The total volume in milliliters of a glucose-water solution is given by the equation below: V = 1001.93 + 111.5282m + 0.64698m2 where m is the molality of the solution. The partial molar volume of glucose ?glucose, is the slope of a V versus m curve, (?V/?m). Find the partial molar volume of glucose in a 0.100m solution of glucose in water

Answer:

a. v' = 24.22 x 10 ⁻³

b. β = 0.825 T

Explanation:

V = 150 V , E = 6.0 x 10 ⁶ N / C

a.

¹/₂ * m * v² = e * V

v = √ 2 * e / m * V

v = √ 2 * 1.76 x 10 ¹¹ * 150 v  = √ 5.28 x 10 ¹³

v = 7.26 x 10 ⁶ m /s

v' / c = 7.26 x 10 ⁶ m /s  / 3.0 x 10 ⁸ = 24.22 x 10 ⁻³

b.  

β = E / V

β = 6.0 x 10 ⁶ / 7.266 x 10 ⁶

β = 0.825 T

c.

When the increasing the accelerating potential speed it doesn't change the up wired the electric force and the magnetic force the electron be beat down more .

Answer:

E

Explanation:

The parallel plate capacitor is being charged by a steady current i. If the constant current is flowing along a straight wire, what happens inside the capacitor, between the plates is that the induced magnetic field strength is zero Tesla near the center of the plates and increases as r increases toward the edges of the plates.  

From Ampère–Maxwell law,  the magnetic field between the capacitor plates assuming that the capacitor is being charged at a constant rate by a steady current is  

                                B = μ₀r /2A [tex]\frac{dQ}{dt} e_{o}[/tex]    

where;

μ₀ is permeability  

Q(t) is the instantaneous charge on the positive plate,

A is the cross-sectional area of a plate and

e₀ is a unit vector

To solve this problem it is necessary to apply the concepts related to the principle of superposition, as well as to constructive interference. From the definition we know that this can be expressed mathematically as

[tex]sin\theta_m = \frac{m\lambda}{d}[/tex]

Where

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

[tex]\lambda[/tex]= Wavelength

d = Distance between slits

From triangle (Watch image below)

[tex]tan\theta_1 = \frac{y_1}{L}[/tex]

[tex]tan\theta_1 = \frac{0.488}{1.64}[/tex]

[tex]\theta_1 = 16.57\°[/tex]

Replacing the angle at the first equation for m=1 we have

[tex]\lambda = d sin \theta_1[/tex]

Each of the distances (d) would be defined by

d = \frac{1}{5300} = 0.0001886

[tex]\lambda = (\frac{1}{5300}) sin(16.57)[/tex]

[tex]\lambda = 538nm[/tex]

Therefore the wavelength of the laser light is 538nm

Answer:

Explanation:

According to Kepler's law a radius vector joining planet and sun swept equal area in equal interval of time thus it can be applied for comets.

when a comet is nearer  to sun it has to swept more area so its velocity is more nearer to the sun ,

The basics of this formula comes from conservation of angular momentum  thus comet moves faster when it approaches the sun.

Answer:

Explanation:

Given

mass of bullet [tex]m=7 gm[/tex]

mass of gun and Puck [tex]M=150 gm[/tex]

speed of gun and Puck is [tex]v=0.53 m/s[/tex]

Let speed of bullet be u

conserving Momentum

initial momentum=Final Momentum

[tex]0=Mv+mu[/tex]

[tex]u=-\frac{M}{m}v[/tex]

[tex]u=-\frac{150}{7}\times 0.53=-11.35 m/s[/tex]          

negative sign indicates that bullet is moving in opposite direction        

Using the law of conservation of momentum, it's possible to calculate that the bullet moves in the opposite direction of the gun and puck at a speed of approximately 113.6 m/s.

This question represents a practical example of the conservation of momentum. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. As the system (gun and bullet) is initially at rest, the total momentum is zero, and it remains zero even after the bullet is fired.

Let's denote the bullet's velocity as v. The momentum of the bullet after firing is its mass times its velocity, or 0.007kg * v. The momentum of the gun and puck is their combined mass times their velocity, or 0.150kg * 0.530m/s. As the total momentum should remain zero, these two quantities must be equal and opposite. Thus, 0.007kg * v = - 0.150kg * 0.530m/s. Solving for v, we find that the bullet's speed is approximately -113.6 m/s. The negative sign indicates that the bullet moves in the opposite direction of the gun and puck.

