Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω

Answers

Answer 1

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω


Related Questions

What is the meant of by renewable energy and non-renewrable with example of each.​

Answers

Answer:

Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.

Non-renewable  energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.

Calculate the buoyant force due to the surrounding air on a man weighing 600 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3.

Answers

Answer:

[tex]F_b= 0.720 N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=600N[/tex]

Average density [tex]\rho=1.20kg/m^3[/tex]

Mass

[tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{600}{9.81}[/tex]

[tex]m=61.22kg[/tex]

Generally the equation for Volume is mathematically given by

[tex]V =\frac{ mass}{density}[/tex]

[tex]V= \frac{61.22}{1000}[/tex]

[tex]V=0.06122 m^3[/tex]

Therefore

Buoyant force [tex]F_b[/tex]

[tex]F_b=\rho*V*g[/tex]

[tex]F_b= rho_air*V*g[/tex]

[tex]F_b= 0.720 N[/tex]

How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V

Answers

a) A charged particle accelerates as it moves from location A to location B. If V A = 170 V and V B = 210 V, what is the sign of the charged particle? positive negative (b) A proton gains electric potential energy as it moves from point 1 to point 2.
Physics
Two identical point charges are fixed to diagonally opposite corners of a square that is 1.5 m on a side. Each charge is +2.0 µC. How much work is done by the electric force as one of the charges moves to an empty corner? I have W(fe)= -EPE=-q[V(f)-V(i)].
physics
Two point charges are placed on the x axis. (Figure 1) The first charge, q1 = 8.00nC , is placed a distance 16.0m from the origin along the positive x axis; the second charge, q2 = 6.00nC , is placed a distance 9.00m from the origin along the negative x

Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?

Answers

Answer:

Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.

En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.

La persona A aplica una fuerza:

Fa = -3N

La persona B aplica una fuerza:

Fb = 5N

La persona C aplica una fuerza Fc, la cual aún no conocemos.

Pero sabemos que la caja está en equilibrio físico, por lo que:

Fa + Fb + Fc = 0N

reemplazando los valores que conocemos, obtenemos:

-3N + 5N + Fc = 0N

Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.

Fc = 0N + 3N - 5N

Fc = -2N

Podemos concluir que la persona C aplica una fuerza horizontal de -2N

A brass road is 2cm long at instance to what is the lense for a temperature rise of 100k, If the expansivity of brass is 18x10^-6/k^-1

Answers

The length of the brass at a temperature rise of 100 K is 2.0036 m

From the question given above, the following data were obtained:

Original length (L₁) = 2 m

Temperature rise (ΔT) = 100 K

Coefficient of linear expansion (α) = 18×10¯⁶ K¯¹

Final length (L₂) =?

The final length of the brass can be obtained as follow:

α = L₂ – L₁ / L₁ΔT

18×10¯⁶ = L₂ – 2 / (2 × 100)

18×10¯⁶ = L₂ – 2 / 200

Cross multiply

L₂ – 2 = 18×10¯⁶ × 200

L₂ – 2 = 0.0036

Collect like terms

L₂ = 0.0036 + 2

L₂ = 2.0036 m

Thus, the length of the brass at a temperature rise of 100 K is 2.0036 m

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Select the correct answer.
What is abstraction?
OA. the concept that software architecture can be separated into modules and that each module can be examined independently
OB. the process of containing information within a module, preventing any crossover or access to Irrelevant information
OC. the process of splitting a program both horizontally and vertically
OD. the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains
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Answers

Answer:

OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains

Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.

What is abstraction?

Abstraction is the practice of removing anything from a set of core features by eliminating or deleting attributes.

One of the three core ideas of object-oriented programming is abstraction order to decrease complexity and maximize efficiency, a programmer uses abstraction to conceal all but the important facts about an object.

Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.

Hence option D is correct.

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You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.

Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?

Answers

Answer:

0.37 m

Explanation:

Given :

Window height, [tex]h_1[/tex] = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex]   seconds.

