Picture isn’t the best but could use the help

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

Answer:

Current

I think The choose (B)

B. Electric current

Answer:

The density at 300 C is [tex]2641.68 kg/m^3[/tex].

Explanation:

density at 0 degree, d = 2700 kg/m3

coefficient of linear expansion, = 24 x 10^{-6} /K

Let the density at 300 C is d'.

Use the formula of the cubical expansion

[tex]d'= d (1-3\alpha \Delta T)\\\\d'= 2700(1- 3 \times 24\times 10^{-6}\times 300)\\\\d' = 2641.68 kg/m^3[/tex]

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

Assumptions:

∴

For a motion under constant acceleration, we can apply the kinematic equation:

[tex]\mathsf{v^2 = u^2 + 2as}[/tex]

where;

From the above equation, making acceleration (a) the subject of the formula:

[tex]\mathsf{v^2 - u^2 =2as }[/tex]

[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]

[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;

v = u + at

where;

v = 0

-u = at

[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]

[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

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Answer:

4.926 L Y 7.389 L

Explanation:

first you calculate the tank volume

V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]

then you convert to liters

[tex]12315 cm^{3}[/tex] = 12.315 l

then you calculate the liters of water

2/5(12.35 l) = 4.926 l

finally we calculate the amount without water

12.315 l - 4.926 l = 7.389 l

HERE IS MORE INFORMATION ON THE SUBJECT. THEY REMOVED THE

ENGLISH SITE BUT YOU CAN USE TRANSLATOR

LINK: https://gscourses.thinkific.com/courses/fisicai

Answer:

The answer is "[tex]143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}[/tex]"

Explanation:

For point a:

Energy balance equation:

[tex]\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\[/tex]

[tex]W=0\\\\Q=0\\\\m_e=0[/tex]

From the above equation:

[tex]\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\[/tex]

because the rate of air entering the tank that is [tex]h_i[/tex] constant.

[tex]\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\[/tex]

Since the tank was initially empty and the inlet is constant hence, [tex]m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\[/tex]

Interpolate the enthalpy between [tex]T = 300 \ K \ and\ T=295\ K[/tex]. The surrounding air  

temperature:

[tex]T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}[/tex]

Substituting the value from ideal gas:

[tex]\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}[/tex]

Follow the ideal gas table.

The [tex]u_2= 298.33\ \frac{kJ}{kg}[/tex] and between temperature [tex]T =410 \ K \ and\ T=240\ K.[/tex]

Interpolate

[tex]\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}[/tex]

Substitute values from the table.

 [tex]\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\[/tex]

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. [tex]\bar{R} \ is\ \frac{R}{M}[/tex] (M is the molar mass of the  gas that is [tex]28.97 \ \frac{kg}{mol}[/tex] and R is gas constant), and T is the temperature.

[tex]n=\frac{pV}{TR}\\\\[/tex]

[tex]=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\[/tex]

For point c:

 Entropy is given by the following formula:

[tex]\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}[/tex]

option a ...

please mark hcfhvx

Answer:

Alternating current at which when entered into the loop cause it to magnetize

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = [tex]\frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150} }{2}[/tex]

       t = [tex]\frac{12.5 \ \pm 27.5}{2}[/tex]

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

Answer:

Final kinetic energies of both astronauts will be the same.

Explanation:

If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.

Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.

Similarly, the final kinetic energies of astronauts will also be the same.

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]

Therefore, the resulting temperature is 23.37 ⁰C

ANSWER; KE=5mv^2 so it is proportional to v^2.

Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.

Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

Answer:

 f = v / 4L

the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Explanation:

In wind instruments the wave speed must meet

          v = λ f

          λ = v / f

from v is the speed of sound that depends on the temperature

          v = v₀ [tex]\sqrt{1+ \frac{T [C]}{273} }[/tex]

where I saw the speed of sound at 0ºC v₀ = 331 m/s  the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation

            (F -32) 5/9 = C

            76ºF = 24.4ºC

            45ºF = 7.2ºC

With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside

at T₁ = 24ºC               v₁ = 342.9 m / s

at T₂ = 7ºC                 v₂ = 339.7 m / s

To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change

              λ / 4 = L

              λ= 4L

              v / f = 4L

              f = v / 4L

Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J

Answer:

a)  [tex]d_g=1.95m[/tex]

b) [tex]d_g'=2.3m[/tex]

Explanation:

From the question we are told that:

Depth [tex]d=2.60[/tex]

a)

Generally the equation for distance to ground is mathematically given by

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

Where

[tex]n=\frac{4}{3}[/tex]

Therefore

[tex]d_g=\frac{d}{n}[/tex]

[tex]d_g=\frac{2.6}{4/3}[/tex]

[tex]d_g=1.95m[/tex]

b)

For when The pool is filled halfway with water

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

[tex]d_g'=\frac{1.3}{4/3}[/tex]

[tex]d_g'=0.98m[/tex]

Therefore

[tex]d_g'=(1.3+0.98)[/tex]

[tex]d_g'=2.3m[/tex]

Answer:

I think Refraction

Explanation:

I think refraction is light which passes through a boundary between 2 transparent media.

Answer:

A denser object will usually have a high index of

refraction.

A denser object will usually have a high index of refraction.

The refractive index is the ratio of the speed of light in vacuum and speed of light in any medium.

n = c/v

The density is greater for the denser medium (water, oil, mercury, etc) then the rarer medium (any gas or air).

When a light ray travels through denser medium, its velocity is reduced and the refracted ray bends towards normal.

As, the index of refraction is inversely proportional to the velocity of light in the medium, index of refraction will be high for denser object.

Thus, denser object have high index of refraction.

Learn more about index of refraction.

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Answer:

No

Explanation:

You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.

Answer:

The car will make the turn perfectly

Explanation:

Given that the centripetal force= mv^2/r

M= mass of the car

v = speed of the car

r= radius

Hence;

F = 1000 × (14)^2/50

F= 3920 N

The frictional force = μmg

μ = coefficient of static friction

m= mass

g = acceleration due to gravity

Frictional force= 0.6 × 1000× 10

Frictional force = 6000 N

The car will not skid off the curve because the frictional force is greater than the centripetal force.

Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

Answer:

3.75 watts

Explanation:

power = volts * current

2.5 * 1.5 =

3.75

Answer:

Explanation:

From the question we are told that:

Number of turns [tex]N=10[/tex]

Area [tex]a=0.23m^2[/tex]

Magnetic field [tex]B=0.947T[/tex]

Generally the equation for maximum flux is mathematically given by

 [tex]\phi=NBa[/tex]

 [tex]\phi=10*0.047*0.23[/tex]

 [tex]\phi=0.1081wbi[/tex]

Therefore induced emf

 [tex]e= \frac{d\phi}{dt}[/tex]

Since

 [tex]t=0[/tex]

Therefore

 [tex]e=0[/tex]

Answer: c. 40m

Explanation:

See picture

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[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]

The answer is...

C. Work and time.

[tex]HopeThisHelps!![/tex]

[tex]AnimeVines[/tex]

Complete Question

A charged particle of mass 0.0040 kg is subjected to a 40-T magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius 0.10 m at a speed of 20m/s, what is the magnitude of the charge on the particle?

Answer:

[tex]q=0.020C[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.004kg[/tex]

Magnetic field [tex]B=40-T[/tex]

Radius [tex]r=0.10m[/tex]

Speed [tex]v= 20m/s[/tex]

Generally the equation for Radius of a circular path is mathematically given by

[tex]r=\frac{p}{qB}[/tex]

Where

[tex]p=mv[/tex]

Therefore

[tex]q=\frac{mv}{rB}[/tex]

[tex]q=\frac{0.004kg*20m/s}{0.1*40}[/tex]

[tex]q=0.020C[/tex]

Answer:

The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".

Explanation:

As the outer spherical shell is conducting, so there is no electric field in side from

⇒ [tex]r_b_1 < r < r_b_2[/tex].

So the electric potential at all points inside the conducting shell that from

⇒ [tex]r_b_1<r<r_b_2[/tex]  

and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:

⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]

Thus the above is the right solution.