Kirchhoff's loop rule for circuit analysis is an expression of which of the following? Conservation of charge Conservation of energy Ampere's law Ohm's law Faraday's law

Yes sirree !  Uh huh, uh huh.  Truer words are rarely spoken.

This experiment shows that like charges repel while unlike charges attract. in the case of the gold leaf electroscope, the repulsion of the leaf is as a sign if the presence of similar charges in the electroscope. positive repels positive while negative repels negative.

These experiments show that like charges repel while unlike charges attract.

The gold leaf electroscope is an instrument that can be used to determine the type of charges carried by a body via induction. The charged body is brought near but does not touch the electroscope.

From electrostatics we know that like charges repel while unlike charges attract. This is demonstrated in these experiments.

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Answer:

200

Explanation:

Melting point of a substance is 200.

The temperature at which a solid and liquid phase can coexist in equilibrium and the point at which matter transforms from a solid to a liquid is known as a substance's melting point. Pure solutions and liquids fall within this definition.

It is important to specify melting point because it relies on pressure. Melting point tables frequently use standard pressures like 100 kPa or 1 atmosphere. The liquefaction point is another name for the melting point.

The temperature at which a liquid changes to a solid (the reverse of melting) is the freezing point or crystallization point. The freezing point and the melting point do not necessarily occur at the same temperature.

Therefore, Melting point of a substance is 200.

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All electromagnetic waves have the same speed when they're all in the same medium.

Note: Sound waves don't belong on this list, because they're not electromagnetic waves. The speed of sound in air is only about 0.0001 percent of the speed of electromagnetic waves in air.

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Answer:

10.6 meters.

Explanation:

We use the law of conservation of energy, which says that the total energy of the system must remain constant, namely:

[tex]\frac{1}{2}mv_i^2+mgh_i-1700j=\frac{1}{2}mv_f^2+mgh_f[/tex]

In words this means that the initial kinetic energy of the roller coaster plus its gravitational potential energy minus the energy lost due to friction (1700j) must equal to the final kinetic energy at top of the second hill.

Now let us put in the numerical values in the above equation.

[tex]m=100kg[/tex]

[tex]h_i=10m[/tex]

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

[tex]v_f=4,6m/s[/tex]

and solve for [tex]h_f[/tex]

[tex]h_f= \frac{\frac{1}{2}mv_i^2+mgh_i-1700j-\frac{1}{2}mv_f^2}{mg} =\boxed{ 10.6\:meters}[/tex]

Notice that this height is greater than the initial height the roller coaster started with because the initial kinetic energy it had.

1. 1.59 s

The period of a pendulum is given by:

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

where L is the length of the pendulum and g the gravitational acceleration.

In this problem,

L = 0.625 m

g = 9.81 m/s^2

Substituting into the equation, we find

[tex]T=2\pi \sqrt{\frac{(0.625 m)}{9.81 m/s^2}}=1.59 s[/tex]

2. 54,340 oscillations

The total number of seconds in a day is given by:

[tex]t=24 h \cdot 60 min/h \cdot 60 s/min =86,400 s[/tex]

So in order to find the number of oscillations of the pendulum in one day, we just need to divide the total number of seconds per day by the period of one oscillation:

[tex]N=\frac{t}{T}=\frac{86,400 s}{1.59 s}=54,340[/tex]

3. 0.842 m

We want to increase the period of the pendulum by 16%, so the new period must be

[tex]T'=T+0.16T=1.16 T = 1.16 (1.59 s)=1.84 s[/tex]

Now we can re-arrange the equation for the period of the pendulum, using T=1.84 s, to find the new length of the pendulum that is required to produce this value of the period:

[tex]L=g(\frac{T}{2\pi})^2=(9.81 m/s^2)(\frac{1.84 s}{2\pi})^2=0.842 m[/tex]

1. The problem statement, all variables and given/known data A parallel-plate capacitor of capacitance C with circular plates is charged by a constant current I. The radius a of the plates is much larger than the distance d between them, so fringing effects are negligible. Calculate B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. 2. Relevant equations When a capacitor is charged, the electric field E, and hence the electric flux Φ, between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell: ∮B⃗ ⋅dl⃗ =μ0(I+ϵ0dΦdt). You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current I is zero.