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To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_f[/tex]

Where,

[tex]m_{1,2}[/tex]= Mass of each object

[tex]v_{1,2}[/tex] = Initial velocity of each object

[tex]v_f[/tex]= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

[tex]m_2v_2 = (m_1+m_2)v_f[/tex]

[tex](0.42)(21) = (90+0.42)v_f[/tex]

[tex]v_f = 0.0975m/s[/tex]

Therefore the velocity right after catching the ball is 0.0975m/s

The receiver's final velocity is approximately 0.098 m/s.

First, we need to find the initial momentum of the ball by multiplying its mass (0.42 kg) by its velocity (21 m/s). The initial momentum of the ball is therefore 8.82 kg·m/s. Next, we need to find the final momentum of the receiver by multiplying his mass (90 kg) by his final velocity. Since he catches the ball at the highest point in his jump, his final velocity is 0 m/s. So the final momentum of the receiver is 0 kg·m/s. According to the law of conservation of momentum, the initial momentum of the ball must be equal to the final momentum of the receiver. Therefore, the final velocity of the receiver is 8.82 kg·m/s divided by his mass (90 kg), which is approximately 0.098 m/s.

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Answer:

Explanation:

Given

Pivot is at h=0.76 L

Where L is the length of String

Conserving Energy at A and B

[tex]mgL=\frac{1}{2}mu^2[/tex]

where u=velocity at bottom

[tex]u=\sqrt{2gL}[/tex]

After coming at bottom the ball completes the circle with radius r=L-0.76 L

Suppose v is the velocity at the top

Conserving Energy at B and C

[tex]\frac{1}{2}mu^2=mg(2r)+\frac{1}{2}mv^2[/tex]

Eliminating m

[tex]u^2=4r+v^2[/tex]

[tex]v^2=u^2-4\cdot gr[/tex]

[tex]v^2=2gL-4g(L-0.76L)[/tex]

[tex]v^2=1.04gL[/tex]

[tex]v=\sqrt{1.04gL}[/tex]          

Answer:

21.73 cm

Explanation:

We have given parameters:

Mass of block, m = 2.0 kg

Force constant of spring, k = 264 N/m

Length of rough area,  L  = 10 cm = 0.1 m

Co-efficient of kinetic friction , [tex]\mu_{k}[/tex] = 0.54

Block's speed after crossing rough area, v = 2.7 m/s

Block's initial speed ( when it was released from compressed spring), [tex]v_{0}[/tex] = 0 m/s

We need to find the distance that the spring was initially compressed, x = ?

Hence, we well apply Work-Energy principle which indicates that,

Work done by the friction = Change in the total energy of block

[tex]- \mu_{k} * mg * L = (\frac{1}{2} * mv^{2} - \frac{1}{2} * mv_{0} ^{2} ) + (0 - \frac{1}{2} * kx^{2})[/tex]

-0.54 * 2 * 9.8 * 0.1 = (1/2 * 2 * [tex]2.7^{2}[/tex] - 1/2 * 2 * 0) + (0 - 1/2 * 264 * [tex]x^{2}[/tex])

x = 0.2173 m = 21.73 cm

Answer:

Approximately [tex]\rm 2.0\; V[/tex].

Approximately [tex]\rm 30 \; mA[/tex]. (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here [tex]R_1= 10\; \Omega[/tex] and [tex]R_2= 100\; \Omega[/tex] are connected in parallel. Their effective resistance would be equal to

[tex]\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega[/tex].

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be [tex]20\; \rm mA = 0.020\; \rm A[/tex]. That should also be the current through the effective [tex]\displaystyle \rm \frac{10}{11} \; \Omega[/tex] resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

[tex]V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V[/tex].

The voltage drop across the entire circuit would equal to

In this case, that value would be equal to [tex]1.83 + 0.182 \approx 2.0\; \rm V[/tex]. That's the voltage that needs to be supplied to the circuit to achieve a current of [tex]20\; \rm mA[/tex] through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

[tex]\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega[/tex].

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

[tex]\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega[/tex].

Again, the current in a serial circuit is the same in all appliances.

[tex]\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA[/tex].

Answer:

14009. 72 kgms^-1

Explanation:

Momentum is the product of an objects mass and velocity

The calculated value of  the magnitude of the car's momentum be  14006.72 kg.m/s or  68739 lb.mph.

