A 100 W electric heater (1 W = 1 J/s) operates for 11 min to heat the gas in a cylinder. At the same time, the gas expands from 1 L to 6 L against a constant external pressure of 3.527 atm. What is the change in internal energy of the gas? (1 L·atm = 0.1013 kJ)

Answer:

Current

Explanation:

Current can refer to the flow of electrons through a conductor of some kind as well as the number of electrons flowing through the conductor.

Final answer:

The amount of electrons flowing through the wire is called current, which is measured in amperes or amps. Current is directly proportional to voltage and inversely proportional to resistance, as stated in Ohm's Law.

Explanation:

The amount of electrons flowing through the wire is called current. Electric current is measured in the unit known as ampere (A), which is defined as the flow of one coulomb of charge through an area in one second. According to Ohm's Law, the current I in a circuit is directly proportional to the voltage V applied across the circuit and inversely proportional to the resistance R of the circuit, which can be expressed by the equation I = V/R. The SI unit of current is ampere, whereas ohm is the SI unit for electrical resistance, which represents how strongly a material opposes the flow of electric current.

Explanation:

Scientist and astronomers have observed that large gas clouds and massive stars are orbiting around in an accelerated manner in center of our galaxy that is milky way. A massive star S2's motion has been studied and it is found that the star is revolving around the center of the galaxy. From this scientist have confirmed presence of super massive black hole of 3 million solar masses lurking around the center of milky way.

Answer:

Part a)

[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]

B) uniform Sphere

[tex]f = \frac{2}{7}[/tex]

C) uniform hollow sphere

[tex]f = \frac{2}{5}[/tex]

D) uniform hollow cylinder with inner radius R/2 and outer radius R

[tex]f = \frac{5}{13}[/tex]

Explanation:

As we know that fraction of total energy as rotational energy is given as

[tex]f = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2}[/tex]

now we have

[tex] f = \frac{mk^2(\frac{v^2}{R^2})}{mv^2 + mk^2(\frac{v^2}{R^2})}[/tex]

[tex]f = \frac{\frac{k^2}{R^2}}{1 + \frac{k^2}{R^2}}[/tex]

now we have

A) uniform Solid cylinder

for cylinder we know that

[tex]\frac{k^2}{R^2} = 0.5[/tex]

[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]

B) uniform Sphere

for sphere we know that

[tex]\frac{k^2}{R^2} = \frac{2}{5}[/tex]

[tex]f = \frac{0.4}{1.4} = \frac{2}{7}[/tex]

C) uniform hollow sphere

for hollow sphere we know that

[tex]\frac{k^2}{R^2} = \frac{2}{3}[/tex]

[tex]f = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}[/tex]

D) uniform hollow cylinder with inner radius R/2 and outer radius R

for annular cylinder we know that

[tex]\frac{k^2}{R^2} = \frac{5}{8}[/tex]

[tex]f = \frac{\frac{5}{8}}{\frac{13}{8}} = \frac{5}{13}[/tex]

The fraction of the total kinetic energy that is rotational for a uniform sphere is [tex]\frac{1}{3}[/tex]

Mathematically, the rotational kinetic energy of an object is given by this formula:

[tex]K.E_{rot}=\frac{1}{2} I\omega^2[/tex]  

Where:

Since we know that half of the total kinetic energy for a hollow, cylindrical shell that is rolling without slipping on a horizontal surface is translational and the other half is rotational. Thus, this fraction is given by this mathematical expression:

[tex]K.E=\frac{\frac{k^2}{R^2} }{1+\frac{k^2}{R^2}}[/tex]

a. For a uniform sphere:

[tex]\frac{k^2}{R^2}=0.5[/tex]

Substituting, we have:

[tex]K.E=\frac{0.5 }{1+0.5}\\\\K.E=\frac{0.5 }{1.5}\\\\K.E =\frac{1}{3}[/tex]

b. For a thin-walled hollow sphere:

[tex]\frac{k^2}{R^2}=\frac{2}{5}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{2}{5} }{1+\frac{2}{5}}\\\\K.E=\frac{0.4 }{1.4}\\\\K.E =\frac{2}{7}[/tex]

c. For a uniform hollow sphere:

[tex]\frac{k^2}{R^2}=\frac{2}{3}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{2}{3} }{1+\frac{2}{3}}\\\\K.E=\frac{0.7 }{1.7}\\\\K.E =\frac{2}{5}[/tex]

d. For a hollow sphere with outer radius (R) and inner radius:

[tex]\frac{k^2}{R^2}=\frac{5}{8}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{5}{8} }{1+\frac{5}{8}}\\\\K.E=\frac{0.625 }{1.625}\\\\K.E =\frac{5}{13}[/tex]

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Answer:

The rotational inertia of the system is 1.8 kg.m².