The electric field E, and consequently the electric flux, between the plates of a capacitor vary as it charges.

Electric flux is defined as the quantity of electric lines of force or electric field lines that cross a specific region is a property of an electric field.  According to this theory, electric field lines begin with positive electric charges and end with negative charges. An electric field's important characteristic is its electric flux. It can be thought of as the quantity of forces interacting in a particular space.

As a function of distance from the axis, the strength of the magnetic field B(r) inside the capacitor is

∫ Bdl = μ₀ I enclosed

B x 2λr =  μ₀ I (λ r² / λ a²)

Br = μ₀ In / 2λa²

Thus, the electric field E, and consequently the electric flux, between the plates of a capacitor vary as it charges.

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It is b because it will always triple

If the number of particles in a sample of gas is doubled, the volume of gas doubles because more number of molecules added to the system.

An increase in the number of gas particles in the container increases the volume of the gas because more space is occupied by the gas particles that enters to the system.

So we can conclude that if the number of particles in a sample of gas is doubled, the volume of gas doubles because more number of molecules added to the system.

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Answer:

The electric potential (voltage) [tex]V[/tex] produced by a point charge [tex]Q[/tex], at any point in space, is given by the following equation:

[tex]V=k\frac{Q}{r}[/tex]

Where:

[tex]k=8.99(10)^{9} Nm^{2}/C^{2}[/tex] is the Coulomb's constant

[tex]r[/tex] is the distance  

The result is a scalar quantity, is defined as the electric potential energy per unit of charge and determines the electric influence exerted by the charge on that point of space.

Answer:

v=k*q/d

Explanation:

The formula v=k*q/r is also correct but in this case v=k*q/d was given as a choice.

Answer:

-3 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum.

In fact, the system squid+water inside is initially at rest, so the total momentum is zero:

[tex]p_i=0[/tex]

while the final momentum is:

[tex]p_f = m_s v_s + m_w v_w[/tex]

where

[tex]m_s=1.5 kg[/tex] is the mass of the squid

[tex]v_s[/tex] is the velocity of the squid

[tex]m_w = 0.10 kg[/tex] is the mass of water

[tex]v_w = 45 m/s[/tex] is the water velocity

Since the total momentum must be conserved,

[tex]p_i = p_f = 0[/tex]

So we have

[tex]m_s v_s + m_w v_w =0\\v_s = -\frac{m_w v_w}{m_s}=-\frac{(0.10 kg)(45 m/s)}{1.5 kg}=-3 m/s[/tex]

where the negative sign means the direction is opposite to that of the water.

The final velocity of the squid after the ejection of the water is 3 m/s backwards.

The given parameters;

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.5(0) + 0.1(0) = 1.5v₁  +  45(0.1)

0 = 1.5v₁  + 4.5

1.5v₁ = -4.5

[tex]v_1 = \frac{-4.5}{1.5} \\\\v_1 = -3 \ m/s[/tex]

Thus, the final velocity of the squid after the ejection of the water is 3 m/s backwards.

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Answer:

The boiling point of water is around 185.9ºF or 85.5ºC

Explanation:

The boiling point of water depends on the atmospheric pressure. So, the higher the elevation, the lower the atmospheric pressure, therefore the boiling point of water is lower as well. Assuming that sea level pressure is constant ,(760 mmHg, 29.92inches Hg or 14.69psi) the boiling point of water is 212ºF (100ºC). Approximately, for every 500 foot (150m) of elevation increase, the boiling point temperature decreases 1ºF (0.55ºC).