Momentum is a property of a body moving in linear motion, It is product of mass and velocity. As velocity is a vector quantity, momentum is also a vector quantity with magnitude and direction. SI unit of momentum be kg.m/s.

Given parameters:

Mass of the car: m =  946.4 kg. = 2083 lb

Speed of the car: v = 14.8 m/s. = 33.0 mph

So, momentum of the car: p = mv = 946.4 × 14.8 kg.m/s = 14006.72 kg.m/s.

or p = mv = 2083 lb×33.0 mph  = 68739 lb.mph.

Hence,  the magnitude of its momentum be  14006.72 kg.m/s or  68739 lb.mph.

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Answer:

scenario A and B

Explanation:

The Doppler effect is the change in frequency of a wave by the relative movement of the source and the observer. It is described by the expression

                 f ’= (v + v₀) / (v - [tex]v_{s}[/tex]) f₀

Where f₀ is the emitted frequency, v the speed of the wave, v₀ and [tex]v_{s}[/tex] the speed of the observer and the source, respectively, the signs are for when they are approaching and in the case of being away the signs are changed.

Consequently, from the above for the Doppler effect to exist there must be a relative movement of the source and the observer.

Let's examine the scenarios

A) True. You agree with the equation shown

B) True. Only the signs should be changed and it is described by the equation shown

C) False. If there is no relative movement there is no Doppler effect

Answer:

True

Explanation:

According to the definition of a closed system, It is true because it's not precisely a closed system. A closed system is a physical system that doesn't let certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.

Answer:

45 N, 30N ,20N

Explanation

torque = perpendicular force× distance

perpendicular force = torque/ distance

1) perpendicular force = 18/0.400 = 45 N

2)  perpendicular force = 18/0.600 = 30N

3) perpendicular force = 18/0.900 = 20N

We have that for the Question " the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque" it can be said that

From the question we are told

You want to apply 18 N ⋅ m of torque to tighten a nut, using a wrench with a length of either 0.400 m, 0.600 m, or 0.900 m. For each wrench, determine the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque.

Generally the equation for the Force is mathematically given as

t=F*d

a)

[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.4}\\\\[/tex]

F=45N

b)

[tex]t=F*d\\\\F=\frac{t}{d}\\\\\F=\frac{18}{0.6}\\\\[/tex]

F=30N

c)

[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.6}\\\\[/tex]

F=20N

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Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

[tex]t = \frac{{2v\sin \theta }}{g}[/tex]

Substitute [tex]9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g and [tex]60^\circ[/tex]  for [tex]\theta[/tex] in above equation.

[tex]t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\[/tex]  

Spitted water will travel [tex]0.80{\rm{ m}}[/tex] horizontally.

Displacement of water in this time period is

[tex]x = vt\cos \theta[/tex]

Substitute [tex]\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}[/tex] for [tex]t\rm 60^\circ[tex] for [tex]\theta[/tex] and [tex]0.80{\rm{ m}}[/tex] for x in above equation.

[tex]\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

[tex]{v_v} = v\sin \theta[/tex]

Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex]  for [tex]\theta[/tex] in above equation.

[tex]\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]

Horizontal component of the velocity is,

[tex]{v_h} = v\cos \theta[/tex]

Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex]  for [tex]\theta[/tex] in above equation.

[tex]\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]

When horizontal [tex]({0.60\;{\rm{m}}}[/tex] distance away from the fish.  

The time of flight for distance (d) is ,

[tex]t = \frac{d}{{{v_h}}}[/tex]

Substitute [tex]0.60\;{\rm{m}}[/tex] for d and [tex]1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_h}[/tex] in equation [tex]t = \frac{d}{{{v_h}}}[/tex]

[tex]\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\[/tex]

Distance of the insect above the surface is,

[tex]s = {v_v}t + \frac{1}{2}g{t^2}[/tex]

Substitute [tex]2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_v}[/tex] and [tex]0.3987{\rm{ s}}[/tex] for t and [tex]- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g in above equation.