Assuming that the speed of the pot just above the window is v then,

[tex]h_2=ut+\frac{1}{2}gt^2[/tex]

[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]

[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]

[tex]$v= 2.69$[/tex] m/s

Initially the pot was dropped from rest. So,  u = 0.

If it has fallen from a height of h above the window then,

[tex]$h = \frac{v^2}{2g}$[/tex]

[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]

h = 0.37 m

derive expression for pressure exerted by gas ​

Answers

From kinetic theory of gases, the pressure exerted by a gas is given by velocity of gas molecules. m = Mass of each molecule of a gas. But by assumptions of the kinetic theory of gases the average kinetic energy of a molecule is constant at a constant temperature.

distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh​

Answers

The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.

Answer:

25 you said ? thats incorecct

Explanation:

What is the feature known as the "Great Dark Spot" of Neptune? It is an apparently permanent feature about five times the size of Earth, similar to the Great Red Spot of Jupiter, near Neptune's south pole. It was a dark hole in the upper atmosphere left by the collision of the comet Shoemaker-Levy 9. It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years. It is a dark surface feature on the surface snow layers caused by radiation discoloration of the older layers. It is a permanent discoloration of the north polar region of Neptune caused by locally prevailing lower surface temperatures there.

Answers

Answer:

It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.

Explanation:

The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.

The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.

An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown

Answers

Answer: [tex]283.2\times 10^{-9}\ nC[/tex]

Explanation:

Given

Cross-sectional area [tex]A=0.4\ cm^2[/tex]

Dielectric constant [tex]k=4[/tex]

Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]

Distance between capacitors [tex]d=5\ mm[/tex]

Maximum charge that can be stored before dielectric breakdown is given by

[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]

Answer:

The maximum charge is 7.08 x 10^-8 C.

Explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]

A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
a. What is the effective spring constant for this motion?
b. How much energy is involved in this motion?

Answers

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first

Answers

Answer:

solid cylinder

Explanation:

the object that arrives first is the object that has more speed, let's use the concepts of energy

starting point. Highest point

         Em₀ = U = m g h

final point. Lowest point

         Em_f = K = ½ mv² + ½ I w²

since the body has rotational and translational movement

how energy is conserved

         m g h = ½ mv² + ½ I w²

linear and angular velocity are related

          v = w r

          w = v / r

we substitute

          m g h = ½ mv² + ½ I (v/r) ²

          mg h = ½ v² (m + I /r²)

          v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]

           

the tabulated moments of inertia for the bodies are

solid cylinder     I = ½ m r²

hollow cylinder  I = m r²

we look for the speed for each body

solid cylinder

          v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]

          v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]

let's call    v₀ = [tex]\sqrt{2gh}[/tex]

         v₁ = 0.816 v₀

hollow cylinder

          v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]

          v₂ = v₀ √½

          v₂ = 0.707 v₀

Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive

Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?

Answers

Answer:

[tex]_{18}^{39} } Ar[/tex]

Explanation:

The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.

[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]

Argon is a stable gas and is found in the group 8 on the periodic table of elements.

Answer:

Answer is below

Explanation:

39 18 Ar

What is the name of the invisible line that runs
down the center of the axial region?

Answers

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Explanation:

The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.

A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.

Answers

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Determine the average acceleration of the car during this time interval.

Answers

Answer:

[tex]a=2.5\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the car, u = 14 m/s

Finally, it comes to rest, v = 0

Time, t = 5.6 s

We need to find the average acceleration of the car during this time interval. We know that,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2[/tex]

So, the acceleration of the car is [tex]2.5\ m/s^2[/tex] in the opposite direction of motion.

If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25

Answers

Answer:

100Hz

Explanation:

In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz

Answer:

HZ 100 is the right answer hope you like it

How to calculate voltage U1 ?

Please help!