(b) is correct option.

Explanation:

Given that,

Mass of disk = 7.0 kg

Radius = 0.65 m

Mass of clay = 2.1 kg

Distance = 0.41 m

We need to calculate the rotational inertia of the system

Using formula of rotational inertia

[tex]I''=I+I'[/tex]

Where, I= the moment of inertia of a solid disk

I'=the moment of inertia of lump of clay

Put the value into the formula

[tex]I=\dfrac{MR^2}{2}+mr^2[/tex]

[tex]I=\dfrac{1}{2}\times7.0\times(0.65)^2+2.1\times0.41^2[/tex]

[tex]I=1.8\ kg.m^2[/tex]

Hence, The rotational inertia of the system is 1.8 kg.m².

Answer:

b. Technician B only

Explanation:

There is a float connected to the variable resistor in a fuel tank unit. The resistance of the variable resistor also changes as the fuel level changes. The tank unit's resistance changes, the dash mount gauge also changes and available on driver's display.  

If the tank transmitter is disconnected, the operation will not take place and the resistance change will not be transmitted to the dash unit. The needle will therefore remain the empty one at all times.Even after being rusty, the ground wire connection to the fuel tank will be able to conduct. Hence Technician B   is correct and Technician A is incorrect.

Technician A is correct in stating that corrosion of the ground wire at the fuel tank sending unit can cause a lower-than-normal fuel gauge reading.

The accuracy of a fuel gauge reading can be affected by the condition of the electrical connections to the fuel tank sending unit. Technician A's statement is correct; if the ground wire's connection to the fuel tank sending unit becomes rusty or corroded, it can cause a higher resistance, leading the fuel gauge to read lower than normal due to insufficient grounding. On the other hand, Technician B's statement is incorrect because if the power lead to the fuel tank sending unit is disconnected and grounded with the ignition on, the fuel gauge should read full, not empty. The fuel gauge is designed so that grounding the sending unit wire to the chassis will mimic the resistance of a full tank, thus moving the gauge needle to the 'Full' position.

The correct answer, therefore, is a. Technician A only.

Answer:

The weight of the bowling ball makes a more significant impact that the ping pong ball so therefore it would take farther to stop the bowling ball

Explanation:

(A)  The time required to stop the ping pong ball will be less than that of bowling ball.

(B)  The distance required to stop the bowling ball will be less than that of ping pong ball.

Given data:

The ping pong ball and bowling ball has the same magnitude of linear momentum.

Same amount of force to be applied on each, to stop.

(A)

With same magnitude of stopping force (F) and linear momentum (p), the time required to stop will be dependent on the mass. Since, mass of ping pong ball is less, so it will be easily stopped with less effort. While the bowling ball will take some extra seconds or minutes to acquire the rest state.

In other words, the ping pong ball has less inertia, due to which the time taken to stop the ping pong pall will be less, comparing to bowling ball.

Thus, we can conclude that the time required to stop the ping pong ball will be less than that of bowling ball.

(B)

The inertia is given as,

[tex]I = mr^{2}[/tex],

Here, m is the mass.

And the distance required to stop is given by third rotational equation of motion as,

[tex]\omega_{2}^{2}=\omega_{1}^{2}+2 \alpha \theta\\\\\theta =\dfrac{ \omega_{2}^{2}-\omega_{1}^{2}}{2 \alpha}[/tex]

Here, [tex]\alpha[/tex] is the angular acceleration.

And angular acceleration is directly proportional to the inertia of object. More the inertia, more will be the angular acceleration and less will be distance required to stop.

Since, ping pong ball has less inertia, so its angular acceleration will be less. So, the distance covered by the ping pong ball will be more, compared to bowling ball.

Thus, we can conclude that the distance required to stop the bowling ball will be less than that of ping pong ball.

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Explanation:

From the first law of thermodynamics,

Δ[tex]Q[/tex]=Δ[tex]U[/tex]+[tex]W[/tex]

Where [tex]Q[/tex] is the heat given to the gas,

[tex]U[/tex] is the internal energy of the gas,

[tex]W[/tex] is the workdone by the gas.

When pressure is constant,

[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]

[tex]V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}[/tex]

When pressure is constant,[tex]W=P[/tex]Δ[tex]V[/tex]

Where [tex]P[/tex] is pressure and [tex]V[/tex] is the volume of the gas.