[tex]\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\[/tex]

Answer:D

Explanation:

Given

mass of object [tex]m=5 kg[/tex]

Distance traveled [tex]h=10 m[/tex]

velocity acquired [tex]v=12 m/s[/tex]

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy[tex]=mgh=5\times 9.8\times 10=490 J[/tex]

Final Energy[tex]=\frac{1}{2}mv^2+W_{f}[/tex]

[tex]=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}[/tex]

where [tex]W_{f}[/tex] is friction work if any

[tex]490=360+W_{f}[/tex]

[tex]W_{f}=130 J[/tex]

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

[tex]\mu[/tex] = Coefficient of friction = 0.4

Energy stored in spring is given by

[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J[/tex]

As the energy in the system is conserved we have

[tex]\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s[/tex]

The speed of the 8 kg block just before collision is 3.258 m/s

The speed of the 8 kg block before the collision is found by considering the work done by friction and the initial potential energy in the spring and using the conservation of energy principle to solve for the speed.

To find the speed of the 8 kg block just before it collides with the 2 kg block, we must calculate the motion of the block under the influence of friction before the collision occurs.

First, we determine the force of friction using the coefficient of friction (μ) and the normal force (N), which in this case equals the weight of the block (m × g) since the surface is horizontal. So, the force of friction is given by Fₙ = μ × m × g = 0.4 × 8 kg × 9.8 m/s² = 31.36 N.

Since there is friction, work done by friction must be considered to find the velocity. The work done by friction (Wₙ) is the force of friction (Fₙ) multiplied by the distance the block moves before colliding with the second block (d), which is not given in this problem. But if we had the distance, we could use the work-energy principle, where the net work done on an object is equal to the change in its kinetic energy (KE).

Assuming no other forces act on the block, then the work done by friction is equal to the initial potential energy stored in the spring. Using Hooke's Law, the potential energy (PE) in the spring is given by PE = ½ kx², where k is the spring constant and x is the compression distance. This energy gets converted entirely into the kinetic energy of the 8 kg block minus the losses due to friction.

Using the conservation of energy:

PE - Wₙ = ½ m v²

½ kx² - Fₙ × d = ½ m v²

We solve this equation for v, the speed of the block, if we know the distance (d).

Answer:

6.13428 rev/s

Explanation:

[tex]I_f[/tex] = Final moment of inertia = 4.2 kgm²

I = Moment of inertia with fists close to chest = 5.7 kgm²

[tex]\omega_i[/tex] = Initial angular speed = 3 rev/s

[tex]\omega_f[/tex] = Final angular speed

r = Radius = 76 cm

m = Mass = 2.5 kg

Moment of inertia of the skater is given by

[tex]I_i=I+2mr^2[/tex]

In this system the angular momentum is conserved

[tex]L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s[/tex]

The rotational speed will be 6.13428 rev/s

Final answer:

To solve this problem, we need to use the principle of conservation of angular momentum. The skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.

Explanation:

To solve this problem, we need to use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of the skater with his arms extended is equal to the final angular momentum when his hands are brought to his chest.

The initial angular momentum is given by the equation:

Li = Iiωi

where Li is the initial angular momentum, Ii is the initial moment of inertia, and ωi is the initial angular velocity. The final angular momentum is given by:

Lf = Ifωf

where Lf is the final angular momentum, If is the final moment of inertia, and ωf is the final angular velocity.

Since the skater is pulling his hands in to his chest, his moment of inertia decreases, and his angular velocity increases. We can set up the following equation:

Iiωi = Ifωf

Substituting the given values, we have:

(5.7 kg.m2)(2π(3.0 rev/s)) = (4.2 kg.m2)ωf

Solving for ωf, we find that the skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.

Answer:

B. Green

Explanation:

The change in wavelength that is caused when one object is moving towards another object from the perspective of the viewer is called the Doppler effect.

When objects move close to one another then wavelength reduces which is called blue shift while the opposite case causes the wavelength to increase which is called red shift.

Here, the color of the traffic light is yellow and you are getting closer to it so the wavelength should blue shift and green should appear.

It takes 0.05 seconds for a part undergoing simple harmonic motion with a frequency of 5.00 Hz to move from x=0 to x=-1.80 cm.

The student is asking how long it takes for a machine part undergoing simple harmonic motion (SHM) with a frequency of 5.00 Hz and amplitude of 1.80 cm to move from its equilibrium position (x=0) to its maximum negative displacement (x=-1.80 cm).

