Tarzan swings on a 30.0-m-long vine initially inclined at an angle of 37 degrees measured from vertical. What is his speed at the bottom of the swings if he starts from rest?

Answer:

the distance between interference fringes increases

Explanation:

For double-slit interference, the distance of the m-order maximum from the centre of the distant screen is

[tex]y=\frac{m \lambda D}{d}[/tex]

where [tex]\lambda[/tex] is the wavelength, D is the distance of the screen, and d the distance between the slits. The distance between two consecutive fringes (m and m+1) will be therefore

[tex]\Delta y = \frac{(m+1) \lambda D}{d}-\frac{m \lambda D}{d}=\frac{\lambda D}{d}[/tex]

and we see that it inversely proportional to the distance between the slits, d. Therefore, when the separation between the slits decreases, the distance between the interference fringes increases.

Answer:

+e

Explanation:

The charge of an up quark is:

[tex]q_u = +\frac{2}{3}e[/tex]

The charge of a down quark, instead, is:

[tex]q_d = -\frac{1}{3}e[/tex]

In this problem, we have a particle consisting of 1 down quark and 2 up quarks, Therefore, the net charge of the particle will be

[tex]q=1(q_d)+2(q_u)=1(-\frac{1}{3}e)+2(\frac{2}{3}e)=-\frac{1}{3}e+\frac{4}{3}e=\frac{3}{3}e=+e[/tex]

So, the particle has a charge of +e.

Answer:

chlorine gas

Explanation:

Answer:

The gas is chlorine gas.

Explanation:

[tex]NaClO + 2 HCl \rightarrow Cl_2 + H_2O + NaCl[/tex]

Chlorine gas is a toxic gas and very reactive inside the human body. Also an irritant to eyes and skin. Exposure to high concentration leads to lung damage or death.

Where as the byproduct formed are water and sodium chloride which an aqueous solution of sodium chloride harmless to humans.

 = 3.73 seconds

Using the equation;

v = u + at

where v is the final velocity, u is the initial velocity and a is the acceleration due to gravity (-g) and t is the time taken.

Therefore;

v = u - gt

Thus;

v = 24.5 m/s, u = 12.1 m/s, a = -g = 9.8 m/s²

-24.5 m/s = 12.1 - 9.8 × t

36.6 = 9.8 t

t = 36.6/9.8

 = 3.73 seconds

the plane was flying at 2,691. 33 m at 5.32°

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

[tex]E=U=\frac{1}{2}kA^2[/tex] (1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

[tex]E=K=\frac{1}{2}mv_{max}^2[/tex] (2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

[tex]\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{(34.9 N/m)(0.05 m)^2}{0.42 kg}}=0.456 m/s[/tex]

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

[tex]E=\frac{1}{2}kx^2+\frac{1}{2}mv^2[/tex] (3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(34.9 N/m)(0.05 m)^2=0.044 J[/tex]

So, solving (3) for v, we find the speed when x=1.5 cm:

[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s[/tex]

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s[/tex]

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

[tex]v=\frac{v_{max}}{2}=\frac{0.456 m/s}{2}=0.228 m/s[/tex]

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

[tex]E=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\x=\sqrt{\frac{2E-mv^2}{k}}=\sqrt{\frac{2(0.044 J)-(0.42 kg)(0.228 m/s)^2}{34.9 N/m}}=0.044 m=4.4 cm[/tex]

The correct answer for part (d) is: [tex]\[ x = \pm\frac{A}{\sqrt{2}} \][/tex] where [tex]\( A \)[/tex] is the amplitude of the motion.

To determine the value of [tex]\( x \)[/tex] at which the speed of the oscillating mass is equal to one-half the maximum value, we first need to establish the relationship between the speed of the mass and its position in the oscillation.

The total mechanical energy [tex]\( E \)[/tex] of a mass-spring system in simple harmonic motion (SHM) is conserved and is given by the sum of its potential energy [tex]\( U \)[/tex] and kinetic energy [tex]\( K \)[/tex]. At the maximum displacement (amplitude [tex]\( A \)[/tex]), all the energy is potential, and the kinetic energy is zero. Conversely, at the equilibrium position, all the energy is kinetic, and the potential energy is zero.