Given [tex]P=1.5\times 10^{5}Pa[/tex]

Δ[tex]V=[/tex][tex]7.8-2.6=5.2m^{3}[/tex]

So,[tex]W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J[/tex]

Given that Δ[tex]U=6\times 10^{5}[/tex]

So,Δ[tex]Q=[/tex][tex]6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J[/tex]

By using the first law of thermodynamics, the ideal gas law, the given parameters and an additional calculation for the work done by the gas, we can calculate the total heat absorbed by the gas.

To calculate the heat absorbed by the gas, we use the first law of thermodynamics which states that the heat absorbed by a system is equal to the change in its internal energy plus the work done by the system on its surroundings, expressed as Q = ΔEint + W. The change in internal energy, ΔEint, is given as +6.0 × 10^5 J.

Since the pressure is constant, the work done by the gas, W, can be calculated using W = PΔV, where P is the pressure and ΔV is the change in volume. The change in volume can be determined using the ideal gas law before and after the change, PV = nRT, thus, ΔV = nR(ΔT)/P. Substituting the given pressure, temperature change from 300 K to 900 K, and ideal gas constant, we can find ΔV.

After plugging ΔV into the equation for work done, we then add this to the change in internal energy to find the heat absorbed.

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1) The temperature of the gas in state B is 1200 K

2) The work done by the gas from A to B is 4000 J

3) The work done by the gas from B to C is -5546 J

Explanation:

1)

The temperature of the gas when it is in state B can be found by using the ideal gas equation:

[tex]\frac{P_A V_A}{T_A}=\frac{P_B V_B}{T_B}[/tex]

where

[tex]P_A = 1\cdot 10^6 Pa[/tex] is the pressure in state A

[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A

[tex]T_A = 600 K[/tex] is the temperature in state A

[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

[tex]T_B[/tex] is the temperature in state B

Solving for [tex]T_B[/tex],

[tex]T_B=\frac{P_B V_B T_A}{P_A V_A}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})(600)}{(1\cdot 10^6)(4\cdot 10^{-3})}=1200 K[/tex]

2)

The work done by a gas during an isobaric transformation (as the one between A and B) is

[tex]W=p\Delta V = p (V_B-V_A)[/tex]

where

p is the pressure (which is constant)

[tex]V_B[/tex] is the final volume

[tex]V_A[/tex] is the initial volume

Here we have:

[tex]p=1\cdot 10^6 Pa[/tex]

[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

Substituting,

[tex]W=(1\cdot 10^6)(8\cdot 10^{-3}-4\cdot 10^{-3})=4000 J[/tex]

3)

During an isothermal expansion, the produce between the pressure of the gas and its volume remains constant (Boyle's law), so we can write:

[tex]P_BV_B = P_C V_C[/tex]

where

[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

[tex]P_C = 2\cdot 10^6 Pa[/tex] is the pressure in state C

[tex]V_C[/tex] is the volume in state C

Solving for [tex]V_C[/tex],

[tex]V_C = \frac{P_B V_B}{P_C}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})}{2\cdot 10^6}=4\cdot 10^{-3} m^3[/tex]

The work done by a gas during an isothermal transformation is given by

[tex]W=nRT ln(\frac{V_C}{V_B})[/tex] (1)

where

n is the number of moles

[tex]R=8.314 J/mol K[/tex] is the gas constant

T is the constant temperature (in this case, [tex]T_B[/tex])

[tex]V_C, V_B[/tex] are the final and initial volume, respectively

The number of moles of the gas can be found as

[tex]n=\frac{p_A V_A}{RT_A}=\frac{(1\cdot 10^6)(4\cdot 10^{-3})}{(8.314)(600)}=0.802 mol[/tex]

So now we can use eq.(1) to find the work done by the gas from B to C:

[tex]W=(0.802)(8.314)(1200) ln(\frac{4\cdot 10^{-3}}{8\cdot 10^{-3}})=-5546 J[/tex]

And the work is negative because the gas has contracted, so the work has been done by the surrounding on the gas.

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The temperature of the gas when it is in state B can be found using the ideal gas law. The work done by the gas in expanding isobarically from A to B can be calculated using a simple equation. The work done on the gas in going from B to C can also be calculated using a different equation.

The temperature of the gas when it is in state B can be found using the ideal gas law, which states that PV = nRT. Since the process is isobaric, the pressure remains constant. We can use the equation:

TB = TA * (VA / VB)

Where TA is the initial temperature (600 K), VA is the initial volume (4 x 10-3), and VB is the final volume (8 x 10-3).

The work done by the gas in expanding isobarically from A to B can be calculated using the equation:

WAB = PA * (VB - VA)

Where PA is the initial pressure (1 x 106) and VA and VB are the initial and final volumes, respectively.

