mass, the velocity of air particles are increased. And since, the velocity is increase the pressure is bound to decrease. Hence, this leads to region of low pressure on the top of the bulb.

Answer:

632.5 MPa

Explanation:

[tex]\sigma_{m}[/tex] = Matrix stress at fiber failure = 10 MPa

[tex]V_f[/tex] = Volume fraction of fiber = 0.25

[tex]\sigma_f[/tex] = Fiber fracture strength = 2.5 GPa

The longitudinal strength of a composite is given by

[tex]\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa[/tex]

The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa

Final answer:

The given question seeks to calculate the longitudinal strength of a carbon fiber-epoxy composite, but lacks sufficient detail or formulae for a complete answer. Typically, this computation would involve using materials science models that consider fiber orientation and other stress-strain interactions between components.

Explanation:

To compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite with a 0.25 volume fraction of fibers, we'd need to consider the following given properties: the average fiber diameter, average fiber length, fiber fracture strength, fiber-matrix bond strength, matrix stress at fiber failure, and matrix tensile strength.

Although the exact method for calculating the longitudinal strength would typically involve applying principles from materials science, such as the rule of mixtures, in combination with the given data points, the actual question does not provide enough information or a specific formula to complete the calculation. For a real-life carbon fiber-epoxy composite, the longitudinal strength could be substantially influenced by the alignment of the fibers, bond quality between the fibers and matrix, and the interaction between the stress and strain of the components.

If we had a suitable model or empirical formula, we would proceed by plugging in the given values to determine the longitudinal strength. However, as the provided data from the question is incomplete for this calculation, it is recommended to refer to a textbook or comprehensive resource on composite material mechanics for the detailed step-by-step methodology and equations.

Answer:

14.43° or 0.25184 rad

Explanation:

v = Speed of sound in air = 343 m/s

f = Frequency = 1240 Hz

d = Width in doorway = 1.11 m

Wavelength is given by

[tex]\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m[/tex]

In the case of Fraunhofer diffraction we have the relation

[tex]dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad[/tex]

The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad

Answer:

Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]

Explanation:

We have given that two long parallel wires each carrying a current of 12 A

So [tex]I_1=I_2=12A[/tex]

Distance between the two conductors d = 9 cm = 0.09 m

We know that magnetic force between two parallel conductors per unit length is given by

[tex]F=\frac{\mu _0I_1I_2}{2\pi d}=\frac{4\times 3.14\times 10^{-7}\times 12\times 12}{2\times 3.14\times 0.09}=3.2\times 10^{-4}N[/tex]

Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]

(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].

Explanation:

Given that,

Speed [tex]v= 3.50\times10^{5}\ m/s[/tex]

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

[tex]U=qV=eV[/tex]

Using conservation of energy

[tex]K_{i}+U_{i}=K_{f}+U_{f}[/tex]

[tex]\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}[/tex]

[tex]]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})[/tex]

[tex]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V[/tex]

[tex]\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}[/tex]

Put the value into the formula

[tex]\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}[/tex]

[tex]\Delta V=639.2=0.639\ kV[/tex]

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

[tex]\Delta U=q\Delta V[/tex]

Put the value into the formula

[tex]\Delta U=1.6\times10^{-19}\times639.2[/tex]

[tex]\Delta U=1.022\times10^{-16}\ J[/tex]

(c). We need to calculate the work done on the proton

Using formula of work done

[tex]\Delta U=-W[/tex]

[tex]W=q(V_{2}-V_{1})[/tex]

[tex]W=-1.6\times10^{-19}(150-100)[/tex]

[tex]W=-8\times10^{-18}\ J[/tex]

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]

(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].

Final answer:

The change in electric potential and potential energy of the proton can be calculated based on the provided potentials. The work done on the proton equals the change in potential energy.

Explanation:

The change in the proton's electric potential: The change in electric potential is the final potential minus the initial potential, thus the change is -150 V - 100 V = -250 V.