The maximum potential energy [tex]\( U_{\text{max}} \)[/tex] occurs at the amplitude [tex]\( A \)[/tex] and is given by:

[tex]\[ U_{\text{max}} = \frac{1}{2}kA^2 \][/tex]

where [tex]\( k \)[/tex] is the spring constant.

The maximum kinetic energy [tex]\( K_{\text{max}} \)[/tex] occurs at the equilibrium position and is equal to the total mechanical energy [tex]\( E \)[/tex]:

[tex]\[ K_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2 = E \][/tex]

Since energy is conserved:

[tex]\[ E = U_{\text{max}} + K_{\text{max}} = \frac{1}{2}kA^2 + \frac{1}{2}mv_{\text{max}}^2 \][/tex]

At any point in the oscillation, the total mechanical energy is:

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

We are given that the speed [tex]\( v \)[/tex] is half the maximum speed [tex]\( v_{\text{max}} \)[/tex], so we can write:

[tex]\[ \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{v_{\text{max}}}{2}\right)^2 \][/tex]

Substituting [tex]\( v_{\text{max}} \)[/tex] from the energy conservation equation:

[tex]\[ \frac{1}{2}m\left(\frac{v_{\text{max}}}{2}\right)^2 = \frac{1}{2}m\left(\frac{\sqrt{\frac{2E}{m}}}{2}\right)^2 = \frac{1}{2}m\left(\frac{1}{2}\sqrt{\frac{2E}{m}}\right)^2 = \frac{1}{2}m\left(\frac{1}{2}\right)^2\frac{2E}{m} = \frac{1}{2}E \][/tex]

Now, we can equate the total mechanical energy at this point to the potential energy, since the kinetic energy is half the total energy:

[tex]\[ \frac{1}{2}kx^2 = \frac{1}{2}E \][/tex]

Substituting [tex]\( E \)[/tex] from the energy conservation equation:

[tex]\[ \frac{1}{2}kx^2 = \frac{1}{2}\left(\frac{1}{2}kA^2\right) \][/tex]

Solving for [tex]\( x \)[/tex]:

[tex]\[ kx^2 = \frac{1}{2}kA^2 \][/tex]

[tex]\[ x^2 = \frac{1}{2}A^2 \][/tex]

[tex]\[ x = \pm\sqrt{\frac{1}{2}A^2} \][/tex]

[tex]\[ x = \pm\frac{A}{\sqrt{2}} \][/tex]

Therefore, the value of [tex]\( x \)[/tex] at which the speed of the oscillating mass is equal to one-half the maximum value is [tex]\( \pm\frac{A}{\sqrt{2}} \)[/tex]. Since the amplitude [tex]\( A \)[/tex] is given as 5.0 cm (0.05 m), the specific values of [tex]\( x \)[/tex] are:

[tex]\[ x = \pm\frac{0.05 \text{ m}}{\sqrt{2}} \approx \pm 0.0354 \text{ m} \][/tex]

The only way of telling about dark energy is our observation of how the universe has been expanding. It basically works the opposite as gravity, pushing things away from it. Thus, the closest answer would be D. The shape of galaxies in cluster galaxies.

Answer:

D. The shape of galaxies in cluster galaxies

Explanation:

When the Universe began to expand, everything present in Universe began to move at a very high rate. This was concluded that some external force is working on it in order to pull these things. It was known  as the dark energy.

Thus the size and shape of galaxies is increasing because of this  dark energy, hence the study of change in shape and size of galaxy gives information about dark energy as it is constantly applying  force thus increasing the size.

Electric energy??? there isn't much information from the question but I can infer it's electricity.

Photovoltaic is the term that describes the process of transforming light energy into electrical energy. This often occurs in solar cells which absorb sunlight and produce electricity. Another instance of light energy conversion is in photosynthesis, where plants convert sunlight into chemical energy.

In the context of the question, the term you're looking for is Photovoltaic. This is a form of energy transformation where light energy is converted into electrical energy. This often happens in devices known as solar cells that are designed to capture sunlight and generate electricity from it.

Consider the example of a solar cell. When sunlight (radiant energy) hits the surface of the cell, the materials within the cell absorb the light and transform it into electrical energy. This process is often used to power electrical devices or store energy for later use.

