difference between freefall and weightlessness.​

Answer:

a is the maybe I am not sure

Answer:

Option 5:

The 60W bulb has a greater resistance and a lower current than the 100 W bulb.

Explanation:

We have to compare the resistance and current of both bulbs.

Bulb A

Power = 60 W

Voltage = 110 V

Power is given as:

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

where V= voltage and R = resistance

[tex]=> 60 = 110^2 / R\\\\R = 201.6 \Omega[/tex]

Power is also given as:

P = IV

where I = current

=> 60 = I * 110

I = 60/110 = 0.54 A

Bulb B

Power = 100 W

Voltage = 110 V

To get resistance:

[tex]100 = 110^2 / R\\\\R = 121 \Omega[/tex]

To get current:

100 = I * 110

I = 100 / 110

I = 0.91 A

Therefore, by comparison, the 60W bulb has a greater resistance and a lower current.

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

Answer:

The charge on the bead is  [tex]q = 6.084 *10^{-7}\ C[/tex]

Explanation:

From the question we are told that

     The  mass of the bead is  [tex]m = 3.6 \ g = 0.0036 \ kg[/tex]

      The  magnitude of the electric field is  [tex]E = 200,000 \ N/C[/tex]

       The  acceleration of the bead is  [tex]a = 24 m/s^2[/tex]

Generally,  the  electric force on the bead is  mathematically represented as  

         [tex]F_ e = q E[/tex]

Where q is the charge on the bead

   Now the gravitational force opposing the upward movement of the bead is  mathematically represented as

       [tex]F_g = mg[/tex]

Generally the net force on the bead is  mathematically represented as

       [tex]F = F_e - F_g = m* a[/tex]

=>    [tex]qE - mg = ma[/tex]

Now substituting values

       [tex]q * 200000 - 0.0036 *9.8 = 0.0036 * 24[/tex]

     [tex]q = 6.084 *10^{-7}\ C[/tex]

Answer:

Moment of inertia of the flywheel is equal to 10.19 kg-m^2

Explanation:

The maximum rotational energy to be stored by the flywheel [tex]E_{r}[/tex] = 2.00 x 10^6 J

Angular speed with which to store this energy ω =  443 rad/s

moment of inertia of the flywheel [tex]I[/tex] = ?

Recall that the energy of a rotating body is gotten from the equation

[tex]E_{r} = Iw^{2}[/tex]

Where [tex]E_{r}[/tex] is the rotational energy of the rotating body

[tex]I[/tex] = moment of inertia of the body

ω = angular speed of the rotating body

imputing the values into the equation, we'll have

2.00 x 10^6 = [tex]I[/tex] x [tex]443^{2}[/tex]

2.00 x 10^6 =  [tex]I[/tex] x 196249

[tex]I[/tex] = (2.00 x 10^6) ÷ 196249 = 10.19 kg-m^2

Answer:

Explanation:

check it out and rate me

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

Complete question:

The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63 kg of this oil from 23 °C to 191 °C?

Answer:

The heat energy required to raise the temperature of this cooking oil is 773220 J

Explanation:

Given;

mass of the cooking oil, m = 2.63 kg = 2630 g

specific heat capacity of the cooking oil, C = 1.75J/g°C

initial temperature of the cooking oil, T₁ = 23° C

final temperature of the cooking oil, T₂ = 191° C

The heat energy required to raise the temperature of this cooking oil;

Q = mcΔT

Q = 2630 x 1.75 x (191 - 23)

Q = 773220 J

Therefore, the heat energy required to raise the temperature of this cooking oil is 773220 J

Answer:

a) 43.98 V

b) E = 21.37 MJ

Explanation:

Parameters given:

Length of cable = 180 km = 180000 m

Diameter of cable = 11 cm = 0.11 m

Radius = 0.11 / 2 = 0.055 m

Current, I = 135 A

a) To find the potential drop, we have to find the voltage across the wire:

V = IR

=> V = IρL / A

where R = resistance

L = length of cable

A = cross-sectional area

ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm

Therefore:

V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)

V = 43.98 V

The potential drop across the cable is 43.98 V

b) Electrical energy is given as:

E = IVt

where t = time taken = 1 hour = 3600 s

Therefore, the energy dissipated per hour is:

E = 135 * 43.98 * 3600

E = 21.37 MJ (mega joules, 10^6)

Answer:

A. 1.6 N/cm

Explanation:

spring constant = 21/13 = 1.6 N/cm

Answer:

The atomic number

Explanation:

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units,  5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters  ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )  

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom  ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter /  grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

The value of gOP7 in the units your architects will use is 5.724 [tex]m/s^2[/tex]

Given that 5.24 flurg = 1 meter,

1 grom = 0.493 second.  