The change in potential energy of the proton: The potential energy change equals the charge of the proton times the change in potential, giving -proton charge x change in potential.

The work done on the proton: The work done is equal to the change in the potential energy of the proton.

Answer:

Velocity will be [tex]v=1.291\times 10^8m/sec[/tex]

Explanation:

We have given magnetic field B = 0.00005 T

Mass m = 238 U

We know that [tex]1u=1.66\times 10^{-27}kg[/tex]

So 238 U [tex]=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg[/tex]

Radius [tex]=R+1.44=6378+1.44=6379.44KM[/tex]

We know that magnetic force is given by

[tex]F=qvB[/tex] which is equal to the centripetal force

So [tex]qvB=\frac{mv^2}{r}[/tex]

[tex]1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}[/tex]

[tex]v=1.291\times 10^8m/sec[/tex]

Answer:x=23.4 cm

Explanation:

Given

mass of block [tex]m=0.5 kg[/tex]

inclination [tex]\theta =30[/tex]

coefficient of static friction [tex]\mu =0.35[/tex]

coefficient of kinetic friction [tex]\mu _k=0.25[/tex]

distance traveled [tex]d=77.3 cm[/tex]

spring constant [tex]k=35 N/m [/tex]

work done by gravity+work done by friction=Energy stored in Spring

[tex]mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}[/tex]

[tex]mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}[/tex]

[tex]0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}[/tex]

[tex]x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}[/tex]

[tex]x=0.234 m[/tex]

[tex]x=23.4 cm[/tex]

An excited atom can return to its ground state by absorbing electromagnetic radiation is false about the electromagnetic radiation.

Option B

Explanation:

In the scope of modern quantum theory, the term Electromagnetic radiation is identified as the movement of photons through space. Almost all the sources of energy that we utilize today such as coal, oil, etc are a product of electromagnetic radiation which was absorbed from the sun millions of years ago.

Various properties of electromagnetic radiations are a directly proportional relationship between the energy and the frequency, Inverse proportionality between frequency and the wavelength, etc. Hence, we can conclude that an "excited atom" can never return to its ground state by assimilating electromagnetic radiation and the 2nd statement is false.

The statement 'An excited atom can return to its ground state by absorbing electromagnetic radiation' is false, as the process actually involves emitting radiation. Electromagnetic radiation's energy increases with frequency, and frequency and wavelength have an inverse relationship.

The false statement among the ones provided is: An excited atom can return to its ground state by absorbing electromagnetic radiation. An excited atom returns to its ground state by emitting radiation, not absorbing it. When it comes to electromagnetic radiation, increasing frequency indeed results in increasing energy. This is because the energy of electromagnetic radiation is directly proportional to its frequency. Additionally, when an electron in the n = 4 state in the hydrogen atom transitions to the n = 2 state, it emits radiation at an appropriate frequency. Lastly, the frequency and the wavelength of electromagnetic radiation are inversely proportional, meaning as one increases, the other decreases.

Answer:

[tex]x|_0^{0.2}=1.59535[/tex]

Explanation:

Given expression of velocity:

[tex]v(t)=10^{0.2t}-1 ;\ \ 0\leq t\leq 5\ s[/tex]

For getting displacement we need to integrate the above function with respect to t.

Given period of integration:

[tex]t_0=0\ s \to t_f=0.2\ s[/tex]

For trapezoidal rule we break the given interval into two parts of 0.1 s each.

∴take n=2

hence, [tex]\Delta t= 0.1[/tex]

[tex]v(0)=0[/tex]

[tex]v(0.1)=1.0471[/tex]

[tex]v(0.2)=1.0965[/tex]

Now, using trapezoidal rule:

[tex]\int_{0}^{0.2}v(t)\ dt=\Delta x[\frac{1}{2}\times v(0)+v(0.1)+\frac{1}{2}\times v(0.2)][/tex]

[tex]\int_{0}^{0.2}v(t)\ dt=0.1 [\frac{1}{2}\times 0+1.0471+\frac{1}{2}\times 1.0965][/tex]

[tex]x|_0^{0.2}=1.59535[/tex]

Note:Smaller the value of sub-interval better is the accuracy.