The work done on the gas in going from B to C can be calculated using the equation:

WBC = -nRT * ln(VC / VB)

Where VC is the final volume (8 x 10-3) and VB is the initial volume (8 x 10-3).

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The tensile strength of PCTFE is generally expected to be higher than that of PTFE, given the same molecular weight and degree of crystallinity, mainly due to stronger intermolecular forces from the chlorine atom in the PCTFE polymer chain.

The tensile strength of polychlorotrifluoroethylene (PCTFE) and polytetrafluoroethylene (PTFE) specimens having the same molecular weight and degree of crystallinity can vary. Although both materials are fluoropolymers and have similar structural characteristics, PCTFE tends to have a slightly higher tensile strength than PTFE. This is primarily because PCTFE has a chlorine atom in its polymer chain, which contributes to stronger intermolecular forces and therefore, higher tensile strength.

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Answer:

[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]

Explanation:

Assuming given,

Mass of wire be M

length of wire be L

small mass at center is = m

Radius of the wire be equal to = R = L/π

mass of small element of the wire

[tex]dM = \dfrac{M}{L}Rd\theta[/tex]

All the force are acting along y- direction

so, for force calculation

[tex]F = \int \dfrac{mdMG}{R^2} sin\theta [/tex]

[tex]F = \int_0^{\pi} \dfrac{m\dfrac{M}{L}RG}{R^2} sin\theta d\theta[/tex]

[tex]F = \int_0^{\pi} \dfrac{mMG}{L\dfrac{L}{\pi}} sin\theta d\theta[/tex]

[tex]F = \pi \dfrac{mMG}{L^2}}\int_0^{\pi}sin\theta d\theta[/tex]

[tex]F = \pi \dfrac{mMG}{L^2}}(-cos \theta)_0^{\pi}[/tex]

[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]

The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]

Given that :

mass of wire = M

length of wire = L

small mass at center = m

radius of wire = L / [tex]\pi[/tex]

First step : express the mass of small element of wire

dM = [tex]\frac{M}{L} Rd[/tex]∅

Since all forces act in the vertical direction the magnitude of the force exerted will be

F = [tex]\int\limits^\pi _o {\frac{m\frac{M}{L} RG}{R^2} } \, sin\beta d\beta[/tex]

[tex]F = \pi \frac{mMG}{L^2} \int\limits^\pi _0 {x} \, sin\beta d\beta[/tex]

Resolving equation above

Therefore F = [tex]2\pi \frac{mMG}{L^{2} }[/tex]

Hence we can conclude that The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]

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Answer:

19.5 m

Explanation:

t = Time taken

u = Initial velocity = 0 (Assumed thrown from rest)

s = Displacement = 81 m

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of linear motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 81=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{81\times 2}{9.81}}\\\Rightarrow t=4.06371\ s[/tex]

The time taken for the stone to reach the river is 4.06371 seconds

Distance = Speed×Time

[tex]Distance=4.8\times 4.06371=19.5\ m[/tex]

The horizontal distance between the raft and the bridge when the child releases the stone should be 19.5 m

Answer:

V = -0.8 m/s

Explanation:

given,

mass of the child (m)= 25 Kg

mass of the boat(M) = 30 Kg

velocity of boat = ?

Assuming Boys throws package of mass(m₁) 6 Kg at the horizontal speed of       10 m/s

using conservation of momentum

(M + m + m₁) V = (M+ m)V + m₁ v

initial velocity V = 0 m/s

(M + m + m₁) x 0 = (M+ m)V + m₁ v

0 = (25+50)V + 6  x 1 0

75 V = -60

V = -0.8 m/s

negative direction shows that velocity in the direction opposite to the motion of package.

Final answer:

The velocity of the boat immediately after, assuming it was initially at rest, is 15.8 m/s.

Explanation:

To calculate the velocity of the boat, we can use vector addition. The boat's velocity relative to the water is perpendicular to the river's velocity. We can use the Pythagorean theorem to find the magnitude of the boat's velocity:

Vboat = sqrt((Vriver)2 + (Vboat)2)

Substituting the given values, we get:

Vboat = sqrt((5.0 m/s)2 + (15.0 m/s)2) = 15.8 m/s

Therefore, the velocity of the boat immediately after is 15.8 m/s.

Answer:

0.2s

Explanation:

SO for the dolphin to hear its echo, the sound wave must travel a distance twice as much as the displacement between the dolphin and the ocean floor. So d = 154 * 2 = 308 m

Since the speed of sound in ocean floor is v = 1533m/s we can find out the time by dividing the distance d by the speed of sound

t = d / v = 308 / 1533 = 0.2s

[tex](t = d / v )[/tex]The time passes before it hears an echo is 0.3secs

A sound wave  servers patterns of disturbance caused by the movement of energy traveling through a medium.