Answer:

[tex]3.14\cdot 10^{-8} m[/tex]

Explanation:

The energy of a photon is related to its wavelength by

[tex]E=\frac{hc}{\lambda}[/tex]

where

E is the energy

h is the Planck constant

c is the speed of light

[tex]\lambda[/tex] is the wavelength

In this problem, we know the energy

[tex]E=6.33\cdot 10^{-18} J[/tex]

So we can solve the previous formula for the wavelength:

[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{6.33\cdot 10^{-18} J}=3.14\cdot 10^{-8} m[/tex]

The wavelength of a photon with an energy of 6.33 × 10-18 joules can be calculated using the equation E = hc/λ, with Planck's constant and the speed of light as known factors.

The wavelength of a photon that has an energy of 6.33 × 10-18 joules can be calculated using the energy-wavelength relationship, E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 × 10-34 J·s), c is the speed of light in a vacuum (approximately 3 × 108 m/s), and λ is the wavelength of the photon.

To find the wavelength λ, we rearrange the equation to λ = hc/E. Substituting the given values, we have λ = (6.626 × 10-34 J·s × 3 × 108 m/s) / (6.33 × 10-18 J), which gives us the wavelength in meters.

1. [tex]-8.78 \cdot 10^5 J[/tex]

The energy lost by the water is given by:

[tex]Q=m C_s \Delta T[/tex]

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

[tex]\Delta T=10.0C-80.0C=-70.0 C[/tex] is the change in temperature

Substituting,

[tex]Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J[/tex]

2. [tex]3.24 \cdot 10^4 J[/tex]

The energy added to the aluminium is given by:

[tex]Q=m C_s \Delta T[/tex]

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

[tex]\Delta T=150.0 C-30.0C =120.0 C[/tex] is the change in temperature

Substituting,

[tex]Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J[/tex]

3a. [tex]-5.6^{\circ}C[/tex]

The temperature change of the water is given by

[tex]\Delta T=\frac{Q}{m C_s}[/tex]

where

[tex]Q = -232 kJ=-2.32\cdot 10^5 J[/tex] is the heat lost by the water

[tex]m=10.0 kg=10000 g[/tex] is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

[tex]\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C[/tex]

3b. [tex]+10.2^{\circ}C[/tex]

The temperature change of the copper is given by

[tex]\Delta T=\frac{Q}{m C_s}[/tex]

where

[tex]Q = 1.96 kJ=1960[/tex] is the heat added to the copper

[tex]m= 500 g[/tex] is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

[tex]\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C[/tex]

4. 42.9 g

The mass of the water sample is given by

[tex]m=\frac{Q}{C_S \Delta T}[/tex]

where

[tex]Q=4300 J[/tex] is the heat added

[tex]\Delta T=39 C-15 C=24C[/tex] is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

[tex]m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g[/tex]

5. 115.5 J

The heat used to heat the copper is given by:

[tex]Q=m C_s \Delta T[/tex]

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

[tex]\Delta T=80.0 C-20.0C =60.0 C[/tex] is the change in temperature

Substituting,

[tex]Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J[/tex]

6. 0.185 J/g•°C

The specific heat of iron is given by:

[tex]C_s = \frac{Q}{m \Delta T}[/tex]

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

[tex]\Delta T=25.0-50.4 C=-25.4 C[/tex] is the change in temperature

Substituting,

[tex]C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC[/tex]

These questions relate to calculating heat energy changes using the concept of specific heat. Specific heat, an intensive property, is used to calculate how much heat is lost or gained during temperature changes of different substances. The formula q=mcΔT, where 'q' is heat energy, 'm' is mass, 'c' is specific heat and 'ΔT' is the temperature change, is used for the calculations.

The specific heat of a substance is an intensive property that defines how much heat energy is needed to raise 1 gram of the substance by 1 degree Celsius.

For the first question, when 3.0 kg of water is cooled from 80.0°C to 10.0°C, we use the formula q=mcΔT, where m is mass, c is specific heat and ΔT is the change in temperature. Substituting the given values, we get q = 3000g * 4.179 J/g°C *(10.0°C - 80.0°C) which gives -881,940 Joules (roughly -882 kJ). The negative sign indicates that heat energy is lost.