In SHM, the period (T) is the time required for one complete cycle of the motion. The period can be calculated using the formula T = 1/f, where f is the frequency. For a frequency of 5.00 Hz, the period is T = 1/5.00 Hz = 0.20 s. In half of a period, the part moves from one extreme to the other, passing through the equilibrium position at the quarter-period. Thus, to go from x=0 to x=-1.80 cm, it takes a quarter of this period, or (1/4)T = 0.05 s.

Therefore, it takes 0.05 seconds for the part to travel from x=0 to x=-1.80 cm while undergoing SHM.

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Answer:

Explanation:

The case relates to interference in thin films , in which we study interference of light waves reflected by upper and lower surface of a medium or glass.

For constructive interference , the condition is

2μt = ( 2n+1)λ/2

μ is refractive index of glass , t is thickness , λ is wavelength of light.

putting the given values

2 x 1.53 x 350 x 10⁻⁹ = ( 2n+1) λ/2

λ = 2142nm /  ( 2n+1)

For n = 2

λ = 428 nm

This wave length will have constructive interference making this light brightest of all .

For n = 1

λ = 714  nm

So second largest  brightness will belong to 700 nm wavelength.

Answer:

It depends on the rate at which the current through it is changing.          

Explanation:

As per the Faraday's law, the induced emf is given by :

[tex]\epsilon=-L\dfrac{di}{dt}[/tex]

Where

L is the inductance of the inductor

[tex]\dfrac{di}{dt}[/tex] is the rate of change of current

So, the self-induced emf of an inductor depends on the rate at which the current through it is changing. Hence, the correct option is (d).

Answer:

Pnitrogen=3.18 kPa, Poxygen=1.62 kPa , Ptotal= 4.80 kPa

Explanation:

partial pressure equation becomes Ptotal = Pnitrogen + Poxygen

Partial pressure of Nitrogen

Pnitrogen= nRT/V

n=no of moles =mass/molar mass

mass of nitrogen=0.6kg

Molar mass of nitrogen gas=28gmol^-1

n=0.6/28=0.0214moles

R=0.2968 kPa·m3/kg·K

T=300k

V=0.6m^3

Pnitrogen=(0.0214 * 0.2968 * 300)/0.6

Pnitrogen=3.18 kPa

Likewise

Poxygen=nRT/V

n=0.4/32=0.0125moles

R=0.2598 kPa·m3/kg·K

T=300k

V=0.6m^3

Poxygen=(0.0125 * 0.2598 * 300)/0.6

Poxygen=1.62 kPa

Ptotal= 3.18+1.62= 4.80 kPa

Using the ideal gas law, the partial pressure for Nitrogen ([tex]N_{2}[/tex]) and Oxygen ([tex]O_{2}[/tex]) is 148.4 kPa and 86.6 kPa respectively. The total pressure of the mixture is the sum of these two partial pressures, equaling 235 kPa.

The pressure exerted by individual gases in a mixture is known as partial pressure. The calculation of the partial pressure of each gas, and the total pressure of the mixture involves using the ideal gas law. In this case, the ideal gas equation is P = mRT/V, where P represents the pressure, m is the mass of gas, R is the gas constant, T is the temperature and V is the volume. Thus, for Nitrogen ([tex]N_{2}[/tex]), the partial pressure is (0.6 kg × 0.2968 kPa·m3/kg·K × 300K) / 0.6 m3 = 148.4 kPa, and for Oxygen ([tex]O_{2}[/tex]), the partial pressure is (0.4 kg× 0.2598 kPa·m3/kg·K × 300K) / 0.6 m3 = 86.6 kPa. The total pressure, then, is the sum of these partial pressures, which equals 148.4 kPa + 86.6 kPa = 235 kPa.

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Answer:

d. 0.043 g/s

Explanation:

Formula for rate of conduction of heat through a bar per unit time is as

follows

Q = k A ( t₁ - t₂ ) / L

A is cross sectional area and L is length of rod ,( t₁ - t₂ )  is temperature

difference . Q is heat conducted per unit time

Putting the values in the equation

Q = (427 x 1 x 10⁻⁴ x 100 )/ 30 x 10⁻²

= 14.23 J/s

mass of ice melted per second

= Q / Latent heat of ice

= 14.23 / 334000

= 0.043 g/s

Final answer:

To find the mass of ice melted per second, the rate of heat transfer is calculated using the thermal conductivity of silver and the latent heat of fusion of ice. The calculated value of 0.426 g/s does not match the provided options, suggesting a possible typo in the options or the need for clarification.