Another instance is in photosynthesis performed by plants. Plants convert light energy (from the sun) into chemical energy that they can use to grow and reproduce. The energy transformation in this case is from light to chemical.

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The higher the frequency, the more energy the photon has. Of course, a beam of light has many photons. This means that really intense red light (lots of photons, with slightly lower energy)

The difference between visible light and gamma rays is their frequency, energy and wavelength. All electromagnetic waves travel at the same speed, which is called the speed of light. ... Visible light has much lower energy and longer wavelengths than gamma radiation.

Doubles and Remains the same

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

(a) 700 Hz

For standing waves on a string, the number of antinodes (n) corresponds to the order of the harmonic. So, three antinodes corresponds to the third harmonic. Also, the frequency of the nth-harmonic is the nth-integer multiple of the fundamental frequency, so we have:

[tex]f_3 = 3 f_1 = 420 Hz[/tex]

where [tex]f_1[/tex] is the fundamental frequency. Solving for f1, we find

[tex]f_1 = \frac{420 Hz}{3}=140 Hz[/tex]

And so now we can find the frequency of the 5th-harmonic:

[tex]f_5 = 5 f_1 = 5 (140 Hz)=700 Hz[/tex]

(b) 56.4 N

The fundamental frequency of a string is given by:

[tex]f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}[/tex]

where we have:

L = 60 cm = 0.60 m is the length of the string

[tex]\mu = 2.0 g/m = 0.002 kg/m[/tex] is the linear density

T = ? is the tension in the string

Solving the formula for T and using the fundamental frequency, f1=140 Hz, we find

[tex]T=\mu (2Lf_1)^2=(0.002 kg/m)(2(0.60 m)(140 Hz))^2=56.4 N[/tex]

The frequency of the fifth harmonic of the string is 700 Hz, and the tension in the string is 56.448 N.

The question pertains to harmonic frequencies and tension in instruments, particularly a guitar string. Using the information given, we will apply the principles of standing waves on the string, wave speed derived from linear mass density and tension, and harmonic frequencies.

(a) If you're looking for the frequency of the fifth harmonic, note that the frequency of a particular harmonic is that harmonic number times the fundamental frequency. In this case, the string shows a standing wave with three antinodes, which refers to the third harmonic. So, the third harmonic frequency is 420 Hz; thus, the fundamental frequency is 420 Hz / 3 = 140 Hz. So, the frequency of the fifth harmonic is 5 times the fundamental frequency = 5*140 Hz = 700 Hz.

(b) To determine the tension in the string, you must first calculate the wave speed. The wave speed (Vw) can be derived from the formula for the frequency of a string, Vw=2Lf, where L is the length of the string in meters and f is the fundamental frequency. Here, L=0.6 m and f= 140 Hz, hence Vw = 2*(0.6m)*(140Hz) = 168 m/s. Given that wave speed is also calculated via the square root of tension (T) divided by linear mass density (μ), you can rearrange this formula to solve for T. This gives you T = (Vw)^2 * μ . The linear density is given as 2.0 g/m, but we need to convert it to kg/m to match the wave speed unit, so μ = 0.002 kg/m. Plugging in the values we find the tension: T = (168 m/s)^2 * 0.002 kg/m = 56.448 N.

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1. 236 kJ

a. The phase (or state of matter) of the substance: from solid state to gas state (sublimation)

b. The enthalphy of sublimation, given by: [tex]\lambda=571 J/g[/tex]

c. The equation to use will be [tex]Q=m\lambda[/tex], where m is the mass of dry ice and [tex]\lambda[/tex] is the enthalpy of sublimation

d. The energy is being absorbed, because the heat is transferred from the environment to the dry ice: as a consequence, the bonds between the molecules of dry ice break and then move faster and faster, and so the substance turns from solid into gas directly.

e. The amount of energy being transferred is

[tex]Q=m\lambda=(412.9 g)(571 J/g)=2.36\cdot 10^5 J=236 kJ[/tex]