Answers

Answer:

he is a baby art and design

a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car

Answers

Answer:

a) T = 0.5 s

b) v = 1.2π m/s ≈ 3.77 m/s

Explanation:

It makes two revolutions in one second so makes one revolution in ½ second

circumference of the circle is

C = 2πr = 0.6π m

which it traverses in one time period

0.6π m / 0.5 s = 1.2π m/s

To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

What is speed?

Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.

Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.

Mathematically, the formula for speed can be given as

speed= distance/time

Values that are given

Time period= 0.5 s

Circumference = 2πr = 0.6π m

substituting all the given values in the above equation, we get

speed     =0.6π m / 0.5 s

On calculations, we get

              = 1.2π m/s

              =3.77 m/s

Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

To learn more about speed, here:

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The force of gravity is an inverse square law. This means that, if you double the distance between two large masses, the gravitational force between them Group of answer choices weakens by a factor of 4. strengthens by a factor of 4. weakens by a factor of 2. also doubles. is unaffected.

Answers

Answer:

the force decreases by a factor of 4

Explanation:

The expression for the law of universal gravitation is

          F = [tex]G \frac{m_1m_2}{r^2}[/tex]

let's call the force Fo for the distance r

          F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]

They indicate that the distance doubles

          r ’= 2 r

we substitute

          F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]

          F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]

         

          F = ¼ F₀

consequently the correct answer is that the force decreases by a factor of 4

If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.

Let the initial force be F

Let the initial distance apart be r

Thus, we can obtain the final force as follow:

Initial force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = 2r

Final force (F₂) =?

F = GM₁M₂ / r²

Fr² = GM₁M₂ (constant)

Thus,

F₁r₁² = F₂r₂²

Fr² = F₂(2r)²

Fr² = F₂4r²

Divide both side by 4r²

F₂ = Fr² /4r²

F₂ = F / 4

From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4

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two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential​

Answers

Answer:

B has greater potential

Explanation:

We know;

Potential Energy (PE) = mgh

where, m=mass of body

g=acceleration due to gravity

h=height of body

From the formula,

PE is directly proportional to the mass of the body

so the body with greater mass has greater potential.

A wire carrying a 23.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.45 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?

Answers

Answer:

3.55 T

Explanation:

Applying,

F = BILsin∅.............. Equation 1

Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)

Substitute these values into equation 2

B = 2.45/(0.03×23×sin90)

B = 2.45/0.69

B = 3.55 T

1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.

Answers

Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

 

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

[tex]I =\frac{q}{t}[/tex]  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]

Therefore,

t = 1.27 x 10⁹ s

Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?

Answers

Answer:

The correct answer would be - Low pitch.

Explanation:

As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)


Solve for a

Answers

Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?

Answers

Answer:

28 j

Explanation:

because when you add you get 28

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)

Answers

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

Urgent please help me

Answers

1433 km

Explanation:

Let g' = the gravitational field strength at an altitude h

[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]

We also know that g at the earth's surface is

[tex]g = G\dfrac{M_E}{R_E^2}[/tex]

Since g' = (2/3)g, we can write

[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]

Simplifying the above expression by cancelling out common factors, we get

[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]

Taking the square root of both sides, this becomes

[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]

Solving for h, we get

[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]

[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]

The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.

Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?

Answers

Answer:

a) required area is 1.1318 m²

b) the maximum potential difference that can be applied across the compactor is 1931.1 V

Explanation:

Given the data in the question;

dielectric constant εr = 2.35

distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m

dielectric strength = 49.5 MV/m

a)

given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F

To find the Area, we use the following the expression.

C = ε₀εrA / d

we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹²  (F/m)

we substitute

0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A  ] /  7.85 × 10⁻⁵

A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]

A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹

A = 1.1318 m²

Therefore, required area is 1.1318 m²

b)

the maximum potential difference that can be applied across the compactor.

We use the following expression;

⇒ 1/2 × dielectric strength × thickness d

we substitute

⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )

1931.1 V

Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V

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