The speed of sound in ocean floor was given as [tex]( v = 1533m/s)[/tex]

To find the time, we can make use if the formula

[tex](t = d / v )[/tex]

Where t= time

v= velocity

d= distance

Then substitute ,we have

[tex]= 308 / 1533 = 0.2s[/tex]

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Among the following, may you do to a vector without changing is to move the vector without changing its orientation.

A vector is a quantity with both magnitude and direction in physics. It is often depicted by an arrow with the same direction as the amount and a length proportionate to the size of the quantity. A vector lacks position, but has magnitude and direction. A vector is therefore unaffected by displacement parallel to itself as far as its length is unaltered.

Scalars are regular variables with a magnitude but no direction, in contrast to vectors. In contrast to speed (the amount of speed), time, or mass, which are scalar values, displacement, speed, and acceleration are all vector quantities.

Therefore, it concludes that option B is correct.

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Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 [tex]m^{3}[/tex]

height through which it falls, h is 150 m

mass of 1 [tex]m^{3}[/tex] of water is 1000 kg

⇒mass of 1250 [tex]m^{3}[/tex] of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 [tex]\frac{m}{sec^{2} }[/tex]

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

        = 1250000×9.8×150 J

        = 1837500000 J

Electrical Energy = [tex]\frac{3}{4}[/tex](KE)

                            = [tex]\frac{3}{4}[/tex]×1837500000

                            = 1378125000 J per second

Answer:

E. The wheel with spokes has about twice the KE.

See explanation in: https://quizlet.com/100717504/physics-8-mc-flash-cards/

Answer:

Explanation:

We have to consider how the location of the mass affects the moment of inertia.

For a solid cylinder, I = mR²

For a hollow cylinder, I = 1/2mR²

Where

I ist the moment of inertia,

m is their masses,

R is the radius of rotation.

Since they have the same mass and radius, it can be seen that a hollow cylinder has twice the moment of inertia as a solid cylinder of the same mass and radius.

We know that the rotational kinetic energy is proportional to the moment of inertia. From;

Rotational KE = 1/2IW²

Where W is the angular speed.

so that at the same angular speed, the wheel with the spokes will have about double the kinetic energy as the solid cylinder. Take note that some of the mass is in the spokes so the moment of inertia is not exactly double.

Answer:

6862.96871 seconds

Explanation:

M = Mass of Planet

G = Gravitational constant

r = Radius

[tex]\rho[/tex] = Density

T = Rotation period

In this system the gravitational force will balance the centripetal force

[tex]G\frac{Mm}{r^2}=mr\omega^2[/tex]

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

[tex]M=\rho v\\\Rightarrow M=\rho \frac{4}{3}\pi r^3[/tex]

[tex]\\\Rightarrow G\frac{Mm}{r^2}=mr\left(\frac{2\pi}{T}\right)^2\\\Rightarrow \frac{G\rho \frac{4}{3}\pi r^3}{r^3}=\frac{4\pi^2}{T^2}\\\Rightarrow T=\sqrt{\frac{3\pi}{G\rho}}[/tex]

Hence, proved

[tex]T=\sqrt{\frac{3\pi}{6.67\times 10^{-11}\times 3000}}\\\Rightarrow T=6862.96871\ s[/tex]

The rotation period of the astronomical object is 6862.96871 seconds

Answer:

         [tex] v_s =7.74\ m/s[/tex]

Explanation:

given,

Speed of sound = 343 m/s

frequency of horn = 260 Hz

the friend is approaching, the frequency is increased by the Doppler Effect. The frequency is 266 Hz

using formula

         [tex]f' = \dfrac{v}{v-v_s}f_0[/tex]

         [tex]266= \dfrac{343}{343 - v_s}(260)[/tex]

         [tex]1.023= \dfrac{343}{343 - v_s}[/tex]

         [tex]343 - v_s = 335.26[/tex]

         [tex] v_s =7.74\ m/s[/tex]

the speed of friends approaching is equal to [tex] v_s =7.74\ m/s[/tex]

Answer:

[tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]

Explanation:

first write the newtons second law:

F[tex]_{s}[/tex]=δma[tex]_{s}[/tex]

Applying bernoulli,s equation as follows:

∑[tex]δp+\frac{1}{2} ρδV^{2} +δγz=0\\[/tex]

Where, [tex]δp[/tex] is the pressure change across the streamline and [tex]V[/tex] is the fluid particle velocity

substitute [tex]ρg[/tex] for {tex]γ[/tex] and [tex]g_{0}-cz[/tex] for [tex]g[/tex]

[tex]dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0[/tex]

integrating the above equation using limits 1 and 2.