2.  165 kJ

a. The phase (or state of matter) of the substance: from gas state to liquid state (condensation)

b. The latent heat of vaporisation of water, given by [tex]\lambda=2260 J/g[/tex]

c. The equation to use will be [tex]Q=m\lambda[/tex], where m is the mass of steam that condenses and [tex]\lambda[/tex] is the latent heat of vaporisation

d. The energy is being released, since the substance turns from a gas state (where molecules move faster) into liquid state (where molecules move slower), so the internal energy of the substance has decreased, therefore heat has been released

e. The amount of energy being transferred is

[tex]Q=m\lambda=(72.9 g)(2260 J/g)=1.65\cdot 10^5 J=165 kJ[/tex]

3. 3.64 kJ

a. Only the temperature of the substance (which is increasing)

b. The specific heat capacity of silver, which is [tex]C_s = 0.240 J/gC[/tex]

c. The equation to use will be [tex]Q=m C_s \Delta T[/tex], where m is the mass of silver, Cs is the specific heat capacity and [tex]\Delta T[/tex] the increase in temperature

d. The energy is being absorbed by the silver, since its temperature increases, this means that its molecules move faster so energy should be provided to the silver by the surroundings

e. The amount of energy being transferred is

[tex]Q=m C_s \Delta T=(39.2 g)(0.240 J/gC)(412.9^{\circ}C-25.9^{\circ}C)=3641=3.64 kJ[/tex]

4. 89 kJ

a. Both the phase of the substance (from solid to liquid) and then the temperature

b. The latent heat of fusion of ice: [tex]\lambda=334 J/g[/tex] and the specific heat capacity of water: [tex]C_s=4.186 J/gC[/tex]

c. The equation to use will be [tex]Q=m\lambda + m C_s \Delta T[/tex], where m is the mass of ice, [tex]\lambda[/tex] the latent heat of fusion of ice, Cs is the specific heat capacity of water and [tex]\Delta T[/tex] the increase in temperature

d. The energy is being absorbed by the ice, at first to break the bonds between the molecules of ice and to cause the melting of ice, and then to increase the temperature of the water

e. The amount of energy being transferred is

[tex]Q=m\lambda +m C_s \Delta T=(156.3 g)(334 J/g)+(156.3 g)(4.186 J/gC)(56.232^{\circ}C-0^{\circ}C)=8.9\cdot 10^4 J=89 kJ[/tex]

D - 1,960 J, 3,720 J

At a height of 50 m from the ground, the potential energy will be;

= 4 × 9.8 × 50

= 1960 Joules

This means that some of the energy possessed by the stone at a height of 145 m was converted to kinetic energy.

Therefore; the energy that was converted to kinetic energy will be;

= 5,680 J - 1960 J

= 3,724 Joules

Approximately kinetic energy is 3,720 Joules

Therefore;

The Potential energy is 1960 Joules and Kinetic energy is 3,720 Joules

(a) 138.2 J

Since the applied force is parallel to the displacement of the block, the work done by the force is given by:

[tex]W=Fd[/tex]

where

F = 46.7 N is the magnitude of the force

d = 2.96 m is the displacement of the block

Substituting the numbers into the equation, we find

[tex]W=(46.7 N)(2.96 m)=138.2 J[/tex]

(b) 45.1 J

In order to calculate the total energy dissipated among the floor and the block as thermal energy, we have to calculate the work done by the frictional force, which is

[tex]W_f = F_f d = (-\mu mg)d[/tex]

where[tex]\mu=0.635[/tex] is the coefficient of friction

m = 4.35 kg is the mass of the block

g = 9.8 m/s^2

d = 2.96 m is the displacement

and the negative sign is due to the fact that the frictional force has opposite direction to the displacement.

Substituting, we find

[tex]W_f =-(0.635)(4.35 kg)(9.8 m/s^2)(2.96 m)=-80.1 J[/tex]

The magnitude of this work is equal to the sum of the thermal energy dissipated between the floor and the block:

[tex]W_f = E_{floor}+E_{block}[/tex]

And since we know

[tex]W_f = 80.1 J\\E_{floor}=35.0 J[/tex]

we find

[tex]E_{floor}=W_f-E_{block}=80.1 J-35.0 J=45.1 J[/tex]

(c) 58.1 J

According to the work-energy theorem, the increase in kinetic energy of the block must be equal to the work done by the applied force minus the work done by friction (which becomes wasted thermal energy):

[tex]\Delta K=W-W_f[/tex]

Substituting

[tex]W=138.2 J\\W_f = 80.1 J[/tex]

We find

[tex]\Delta K=138.2 J-80.1 J=58.1 J[/tex]

B.A heat engine that uses work to move heat

I believe A. I'm so sorry if it's wrong! but it's the most best answer to me since it makes sense to me.