Answer:

 [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]

Explanation:

given,

diameter of the beam (d)= 5.85 cm

                                        = 0.0585 m

average wavelength of the(λ) = 580 n m

angle of of spreading = ?

according to the Rayleigh Criterion the minimum angular spreading, for a circular aperture, is

                [tex]\theta_{min} = 1.22\ \dfrac{\lambda}{d}[/tex]

                [tex]\theta_{min} = 1.22\ \dfrac{580 \times 10^{-9}}{0.0585}[/tex]

                [tex]\theta_{min} = 1.22\times 9.145 \times 10^{-6}[/tex]

               [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]

the minimum angle of spreading is [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]

Explanation:

Given that,

The disintegration constant of the nuclide, [tex]\lambda=0.0178\ h^{-1}[/tex]

(a) The half life of this nuclide is given by :

[tex]t_{1/2}=\dfrac{ln(2)}{\lambda}[/tex]

[tex]t_{1/2}=\dfrac{ln(2)}{0.0178}[/tex]

[tex]t_{1/2}=38.94\ h[/tex]

(b) The decay equation of any radioactive nuclide is given by :

[tex]N=N_oe^{-\lambda t}[/tex]

[tex]\dfrac{N}{N_o}=e^{-\lambda t}[/tex]

Number of remaining sample in 4.44 half lives is :

[tex]t_{1/2}=4.44\times 38.94[/tex]

[tex]t_{1/2}=172.89\ h^{-1}[/tex]

So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 172.89}[/tex]

[tex]\dfrac{N}{N_o}=0.046[/tex]

(c) Number of remaining sample in 14.6 days is :

[tex]t_{1/2}=14.6\times 24[/tex]

[tex]t_{1/2}=350.4\ h^{-1}[/tex]

So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 350.4}[/tex]

[tex]\dfrac{N}{N_o}=0.0019[/tex]

Hence, this is the required solution.

Answer:

The path difference between the two waves should be one-half of a wavelength

Explanation:

When two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P , then the condition for it is that the path difference of two beams must be odd multiple of half wavelength. Symbolically

path difference = ( 2n+1 ) λ / 2

So path difference may be λ/2 , 3λ/ 2,  5λ/ 2 etc .

Hence right option is

The path difference between the two waves should be one-half of a wavelength.

The path difference between the two waves should be one-half of a wavelength.

This can be defined as the distance between successive crests or troughs and the path difference is denoted below:

Path difference = ( 2n+1 ) λ / 2 which could  be λ/2 , 3λ/ 2 etc.

Hence , the path difference between the two waves should be one-half of a wavelength

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Final answer:

a) The wavelength of the radio station's broadcast is approximately 272.73 meters. b) At a time of 3.1 μs, the value of the radio wave's electric field is approximately 0.619 N/C.

Explanation:

a) To calculate the wavelength of the radio station's broadcast, we can use the formula λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency. Plugging in the given frequency of 1100 kHz (or 1100 x 10^3 Hz), we get: λ = (3 x 10^8 m/s) / (1100 x 10^3 Hz) = 272.73 m

b) To find the value of the radio wave's electric field at a specific time, we can use the given time dependence equation E(t) = E0 sin(2πft), where E0 is the amplitude, f is the frequency, and t is the time. Plugging in the given amplitude of E0 = 0.62 N/C, frequency of 1100 kHz (or 1100 x 10^3 Hz), and time of 3.1 μs (or 3.1 x 10^-6 s), we get: E(t) = 0.62 sin(2π x 1100 x 10^3 x 3.1 x 10^-6) ≈ 0.619 N/C

Answer:2.74 Hz

Explanation:

Given

mass Puck [tex]m=0.51 kg[/tex]

length of cord [tex]L=0.827 m[/tex]

Maximum Tension in chord [tex]T=126 N[/tex]

as the Puck is moving in a horizontal circle so maximum Tension in the string will be equal to centripetal force

[tex]F_c=m\omega ^2L=T[/tex]

[tex]126=0.51\times (\omega )^2\times 0.827[/tex]

[tex]\omega =\sqrt{298.74}[/tex]

[tex]\omega =17.28 rad/s[/tex]

[tex]\omega =2\pi f[/tex]

[tex]f=\frac{2\pi }{\omega }[/tex]

[tex]f=2.74 Hz[/tex]

To find the highest frequency at which the puck can go around the circle without the cord breaking, we use the formula for tension in a circular motion and solve for velocity. Then we use the velocity to find the frequency. The highest frequency is approximately 2.18 Hz.

To determine the highest frequency at which the puck can go around the circle without the cord breaking, we need to find the maximum tension in the cord.

Since the tension in the cord is equal to the centripetal force required to keep the puck moving in a circle, we can use the formula:

Tension = mass × velocity² / radius

Substituting the given values, we get:

126N = 0.51kg × v² / 0.827m

Now, solving for v, we find:

v² = (126N × 0.827m) / 0.51kg
v² = 204.2 m²/s²

v = √(204.2 m²/s²) = 14.29 m/s

Since the frequency of an object moving in a circle is equal to its velocity divided by the circumference of the circle, we can calculate the highest frequency as:

Frequency = v / (2πr)
Frequency = 14.29 m/s / (2π × 0.827m)
Frequency ≈ 2.18 Hz

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Answer:

6.87 ft/s is the rate at which the top of ladder slides down.

Explanation:

Given:

Length of the ladder is, [tex]L=20\ ft[/tex]

Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.

Now, from triangle ABC,

AB² + BC² = AC²

[tex]h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1[/tex]

Differentiating the above equation with respect to time, 't'. This gives,

[tex]\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2[/tex]

In the above equation the term [tex]\frac{dh}{dt}[/tex] is the rate at which top of ladder slides down and [tex]\frac{db}{dt}[/tex] is the rate at which bottom of ladder slides away.

Now, as per question, [tex]h=8\ ft, \frac{db}{dt}=3\ ft/s[/tex]

Plug in [tex]h=8[/tex] in equation (1) and solve for [tex]b[/tex]. This gives,

[tex]8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft[/tex]

Now, plug in all the given values in equation (2) and solve for [tex]\frac{dh}{dt}[/tex]

[tex]8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s[/tex]

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

Answer:

E.

Explanation:

Mercury is the first and the smallest planet of the solar system. It has the smallest radius of rotation. And temperature in the planet is quite high.  Mercury has so many tremendous cliffs because they were probably formed by tectonic stresses when the entire planet shrank as its core cooled. So its crust mus have contracted.

Final answer:

Mercury's tremendous cliffs were likely formed as the planet shrank due to its cooling core, leading to compression and wrinkling of the crust, rather than by volcanic activity, running water, or a sequence of impacts.

Explanation:

The tremendous cliffs seen on Mercury were likely formed as a result of tectonic stresses when the planet shrank due to the cooling and solidification of its core over time. There is no evidence of plate tectonics on Mercury, but the existence of long scarps suggests that at some point the planet underwent compressional forces leading to the formation of these cliffs. The scarps cut across craters, indicating they are younger than the craters themselves and thus were not formed by running water, volcanic activity, or a sequence of impacts.

Discovery Scarp, a prominent feature on Mercury that is nearly 1 kilometer high and more than 100 kilometers long, provides critical evidence of these events, giving us an insight into the chaotic early solar system where impacts played a major role in shaping planetary surfaces. These cliffs are geological evidence of Mercury's dynamic past and are part of the wrinkling observed on its surface due to the shrinkage of the planet.