For the second question, we apply the same formula, with m=0.30 kg, c=0.900 J/g°C and ΔT = 150.0°C - 30.0°C, giving 32.4 kJ heat energy needed.

For the third question, we reverse the equation to solve for temperature. For 10.0 kg of water losing 232 kJ, ΔT = q / (mc), or 232,000J / (10,000g * 4.179 J/g°C), resulting in an approximately 5.54°C temperature loss. For 500g of copper gaining 1.96 kJ of heat, the equation gives a temperature change of about 10°C.

The same principles are applied to answer the remaining questions.

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Final answer:

To achieve the same electric field that sharks sense, you would have to place the poles of a 1.5 V battery 1.5×10^4 meters apart.

Explanation:

The question asks how far apart you would have to place the poles of a 1.5 V battery to achieve the same electric field that sharks and related fish can sense, which is a voltage gradient of 1 µV across 1 cm of seawater. To find the distance, we use the formula for electric field strength, E = V/d, where E is the electric field strength, V is the voltage, and d is the distance between the poles. Given that sharks can detect a field of 1 µV/cm, for a 1.5 V battery, the distance d to achieve the same field strength of 1 µV/cm would be d = V/E = 1.5/(1 × 10^-6) = 1.5 × 10^6 cm, which equals 15,000 meters or 1.5 × 10^4 meters. Hence, the correct answer is 1.5×10^4 m.

The imaginary second level is 60 dB more intense than the real level of the caller.

60 dB means a multiplication of  10⁶  = 1 million.

That's how many equally-talented callers it would take to be 60 dB louder than him.

The number of callers would be required to reach the pain level of 120 db is 10⁶ , that is, one million.

Decibel is the unit of sound. Its symbol is ‘dB’. We can define decibel as:

A unit of measurement used to express the ratio of one value of a power or field quantity to another on a logarithmic scale, the logarithmic quantity being called the power level or field level, respectively.

60 dB = 10 log₁₀ (P₁/P)

120 dB =  10 log₁₀( P₂/P₀)

The man have to produce square of his intensity level to reach the pain level of 120 dB.

The number of callers would be required to reach the pain level of 120 db is 10⁶ , that is, one million.

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17 protons make up a chlorine ion

17 protons, 18 electrons.

a -1 charge means there is an extra electron. if it was +1 charge there would be 16 electrons

Answer:

D) Roger is incorrect. Only a chemical change is taking place as evidenced by the light and heat of the burning candles.

Explanation:

- A physical change is a change in which there is no formation of new substances. Examples of physical changes are the melting or the evaporation of a substance (all phase transitions are examples of physical changes): in such cases, there is no formation of new substances.

- A chemical change is a change in which new substances form. Examples of chemical changes are the chemical reactions: for instance, when a candle burns, a reaction is taken place (oxygen is burnt, transforming into carbon dioxide + heat + light, so this is an example of chemical change).

Therefore, the correct answer is

D) Roger is incorrect. Only a chemical change is taking place as evidenced by the light and heat of the burning candles.

Answer:

Its c rod is correct...

Answer:

3.12 m

Explanation:

The fundamental frequency of an open-closed tube is given by

[tex]f_1 = \frac{v}{4L}[/tex]

where

f1 is the fundamental frequency

v is the speed of sound in air (343 m/s)

L is the length of the tube

If we use the model mentioned, we can consider L to be the effective length of the instrument. This means we can re-arrange the formula and use the fundamental frequency, f1 = 27.5 Hz, to find L:

[tex]L=\frac{v}{4 f_1}=\frac{343 m/s}{4(27.5 Hz)}=3.12 m[/tex]

(a) [tex]9.1 \cdot 10^{-21} kg m/s[/tex]

The magnitude of the linear momentum of an object is given by

[tex]p=mv[/tex]

where

m is the object's mass

v is its speed

In this case, we have

[tex]m=1.67\cdot 10^{-27} kg[/tex] (mass of the proton)