Speed = (distance covered) / (time to cover the distance)

Speed = (60 km) / (3.5 hr)

Speed = (60 / 3.5) (km/hr)

Speed = 17.14 km/hr

Answer:

C. red main-sequence star

Explanation:

Which of these stars has the hottest core?A)a blue main-sequence star. B)a red supergiant. C)a red main-sequence star.

Stars are basically exploding gases usually hydrogen and helium gases, held together by gravity. The closest to us is the sun. which contains gases

a blue main-sequence star temperature is between 10,000 Kelvin to 40000kelvin

red main-sequence star. can about 10million kelvin

while

B)a red supergiant is 3500-4500 K

Blue stars are generally the hottest, so a blue main-sequence star likely has the hottest core. The star's color provides insight into its core temperature.

The temperature of a star is typically indicated by its color, which can be used to infer information about its core. In general, blue stars are the hottest and red stars are the coolest. Therefore, of the options given, a blue main-sequence star likely has the hottest core. It's interesting to note that a star's core temperature can alter its life cycle stages, hence why we classify stars into different categories like main-sequence, supergiant and so on.

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Answer:

[tex]8.1\cdot 10^{-4} C^{-1}[/tex]

Explanation:

The volumetric expansion of the liquid is given by

[tex]\Delta V=\alpha V_0 \Delta T[/tex]

where

[tex]\alpha[/tex] is the coefficient of volume expansion

[tex]V_0[/tex] is the initial volume

[tex]\Delta T[/tex] is the change in temperature

For the liquid in this problem,

[tex]V_0 = 2.35 m^3\\\Delta T=48.5^{\circ}C\\\Delta V=0.0920 m^3[/tex]

So we can solve the equation to find [tex]\alpha[/tex]:

[tex]\alpha=\frac{\Delta V}{V_0 \Delta T}=\frac{(0.0920 m^3)}{(2.35 m^3)(48.5^{\circ}C)}=8.1\cdot 10^{-4} C^{-1}[/tex]

Answer:

5.33 cm

Explanation:

The lens equation states that:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem,

p = 8 cm

q = 16 cm ( the sign is positive since the image is real, which means it is formed on the other side of the lens)

Substituting into the equation,

[tex]\frac{1}{f}=\frac{1}{8 cm}+\frac{1}{16 cm}=\frac{3}{16 cm}[/tex]

[tex]f=\frac{16}{3}cm=5.33 cm[/tex]

Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

2.063d = 10.63  

d = 5.15m  

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

If you are using PLATO, which I'm sure you are cause I've had the same question, the answer it the following:

Transformers increase and decrease voltage as needed.

The alternating current more effective at long–distance travel than direct current because transformers increase and decrease voltage as needed.

Alternating current is an electric current which periodically reverses direction and changes its magnitude continuously with time.

The voltage difference developed through transmission lines is very high,. This reduces the current and minimizes the energy lost through transmission. So, alternating current is chosen over direct current for transmitting electricity.  It is much cheaper to change the voltage of an alternating current.

Thus, the alternating current more effective at long–distance travel than direct current because transformers increase and decrease voltage as needed

Learn more about alternating current.

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Less than

For the parallel circuit in the previous part (with the switch closed), the current through the 20-ω resistor is less than the current through 10-ω resistor.

According to Ohm's Law, the current through a conductor is directly proportional to the potential difference at constant pressure and temperature.

Therefore; V α I

Mathematically;  V = IR ; Where R is the resistance of a device or a conductor.

From the relationship; I = V/R ; which means current and resistance have inverse relationship such that an increase in resistance causes a decrease in electric current.

Therefore; the current through the 20-ω resistor is less than the current through 10-ω resistor.