First, we will convert the length units:

[tex]7.29 flurg / grom^2 / 5.24 flurg / meter = 1.39 meter / grom^2[/tex]

Now we convert the time units:

[tex]1.39 meter / grom^2 / 0.493 second / grom = 2.82195 meter / grom * second[/tex]

 [tex]2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2[/tex]

The value of gOP7 in the units the architects will use is [tex]5.724m/s^2[/tex]

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Answer:

 e = 50.27 give / s

Explanation:

The expression for simple harmonic motion is

    x = A cos (wt + Ф)

in this case they give us the amplitude A = 3.9 cm and frequency f = 8.0 Hz

The angular and linФear variables are related

      e = 2π d

      e = 2π 8

      e = 50.27 give / s

let's look for the constant fi

       so let's find the time to have the maximum displacement

       v = dx / dt

       v = -A w sin (wt +Ф)

for the point of maximum displacement the speed is I think

        0 = - sin (0 + Ф)

therefore fi = 0

Let's put together the equation of motion

          x = 0.039 sin (50.27 t)

          v = 0.039 50.27 sin (50.27 3)

           v = 1.96 50 0.01355

            v = 0.0266 m / s

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

Answer:

(1) Percent Difference = 2.47%

(2) Percent Error (9.96 m/s²) = 1.63 %

    Percent Error (9.72 m/s²) = 0.82 %

(3) Percent Error (Mean) = 0.41 %

Explanation:

(1)

Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %

Percent Difference = 2.47%

(2)

Percent Error = (|Measured Value - Original Value|/Original Value)*100%

Therefore,

Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (9.96 m/s²) = 1.63 %

Now,

Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (9.72 m/s²) = 0.82 %

(3)

First we need to find the mean of values:

Mean = (9.96 m/s² + 9.72 m/s²)/2

Mean = 9.84 m/s²

Therefore,

Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

Percent Error (Mean) = 0.41 %

The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).

So if a wave completes one cycle in 2 seconds, then that is its period.

Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Answer:

Second image in your list of possible answers

Explanation:

The second option is what you would expect from modulating a sinusoidal carrier wave of higher frequency after being modulated by a square pulse of lower frequency that allows part of the carrier signal to travel during the time the square signal is constant different from zero, and be absent (flat) during the time the square pulse signal has amplitude zero.

The second line is the best picture of a pulse-modulated carrier.

It would be easy to build a circuit where each pulse ... when it comes along ... just switches the carrier OFF for as long as the pulse lasts.

Answer:

Choice d. The set of vectors: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex] isn't a vector space over [tex]\mathbb{R}[/tex].

Explanation:

Let a set of vectors [tex]V[/tex] to be a vector field over some field [tex]\mathbb{F}[/tex] (for this question, that "field" is the set of all real number.) The following must be true:

Note that in the general form of a vector in [tex]V[/tex], the second component is a always non-zero. Because of that non-zero component,

Assume by contradiction that [tex]V[/tex] is indeed a vector field. Therefore, it should contain a zero vector. Let [tex]\mathbf{0}[/tex] denote that zero vector. For all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].

Using the definition of set [tex]V[/tex]: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex], there exist real numbers [tex]a[/tex] and [tex]b[/tex], such that:

[tex]\displaystyle \mathbf{v} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].

Hence, [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex] is equivalent to:

[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} + \mathbf{0} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].

Apply the third property that [tex]V[/tex] is closed under scalar multiplication. [tex]-1[/tex] is indeed a real number. Therefore, if [tex]\mathbf{v}[/tex] is in

Therefore:

[tex]\displaystyle -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} \in V[/tex].

Apply the second property and add [tex]\displaystyle - \mathbf{v} = -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex] to both sides of [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex]. The left-hand side becomes:

[tex]\mathbf{v} - \mathbf{v} + \mathbf{0} = \mathbf{0}[/tex].

The right-hand side becomes:

[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} = \begin{bmatrix}a - 4\, b - (a - 4\, b) \\ 5 - 5 \\ 4\, a+ b-(4\, a+ b)\\ -a -b - (-a -b)\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].

Therefore:

[tex]\displaystyle \mathbf{0} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].

However, [tex]\mathbf{0} = \displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex] isn't a member of the set [tex]\displaystyle V = \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex]. That's a contradiction, because [tex]\mathbf{0}[/tex] was supposed to be part of [tex]V[/tex].