[tex]v=5.45\cdot 10^6 m/s[/tex] (speed of the proton)

So, the momentum is

[tex]p=(1.67\cdot 10^{-27} kg)(5.45\cdot 10^6 m/s)=9.1 \cdot 10^{-21} kg m/s[/tex]

b) 7.0 kg m/s

In this case, we have

m = 16.0 g = 0.016 kg (mass of the bullet)

v = 435 m/s (speed of the bullet)

By applying the same formula, the linear momentum is

[tex]p=(0.016 kg)(435 m/s)=7.0 kg m/s[/tex]

c) 797.5 kg m/s

In this case, we have

m = 72.5 kg (mass of the sprinter)

v = 11.0 m/s (speed of the sprinter)

By applying the same formula, the linear momentum is

[tex]p=(72.5 kg)(11.0 m/s)=797.5 kg m/s[/tex]

d) [tex]1.8\cdot 10^{29} kg m/s[/tex]

In this case, we have

[tex]5.98\cdot 10^{24} kg[/tex] (mass of the Earth)

[tex]v=2.98\cdot 10^4 m/s[/tex] (speed of the Earth)

By applying the same formula, the linear momentum is

[tex]p=(5.98\cdot 10^{24} kg)(2.98\cdot 10^4 m/s)=1.8\cdot 10^{29} kg m/s[/tex]

Bending your knees when you land from a jump increases the impact time, reducing the average force on your body and preventing injuries such as bone fractures.

When you jump from an elevated position and bend your knees upon reaching the ground, you increase the time of impact, which decreases the average force your body experiences compared to landing stiff-legged. This is due to the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it. The impulse is the product of the average force and the time over which that force acts. By bending the knees, the stopping distance is increased, which spreads the force over a longer time period and reduces the strain on the body, particularly the joints and bones. This is crucial because bones in a body can fracture if the force on them is too great. An example of this concept in action is with kangaroos, where the shock of hopping is cushioned by the bending of their hind legs.

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Runners breathe heavily post a race in an attempt to refuel their oxygen supply while eradicating unwanted carbon dioxide that has accumulated as a waste product.  

• In the process of breathing, inhaling brings oxygen within the body, and discharges carbon dioxide and other gases out of the body.  

• At the time of running, humans seem to take shallow breaths due to which adequate oxygen does not get enter into the respiratory system to fit the requirements of cells and blood.  

• Apart from this, at the time of running, accumulation of carbon dioxide also takes place, this results in the phenomenon known as shortness of breath.  

• Therefore, post breathing, runners breath heavily as they try to refuel the supply of oxygen and at the same time eliminates the unwanted carbon dioxide from the body that got accumulated.  

Thus, to take oxygen and to remove unwanted carbon dioxide from the body, the runners breath heavily post a sprint race.  

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Answer:

0.148

Explanation:

For a car travelling on a flat curve, the centripetal force is provided by the frictional force between the tires and the road:

[tex]m\frac{v^2}{r}=\mu mg[/tex]

where

m is the car's mass

v is the tangential speed

r is the radius of the curve

[tex]\mu[/tex] is the coefficient of static friction

g is the gravitational acceleration

Dividing by m (the mass), we obtain an expression for the centripetal acceleration:

[tex]a_c = \frac{v^2}{r}=\mu g[/tex]

Therefore, by substituting a = 1.45 m/s^2 and g =9.8 m/s^2, we can solve the formula to find the coefficient of static friction:

[tex]\mu=\frac{a_c}{g}=\frac{1.45 m/s^2}{9.8 m/s^2}=0.148[/tex]

The coefficient of static friction between the car and track is about 0.465

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

tangential acceleration = a_t = 1.45 m/s²

initial angular velocity = ωo = 0 rad/s

position angle = θ = ¹/₄ rev = ¹/₂π rad

Asked:

coefficient of static friction = μ = ?