[tex]\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]

there the bernoulli equation for this flow is [tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]

note: [tex]ρ[/tex]=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

Answer:

The wavelength of the station’s signal is 2.9 meters

The energy of a photon of the station’s electromagnetic signal is [tex]6.9\times10^{-26}J [/tex]

Explanation:

Wavelength [tex] \lambda [/tex] is inversely proprtional to frequency (f) and directly proportional to velocity of the wave (v).

[tex]\lambda=\frac{v}{f} [/tex] (1)

But electromagnetic waves as radio signals travel at speed of light so using this on (1):

[tex]\lambda=\frac{c}{f}=\frac{3.0\times10^{8}}{104.5\times10^{6}}\approx2.9\,m [/tex]

Albert Einstein discovered that energy of electromagnetic waves was quantized in small discrete packages of energy called photons with energy:

[tex] E=hf=(6.6\times10^{-34})(104.5\times10^{6})\approx6.9\times10^{-26}J[/tex]

with h the Planck's constant.

The wavelength of a 104.5 MHz FM radio signal is approximately 2.87 meters, and the energy of a photon of this radio signal is approximately 6.92 * 10^-26 Joules.

The subject of this question is the relationship between the frequency, wavelength, and energy of radio waves, specifically those used in FM radio broadcasting.

To calculate the wavelength of the radio signal, one can employ the wave equation: velocity = frequency * wavelength. Since the velocity of electromagnetic waves, which include radio waves, is the speed of light (3 * 10^8 m/s), the wavelength can be obtained by rearranging the equation to: wavelength = velocity / frequency. Using your FM station's frequency of 104.5 MHz (or 104.5 * 10^6 Hz), the wavelength of the station's signal would be approximately 2.87 meters.

The energy of a photon from this radio signal could be found through the photon energy equation: E = h * f, where h is Planck's constant (6.62607004 × 10^-34 m^2 kg / s) and f is the frequency in Hz. Thus, the energy of a radio signal photon at 104.5 MHz would be approximately 6.92 * 10^-26 J.

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Answer:

(1) Elliptical galaxies

(2) Spiral galaxies

(3) Irregular galaxies

(4) S0 galaxies

Explanation:

(1) Elliptical galaxies

Elliptical galaxies

These systems exhibit certain characteristic properties. They have complete rotational symmetry; i.e., they have figures of revolution with two equal principal axes. Third smaller axis is presumed axis of rotation. The surface brightness of elliptical at optical wavelengths decreases monotonically outward from a maximum value at the centre, following a common mathematical law of the form:

I = I0( r/a +1 )−2,

where I is the intensity of the light, I0 is the central intensity, r is the radius, and a is a scale factor.

(2) Spiral galaxies

Spiral galaxies are classified into two groups; ordinary and barred. The ordinary group is designated by S or SA, and the barred group by SB. In normal spirals, the arms originate directly from the nucleus, or bulge, where in the barred spirals, there is a bar of material that runs through the nucleus that the arms emerge from. Both types are given a classification according to how tightly their arms are wound. The classifications are a, b, c, d ... with "a" having the tightest arms. In type "a", the arms are usually not well defined and form almost a circular pattern. Sometimes you will see the classification of a galaxy with two lower case letters. This means that the tightness of the spiral structure is halfway between those two letters.

(3) Irregular galaxies:

Irregular galaxies have no regular or symmetrical structure. They are divided into two groups, Irr I and IrrII. Irr I type galaxies have HII regions, which are regions of elemental hydrogen gas, and many Population I stars, which are young hot stars. Irr II galaxies simply seem to have large amounts of dust that block most of the light from the stars. All this dust makes is almost impossible to see distinct stars in the galaxy.

(4) S0 galaxies

These systems exhibit some of the properties of both the elliptical and the spirals and seem to be a bridge between these two most common galaxy types. Hubble introduced the S0 class long after his original classification scheme had been universally adopted, largely because he noticed the dearth of highly flattened objects that otherwise had the properties of elliptical galaxies.

Answer:

[tex]g=0.20\ m/s^2[/tex]    

Explanation:

It is given that,

Mass of the asteroid Ceres, [tex]m=6.797\times 10^{20}\ kg[/tex]

Radius of the asteroid, [tex]r=472.9\ km=472.9\times 10^3\ m[/tex]

The value of universal gravitational constant, [tex]G=6.67259\times 10^{-11}\ N.m^2/kg^2[/tex]

We know that the expression for the acceleration due to gravity is given by :

[tex]g=\dfrac{Gm}{r^2}[/tex]

[tex]g=\dfrac{6.67259\times 10^{-11}\times 6.797\times 10^{20}}{(472.9\times 10^3)^2}[/tex]

[tex]g=0.20\ m/s^2[/tex]

So, the value of acceleration due to gravity on that planet is [tex]0.20\ m/s^2[/tex]. Hence, this is the required solution.