Hence, [tex]V[/tex] isn't a vector space by contradiction.

Answer:

Wavelength is 0.359 m

Explanation:

Given that,

Magnetic field, B = 0.547 T

We need to find the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.

The frequency of revolution of proton in the cyclotron is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

m is mass of proton

q is charge on proton

So,

[tex]f=\dfrac{1.6\times 10^{-19}\times 0.547}{2\pi \times 1.67\times 10^{-27}}\\\\f=8.34\times 10^6\ Hz[/tex]

We know that,

Speed of light, [tex]c=f\lambda[/tex]

[tex]\lambda[/tex] = wavelength

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{834\times 10^6}\\\\\lambda=0.359\ m[/tex]

So, the wavelength of the radiation produced by a proton is 0.359 m.

Answer:

Explanation:

this is the answer to your question

The electron is going with a velocity of 1.25 × 10⁷ m/s when it hits the positive plate.

According to the law of the conservation of mechanical energy, the total mechanical energy is always conserved by an electron. We can say that the sum of potential energy (U) and kinetic energy (K) is always constant.

K + U = E

Given, the distance between the two parallel plates = 1.5 mm

The potential difference between the plates, V = 450V

The charge on an electron, q = [tex]-1.6\times 10^{-19} C[/tex]

The mass of an electron, m = 9.1× 10⁻³¹ Kg

The change in the potential energy of the charge moving through the potential difference of 450V.

ΔU = qΔV = (-1.6× 10⁻¹⁹)(450)  = -7.2 × 10⁻¹⁷J

From the law of the conservation of mechanical energy, we can write:

K + U = E

ΔK + ΔU = 0

ΔK = -ΔU

1/2mv² = -ΔU

v² = -2ΔU/m

[tex]v^2 =\frac{-2\times (-7.2\times 10{-17})}{9.1\times 10^{-31}}[/tex]

[tex]v=\sqrt{1.58\times 10^{14}}[/tex]

v = 1.25 × 10⁷ m/s

Therefore, the electron is going with the speed of 1.25 × 10⁷ m/s  when it hits the positive plate.

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Answer:

The resistivity is  [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]

Explanation:

From the question we are told that

    The magnitude of the electric field is  [tex]E = 6.2 V/m[/tex]

     The  current density is  [tex]J = 2.4 *10^{8} \ A/m^2[/tex]

Generally  the resistivity is mathematically represented as

         [tex]\rho = \frac{E}{J}[/tex]

substituting values

        [tex]\rho = \frac{6.2}{2.4 *10^{8}}[/tex]

        [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]

Answer:

There are two possible solutions.

M1 = 4.68*10^30kg,  M2 = 5.53*10^30kg

M1 = 5.53*10^30kg,  M2 = 4.7*10^29kg

Explanation:

In order to find the mass of each black hole, you take into account the gravitational force between them and the sum of their masses.

You use the formula for the gravitational force between two masses:

[tex]F_g=G\frac{M_1M_2}{r^2}[/tex]              (1)

G: Cavendish's constant = 6.674*10^-11 m^3/kg.s^2

M1, M2: mass of each black hole = ?

r: distance between the black holes = 10.0 AU = 10.0(1.50*10^11m) = 1.5*10^12m

Fg: gravitational force between the black holes = 7.70*10^25N

Furthermore, you take into account that the sum of the masses M1 and M2 is:

M1 + M2 = 6.00*10^30 kg        (2)

You solve the equation (2) for M2.

[tex]M_2=6.00*10^{30}-M_1[/tex]

Next, you replace the obtained expression for M2 into the equation (1) and solve for M1, as follow (for simplicity, you do not add the units):

[tex]F_g=G\frac{M_1(6.00*10^{30}-M_1)}{r^2}\\\\\frac{r^2F_g}{G}=6.00*10^{30}M_1-M_1^2\\\\\frac{(1.5*10^{12})^2(7.70*10^{25})}{(6.674*10^{-11}}=6.00*10^{30}M_1-M_1^2\\\\2.59*10^{60}=6.00*10^{30}M_1-M_1^2\\\\M_1^2-6.00*10^{30}M_1+2.59*10^{60}=0[/tex]

Then, you have obtained a quadratic polynomial. You solve it with the quadratic formula:

[tex]M_1=\frac{-(-6.00*10^{30})\pm \sqrt{(-6.00*10^{30})^2-4(1)(2.59*10^{60}))}}{2(1)}\\\\M_1=\frac{6.00*10^{30}\pm 5.06*10^{30}}{2}\\\\M_1=4.68*10^{29}\\\\M_1=5.53*10^{30}[/tex]

Both results are consistent, then the mass of one black hole can be 4.68*10^30kg and also 5.53*10^30kg.