Solution:

Firstly, let's find the final angular velocity by using following formula:

[tex]\omega^2 = \omega_o^2 + 2\alpha \theta[/tex]

[tex]\omega^2 = \omega_o^2 + 2(\frac{a_t}{R}) \theta[/tex]

[tex]\omega^2 = 0 + 2(\frac{1.45}{R}) (\frac{1}{2}\pi})[/tex]

[tex]\omega^2 = \frac{1.45\pi}{R}[/tex]

[tex]\texttt{ }[/tex]

Next , we could use the centripetal acceleration as follows:

[tex]\Sigma F = ma[/tex]

[tex]f = ma[/tex]

[tex]\mu N = m \omega^2 R[/tex]

[tex]\mu mg = m \omega^2 R[/tex]

[tex]\mu g = \omega^2 R[/tex]

[tex]\mu = \omega^2 R \div g[/tex]

[tex]\mu = \frac{1.45\pi}{R} R \div g[/tex]

[tex]\mu = 1.45 \pi \div g[/tex]

[tex]\mu = 1.45 \pi \div 9.8[/tex]

[tex]\mu \approx 0.465[/tex]

[tex]\texttt{ }[/tex]

The coefficient of static friction between the car and track is about 0.465

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer:

c) quadruple in magnitude

Explanation:

The power dissipated in the circuit is given by:

[tex]P=I^2 R[/tex]

where

I is the current in the circuit

R is the total resistance of the circuit

In this problem:

- The current is doubled: I' = 2 I

- The resistance is kept constant: R' = R

So, the power dissipated is

[tex]P' = (I')^2 R' = (2I)^2 R=4 I^2 R=4 P[/tex]

so, the power dissipated increase by a factor 4 (quadruples).

Final answer:

When the current through a resistor with constant resistance is doubled, the power dissipated will quadruple instead of just doubling because power is proportional to the square of the current. Correct answer is (c).

Explanation:

Understanding Power Dissipation in Circuits

When we talk about power dissipation in a circuit, we are often referring to the power consumed by a resistor. The power dissipated by a resistor is given by the formula P = I² * R, where P stands for power, I is the current through the resistor, and R is the resistance. If the current flowing through a circuit with constant resistance is doubled, the power dissipation is affected as follows:

Original power dissipation: P = I² * R

After doubling the current: [tex]P_{new[/tex] = (2I)² * R = 4 * I² * R

Thus, the power dissipated by the circuit after doubling the current will quadruple in magnitude, since 4 times the original I² * R would be 4 * P. Therefore, the correct answer is (c) quadruple in magnitude.

To determine the compression of the spring when a 4.0 kg object is dropped from 1.0 meter, we use the conservation of energy principle. The gravitational potential energy of the dropped object is converted into the spring's elastic potential energy. Calculations show the spring will compress approximately 9.0 cm.

We can solve the problem of the 4.0 kilogram object being dropped onto a spring by considering the conservation of energy. The gravitational potential energy at the height of 1.0 meter will convert into the elastic potential energy of the spring once the spring is compressed and the system momentarily comes to rest.

The gravitational potential energy (PE) can be calculated using the formula PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s2), and h is the height from which the object is dropped. Therefore, PE = 4.0 kg * 9.8 m/s2 * 1.0 m = 39.2 Joules.

The elastic potential energy stored in a compressed spring is given by the formula PEspring = 1/2 * k * x2, where k is the spring constant and x is the compression distance. Setting the gravitational potential energy equal to the spring's elastic potential energy gives 39.2 J = 1/2 * 850 N/m * x2. Solving for x gives x = sqrt((2 * 39.2) / 850) which approximately equals 0.09 meters or 9.0 cm.

C. The molecules will move more slowly

B because it is the only evidence that displays an effect from the field without contact

Hey i am here to help you.

The formula is speed= distance/ time

Time=45 minutes we have to convert it to seconds, therefore 45*60=2700 seconds .

Distance= 4km

S=45/2700 when we divide it we will get 675

therefore the answer is 675

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It makes them be three forces