Final answer:

To calculate the surface gravity of Ceres, apply Newton's law of universal gravitation and rearrange it to solve for 'g.' The result is approximately g = 0.28 m/s².

Explanation:

Calculating Gravitational Acceleration on Ceres

To find the acceleration due to gravity ‘g’ on the surface of Ceres, use Newton’s law of universal gravitation:

F = G * (m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant, m1, and m2 are the masses of the two objects (in this case, a mass on the surface of Ceres and Ceres itself), and r is the distance between the centers of the two masses (the radius of Ceres in this scenario).

Since we are interested in 'g,' we rearrange this formula to solve for F/m2 (where m2 is a mass on Ceres’ surface and F/m2 equals g):

g = G * m1 / r^2

Plugging in the given values:

G = 6.67259 × 10⁻¹¹ N·m²/kg²

m1 (mass of Ceres) = 6.797 × 10²° kg,

r (radius of Ceres) = 472.9 × 10³ m,

The calculation is:

g = (6.67259 × 10⁻¹¹ N·m²/kg² * 6.797 × 10²° kg) / (472.9 × 10³ m)^2

After performing the calculation, ‘g’ on the surface of Ceres is found to be approximately 0.28 m/s²

Answer:

1.6 miles/min..

Explanation:

Let y be the height of the balloon at time t. Our goal is to compute the balloon's velocity at the moment .

balloon's velocity dy/dt when θ =π/3 radian

so we can restate the problem as follows:

Given dθ/dt = 0.1 rad/min at θ = π/3

from the figure in the attachment

tanθ = y/5

Differentiating w.r.t "t"

sec^2 θ×dθ/dt = 1/4(dy/dt)

⇒ dy/dt = (4/cos^2 θ)dθ/dt

At the given moment θ =π/3 and dθ/dt = 0.1 rad/min.

therefore putting the value we get

[tex]\frac{dy}{dt} = \frac{4}{\frac{1}{2}^2 }\times0.1[/tex]

solving we get

= 1.6 miles/min

So the balloon's velocity at this moment is 1.6 miles/min.

Answer:

2.92 * 10³ rad/s

Explanation:

Given:

Initial Radius of Original Star (Ri) = 6.0 * 10^5 km

Final Radius of Neutron Star (Rf) = 16km

Angular Speed = 1 revolution in 35 days

We need to convert this to rad/s

To do that, we first convert to rad/day

i.e (1 rev/35 days) * (2π rad/ 1 rev)

We then convert the days to hour

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours)

Finally, we convert the hour to seconds (3600 seconds makes an hour)

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours) * (1 hour/ 3600 sec)

Angular Speed = 2π rad/ 3024000 secs

Angular Speed (wi) = 2.079 * 10^-6rad/s

From the question, we're asked to calculate the angular speed of the neutron star (wf)

Applying law of conservation of angular momentum to a system whose moment of Inertia changes, we have

Ii*wi = If * wf ----------------- Formula

Where Ii and If are the initial and final Inertia of the star

The relationship between Inertia and Radius of each object is I = 2/5MR²

So, Ii = 2/5(MRi²) and If = 2/5(MRf²)

Substitute the above in the formula quoted

We have 2/5(MRi²)wi = 2/5(MRf²)wf ---------------- Divide through by 2M/5

We are left with, Ri²wi = Rf²wf

Make wf the subject of the formula

wf = wi * (Ri/Rf)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km) * (6.0 * 10^5 km/16km)

wf = 2.92 * 10³ rad/s

Answer:

ground state

Explanation:

  Lets take  

n=3 ,n=2 ,n=1 are the energy level.

Energy level n=1 is the ground energy level.

The energy from 3 to 1 = hν

The energy from 3 to 2 = hν₁

The energy from 2 to 1 = hν₂

We can say that

hν = hν₁ +  hν₂

If the electron were de-excitation from the third level to ground level then the sum of emitted frequency will be equal to the frequency of a single electron.

Therefore the answer is ground state.