In a parallel circuit, the current through a 20-ω resistor is half of the current through a 10-ω resistor. This is because in a parallel circuit, resistors with lower resistance will have more current flowing through them, which is a consequence of Ohm's Law.

For a parallel circuit with a closed switch, the current through the resistors is influenced by their resistances.

According to Ohm's law, the current is inversely proportional to the resistance. Therefore, for the resistors of 20-ω and 10-ω, the current through the 20-ω resistor is half the current through the 10-ω resistor.

It is important to understand that in a parallel circuit, the voltage across each component is the same, and the total current in the circuit is the sum of the currents through each component. This implies that lower resistance components will draw more current from the circuit. The reason for this is represented mathematically in Ohm's law, which states that I = V/R.

Thus, when we speak about the 20-ω and 10-ω resistors, since they are assumed to have the same voltage across them, the current in the 20-ω resistor would be V/20, and the current in the 10-ω resistor would be V/10. This essentially means the current through the 20-ω resistor is half of the current through the 10-ω resistor given the same voltage.

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After 9 years of spectacular findings, NASA shut down the Kepler space telescope at the end of last October (2018).

From 2009 to 2018, the Kepler mission found more than 2,600 planets in other "solar systems" ... in orbit around other stars besides our Sun.  AND, at the time the mission ended, (5 months before the night I'm writing this), there were still more than 2,000 sets of data still to be analyzed.

The results suggest that having a family of planets is really not an unusual thing for a star.  Indeed, the rare, unusual stars may be the few that don't have planets orbiting them.

Planets around other stars are probably so common that they are the rule, and not the exception to the rule.

Explanation:

The law of repulsion is given by Coulomb. The mathematical form of Coulomb law is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]...............(1)

Where

F is the force

k is the electrostatic constant

[tex]q_1\ and\ q_2[/tex] are electric charges

r is the distance between charges

The Newton's law of universal gravitation is given by :

[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]..............(2)

G is the universal gravitational constant

From equation (1) and (2) it is clear that both law obeys inverse square law and both are of same type. So, the law of repulsion by Coulomb agrees with the Newton's law of universal gravitation.

The Law of Coulomb agrees with the inverse-square law and Newton's law of universal gravitation. It doesn't directly correspond with Newton's laws of motion or the findings of Gilbert on magnetism.

The Law of Coulomb is closely related to and agrees with the inverse-square law, which states that a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical property. This rule is also employed in Newton's law of universal gravitation.

However, the Law of Coulomb does not directly correspond or agree with Newton's laws of motion or the findings of Gilbert. Newton's laws of motion govern the relationship between the motion of an object and the forces acting upon it, while Gilbert's findings primarily concern magnetism.

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Substitute your values into the formula:

W = Work done = 288

[tex]Q_{in}[/tex] = 360

Solve to find e:

e = 288 ÷ 360 = 0.8

Convert e to a percentage by multiplying by 100.

0.8 × 100 = 80

Final answer:

Observational characteristics that could indicate a white dwarf or neutron star include being surrounded by a planetary nebula, emitting most strongly in visible and ultraviolet, and being in a binary system that undergoes nova explosions.

Explanation:

The observational characteristics that could indicate that an object is either a white dwarf or a neutron star are:

May be surrounded by a planetary nebula: A white dwarf can be surrounded by a planetary nebula, which is a glowing shell of gas and dust expelled by the dying star.

Emits most strongly in visible and ultraviolet: Both white dwarfs and neutron stars emit most of their energy in the visible and ultraviolet parts of the electromagnetic spectrum.

May be in a binary system that undergoes nova explosions: Some white dwarfs can be in a binary system where the companion star transfers material to the white dwarf, causing periodic nova explosions.

Answer:

Yes

Explanation:

When a sound wave moves through the air, a point in the air undergoes alternative changes in density called compressions and rarefactions.

A sound wave is a longitudinal waves, which means that the vibrations of the particles in the medium occur in a direction parallel to the direction of motion of the wave. Longitudinal waves creates two different regions in the medium:

- Compressions: these are regions where the density of the particles of the medium (in this case, air particles) are higher

- Rarefactions: these are regions where the density of the particles of the medium (in this case, air particles) are lower