The other black hole has a mass of:

[tex]M_2=6.00*10^{30}kg-4.68*10^{29}kg=5.53*10^{30}kg\\\\M_2=6.00*10^{30}kg-5.53*10^{30}kg=4.7*10^{29}kg[/tex]

Hence, you have a pair of solutions:

M1 = 4.68*10^30kg,  M2 = 5.53*10^30kg

M1 = 5.53*10^30kg,  M2 = 4.7*10^29kg

Answer:

clockwise

Explanation:

when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.

Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same  polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.

The direction of the induced current in the bottom ring is in the clockwise direction.

The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.

Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.

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Answer:

U =-2.39*10^-18 J

Explanation:

In order to calculate the electric potential energy of the electron you use the following formula:

[tex]U=k\frac{q_1q_2}{r}[/tex]           (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between charges

In this case the electron is at point midway between two charges, then the electric potential energy is the sum of two contributions:

[tex]U=U_1+U_2=k\frac{eq_1}{r}+k\frac{eq_2}{r}=\frac{ke}{r}[q_1+q_2][/tex]

e: charge of the electron = 1.6*10^-19C

q1: charge 1 = 3.00nC = 3.00*10^-9C

q2: charge 2 = 2.00nC = 3.00*10^-9C

r: distance to each charge = 60.0cm/2 = 30.0cm = 0.3m

If you consider that the electron is at the origin of coordinates, with the first charge in the negative x axis, and the other one in the positive x axis, you have:

[tex]U=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{0.6m}[-3.0*10^{-9}C+2.0*10^{-9}C]\\\\U=-2.39*10^{-18}J[/tex]

The electric potential energy of the electron is -2.39*10^-18 J

Answer:

did you mean released?

Explanation:

If so the process is called respiration

Answer:

power dissipated in the target is 100 W

Explanation:

given data

electrons = 1 mev = [tex]10^{6}[/tex] eV

1 eV = 1.6 × [tex]10^{-19}[/tex] J

current =  100 microampere = 100 × [tex]10^{-6}[/tex] A

solution

when energy of beam strike with 1 MeV so energy of electron is

E = e × v   ...................1

e is charge of electron and v is voltage

so put here value and we get voltage

v = 1 ÷ 1.6 × [tex]10^{-19}[/tex]

v =  [tex]10^{6}[/tex] volt

so power dissipated in target

P = voltage × current   ..............2

put here value

P =  [tex]10^{6}[/tex]  × 100 × [tex]10^{-6}[/tex]

P = 100 W

so power dissipated in the target is 100 W

Answer:

  F ’= F 0.25

Explanation:

This problem refers to the electric force, which is described by Coulomb's law

        F = k q₁ q₂ / r²

where k is the Coulomb constant, q the charges and r the separation between them.

The initial conditions are

       F = k q_A q_B / d²

they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d

       F ’= k (q / 4 q / 4) / (0.5 d)²

       F ’= k q / 16 / 0.25 d²

       F ’= k q² / d²    0.0625 / 0.25

       F ’= F 0.25

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.

Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]

where,

If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:

[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

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Answer:

The final temperature of the metals will be 384.97 K

Explanation:

For the gold;

mass = 1.3 kg

temperature = 300 K

specific heat =  126 J/kg-K

For the copper;

mass = 2.4 kg

temperature = 400 K

specific heat = 386 J/kg-K

Firstly, we will have to calculate for the thermal energy possessed by each of the metal.

The heat possessed by a body =  mcT

Where,

m is the mass of the body

c is the specific heat of the body, and

T is the temperature of the body at that instance

so we calculate for the thermal energy of the gold and the copper below

For gold;

heat energy = mcT = 1.3 x 126 x 300 = 49140 J

For copper;

heat energy = mcT = 2.4 x 386 x 400 = 370560 J

When the two metal come in thermal contact, this heat is evenly distributed between them.

The total heat energy = 49140 J + 370560 J = 419700 J

At thermal equilibrium, the two metals will be at the same temperature, to get this temperature, we equate the total thermal energy to the heat energy that will be possessed by the metals at equilibrium.

419700 = (1.3 x 126 x T) + (2.4 x 386 x T) = 163.8T + 926.4T

419700 = 1090.2T

T = 419700/1090.2 = 384.97 K

The final temperature of the metals will be 384.97 K