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

[tex]P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m[/tex]

The spring constant of the net is 20130.76 N

From Hooke's Law

[tex]F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m[/tex]

The net would strech 0.03167 m

If h = 35 m

From energy conservation

[tex]65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0[/tex]

Solving the above equation we get

[tex]x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45[/tex]

The compression of the net is 1.52 m

29.62 feet far, to the nearest tenth of a​ foot, does the balloon rise during this​ period

Explanation:

Hot-air balloon is rising vertically, the angle of elevation from a point on level ground = 122 feet (from the balloon)

The passenger compartment changes from = 17.9 degrees to 29.5 degree

              [tex]\tan \left(17.9^{\circ}\right)=\frac{a}{122 \mathrm{ft}}[/tex]

              [tex]a=122 \times \tan 17.9^{\circ}=122 \times 0.32299=39.4[/tex]

Similarly,

              [tex]\tan \left(29.5^{\circ}\right)=\frac{b}{122 f t}[/tex]

              [tex]b=122 \times \tan \left(29.5^{\circ}\right)=122 \times 0.56577=69.02[/tex]

The nearest tenth of a foot, the balloon rise during the period, as below

               69.02 – 39.4 = 29.62 ft.

Final answer:

To calculate the moment of inertia and angular velocity of the rod-ball system after the collision, use the conservation of angular momentum, considering both the inertia of the rod and the putty ball. The total moment of inertia is the sum of each component's moment of inertia, and the angular velocity is determined by the ratio of initial angular momentum to the total moment of inertia.

Explanation:

When a putty ball strikes and sticks to a rotating rod in a collision with rotating rod, the conservation of angular momentum applies. To solve for the moment of inertia (I) and angular velocity (ω), we need to consider both the moment of inertia of the rod and the additional inertia from the putty ball at the point it sticks to the rod.

For the rod and ball system:

The angular speed ω can be found by equating the initial angular momentum of the ball before the collision to the final angular momentum of the system after the collision:

The rotational kinetic energy of the system after collision can be calculated using the expression K_{rot} = \frac{1}{2}Iω^2.

Answer:

1.1

Explanation:

Cosmic rays are particles traveling at extreme speeds through intergalactic space. Many are launched by exploding supernovae. From the question, we know that slowest cosmic rays travel at 43% of the speed of light that is 43/100 multiply by the speed of light(speed of light= 3.0×10^8 m/s).

Gamma factor can be calculated using the formula below;

Gamma factor= 1/(1- v^2/c^2)^1/2 --------------------------------------------(1).

Gamma factor= 1/(1-3.0×10^8 metre per seconds ×43/100÷ 3.0×10^8 metre per seconds) ^1/2

Gamma factor= 1/(1-1.6641×10^16/3.0×10^2)^1/2

Gamma factor= 1/(1-1.16641×10^16/9×10^16)^1/2

Gamma factor= 1/(1-0.1296)^1/2

Gamma factor= 1/(0.87)^1/2

Gamma factor= 1/0.9327

Gamma factor= 1.072112535

Gamma factor= ~ 1.1

Cosmic rays are high-speed particles that travel through space close to the speed of light. The gamma factor for cosmic rays traveling at 43% of the speed of light is  1.17.

Cosmic rays are high-speed particles that travel through interstellar space at speeds close to the speed of light. When calculating the gamma factor for cosmic rays traveling at 43% of the speed of light, we can use the formula: γ = 1 / √(1 - v^2/c^2), where v is the velocity of the particles and c is the speed of light. Substituting the values, the gamma factor for cosmic rays traveling at 43% of the speed of light is approximately 1.17.

Answer:

The height of the bag will be 0.133 m

Explanation:

We have given gauge pressure [tex]P=1.33\times 10^3Pa=1330Pa[/tex]

Density of solution [tex]\rho =1.02\times 10^3=1020kg/m^3[/tex]

We have to find the height of the bag

We know that gauge pressure is given by P=\rho gh

[tex]1330=1020\times 9.8\times h[/tex]

h=0.133m

So the height of the bag will be 0.133 m

Final answer:

The minimum height of the bag should be 0.245 meters to infuse glucose into the vein.

Explanation:

In order for the fluid to enter the vein, its pressure at entry must exceed the blood pressure in the vein. To find the height of the fluid, we need to convert the blood pressure in mm Hg to SI units. Since 1.0 mm Hg = 133 Pa, the blood pressure in the vein is 18 mm Hg above atmospheric pressure, which is equivalent to (18 mm Hg)(133 Pa/mm Hg) = 2386 Pa.

Now we can calculate the height of the fluid using the formula:

h = P/(ρg)

Where:

Substituting the given values into the formula, we find:

h = (2386 Pa)/((1.02 x 10^3 kg/m^3)(9.8 m/s^2)) = 0.245 m

The minimum height of the bag should be 0.245 meters in order to infuse glucose into the vein.