Explanation:

The question is about calculating the amount of ice that is melted per second when a silver bar conducts heat from a hot reservoir to a block of ice. To answer this question, we will use the given thermal conductivity of silver (k = 427 J/s·m·°C), the cross-sectional area of the silver bar (1.0 cm2), and the latent heat of fusion for ice (Lf = 334,000 J/kg).

First, we need to calculate the rate of heat transfer (Q) from the 100°C reservoir through the silver bar to the 0°C ice using the formula:

Q = k · A · (ΔT/L),

where A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the bar. Converting A to meters squared (A = 1.0 x 10-4 m2) and L to meters (L = 0.30 m), and plugging in the values:

Q = 427 J/s·m·°C · 1.0 x 10-4 m2 · 100°C / 0.30 m = 142.333 J/s.

Now, using the formula Q = mLf to find the mass of ice melted per second (m), where Q is the rate of heat transfer and Lf is the latent heat of fusion:

m = Q / Lf = 142.333 J/s / 334,000 J/kg = 0.000426 kg/s.

Converting this mass to grams (since 1 kg = 1000 g), we get:

m = 0.000426 kg/s · 1000 g/kg = 0.426 g/s.

This value is not exactly matching any of the options provided (a-d), but with rounding and considering significant figures, the closest answer would be 0.43 g/s, which is not listed amongst the options provided.

Answer:

1.  2176 kilograms of plastic  2. $15,232  3. $397 (U)     4. $947 (F)   $1,344  (U)

Explanation:

Generally, cost variance analysis can be used to estimate the difference between the actual cost and the expected costs. However, if the actual cost is more than the expected cost, then the variance is said to be unfavorable and vice versa.

1. The standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets?

We know that each helmet requires 0.68 kilograms of plastic. Thus, to make 3200 helmets, we will need 0.68*3200 = 2176 kilograms of plastic

2. The standard materials cost allowed (SQ × SP) to make 3,200 helmets.

We also know that each helmet costs $7.00 per kilogram. Therefore, to make 3200 helmets, the standard materials cost = $7.00*2176 = $15,232

3. The material spending variance = difference between the actual cost and the standard cost = $15,629 - $15,232 = $397  U

4.  The materials price variance and the materials quantity variance?

The materials price variance  is the actual cost- (standard cost per kilogram x actual number of plastic used) . Therefore:

Materials price variance = $15,629 - ($7 x 2,368 kg)

Materials price variance = $15,629 - $16,576 = ($947) F

Since the budgeted cost is relatively higher than the actual cost, the materials price variance is favorable (F) by $947.

The materials quantity variance = (Actual number of plastic used x Standard cost per kilogram) - Standard cost

Materials quantity variance = ($7 x 2,368 kg) - $15,232

Materials quantity variance = $16,576 - $15,232 = $1,344  U

Since the budgeted cost is relatively higher than the standard cost, the materials quantity variance is unfavorable (U) by $1,344.

To find the standard quantity of plastic and standard materials cost allowed, multiply the standard quantity per helmet by the number of helmets produced and multiply the standard quantity by the standard price per kilogram, respectively. The materials spending variance is the difference between the actual cost and the standard cost that should have been incurred. The materials price variance is the difference between the actual quantity of plastic used multiplied by the standard price per kilogram and the actual cost, while the materials quantity variance is the difference between the standard quantity of plastic allowed multiplied by the standard price per kilogram and the actual cost.

To calculate the standard quantity of plastic (SQ) allowed to make 3,200 helmets, multiply the standard quantity per helmet by the number of helmets produced. In this case, SQ = 0.68 kg/helmet * 3,200 helmets = 2,176 kg. To calculate the standard materials cost allowed (SQ × SP), multiply the standard quantity by the standard price per kilogram. In this case, the materials cost allowed is $15,232 (2,176 kg * $7.00/kg).

To calculate the materials spending variance, subtract the actual cost from the standard cost that should have been incurred. In this case, the materials spending variance is $397 (Standard Cost - Actual Cost = $15,232 - $15,629). The materials price variance is calculated by subtracting the actual quantity of plastic used multiplied by the standard price per kilogram from the actual cost.

The materials quantity variance is calculated by subtracting the standard quantity of plastic allowed multiplied by the standard price per kilogram from the actual cost. In this case, the materials price variance is $7,774 (Actual Cost - Actual Quantity * Standard Price = $15,629 - 2,368 kg * $7.00/kg) and the materials quantity variance is -$2,145 (Actual Quantity * Standard Price - Standard Cost = 2,368 kg * $7.00/kg - $15,232).

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Answer:

The time required for sucrose transportation through the tube is 8.4319 sec.

Explanation:

Given:

L = 0.025 m

A = 6.5×10^-4 m^2

D = 5×10^-10 m^2/s

ΔC = 5.2 x 10^-3 kg/m^3

m = 5.7×10^-13 kg

Solution:

t = m×L / D×A×ΔC

t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)

t = 8.4319 sec.

This question is wrong because of not correct values.The Correct question is

A gas in a cylinder is held at a constant pressure of 1.80×10⁵Pa and is cooled and compressed from 1.70m³ to 1.20m³. The internal energy of the gas decreases by 1.40×10⁵J.

(a) Find the work done by the gas.

(b) Find the absolute value of the heat flow, |Q|, into or out of the gas, and state the direction of the heat flow.

(c) Does it matter whether the gas is ideal? Why or why not?

Answer:

(a) W= -9×10⁴J

(b) |Q|=2.3×10⁵

(c) It does not matter whether gas is ideal or non-ideal

Explanation:

Given

V₁=1.70m³

V₂=1.20m³

p=1.8×10⁵ pa

ΔU= -1.40×10⁵J

For (a) work done by gas

[tex]W=p(V_{2}-V_{1} )\\W=1.8*10^{5}(1.2-1.7)\\ W=-9*10^{4}J[/tex]

For (b) Heat flow |Q|

|Q|=ΔU+W

[tex]Q=(-1.4*10^{5}) +(-9*10^{4})\\Q=-2.3*10^{5}J\\So\\|Q|=|-2.3*10^{5}|\\|Q|=2.3*10^{5}J[/tex]

For (c) part

It does not matter whether the gas is ideal or not because the first law of thermodynamics which applied in our solution could applied to any material ideal or non ideal  

For the calculation of resistance there are generally two paths. The first is through Ohm's law and the second is through the relationship

[tex]R = \frac{pL}{A}[/tex]

Where

p = Specific resistance of material

L = Length

A = Area

The area of nerve axon is given as

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

[tex]A = \pi (5*10^{-6})^2[/tex]

[tex]A = 7.854*10^{-11}m^2[/tex]

The rest of values are given as

[tex]p= 2 \Omega\cdot m[/tex]

[tex]L = 2cm = 0.02m[/tex]

Therefore the resistance is

[tex]R = \frac{pL}{A}[/tex]

[tex]R = \frac{2*0.02}{7.854*10^{-11}}[/tex]

[tex]R = 509.3*10^6\Omega[/tex]

[tex]R = 509.3M\Omega[/tex]

Answer:

[tex]W=-3.2\times 10^4\ J[/tex]

Explanation:

Given:

Process 1:

Process 2:

We know that the work done by an ideal gas is given as:

[tex]W=P\times (V_f-V_i)[/tex]

Now for process 1:

[tex]W_1=0\ J[/tex]

∵there is no change in volume in this process.

For process 2:

[tex]W_2=4\times 10^5\time (0.12-0.2)\ J[/tex]

[tex]W_2=-3.2\times 10^4\ J[/tex]

∵Negative , sign indicates that the work is being done on the gas here since the gas is being compressed.

Hence the total work done by the gas during this two step process is :

[tex]W=W_1+W_2[/tex]

[tex]W=0-3.2\times 10^4[/tex]

[tex]W=-3.2\times 10^4\ J[/tex] is the work done by the gas.

Answer:

I = 0.0674 kg.m²

Explanation:

given,

mass = 13.5 Kg

torsion constant = k = 0.618 N.m

number of cycle = 28

time = 58.1 s

Time of one cycle

[tex]T = \dfrac{58.1}{28}[/tex]

[tex]T =2.075\ s[/tex]

we know,

[tex]T = 2\pi\sqrt{\dfrac{I}{k}}[/tex]

[tex]I = k (\dfrac{T}{2\pi})^2[/tex]

[tex]I =0.618\times \dfrac{T^2}{4\pi^2}[/tex]

[tex]I =0.618\times \dfrac{2.075^2}{4\pi^2}[/tex]

      I = 0.0674 kg.m²

the rotational inertia of the object is equal to  I = 0.0674 kg.m²