A 12 V battery is connected across a device with variable resistance. As the resistance of the device increases, determine whether the following quantities increase, decrease, or remain unchanged. HINT (a) The current through the device. increases decreases remains unchanged (b) The voltage across the device. increases decreases remains unchanged (c) The power consumed by the device. increases decreases remains unchanged

Answers

Answer 1

Answer:

a) DECREASE, b) INCREASE, c)  power remains constant

Explanation:

The resistance and the battery are connected in series.

a) How the current changes when the resistance changes

               V = I R

               I = V / R

               I = 12 / R

if the resinification increases, the current must DECREASE

b) when you check the expression

                V= I R

, if R increases the voltage of the INCREASE

c) the power in an electric circuit is

               P = I V

               

Let's analyze this expression, the voltage increases linearly with the increase in resistance and the current decreases linearly with the increase in r, for which the two effects are compensated and the dissipated power remains constant


Related Questions

You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a battery. The first piece has a length L and cross-sectional area A. The second is twice as long as the first, but has the same thickness. The third is the same length as the first, but has twice the cross-sectional area. Rank the wires in order of which carries the most current (has the lowest resistance) when connected to batteries with the same voltage difference.

Rank the wires from most current (least resistance) to least current (most resistance).
a. Wire of Lenght L and area A
b. Wire of Lenght 2L and area A
c. Wire of Lenght L and area 2A

Answers

Answer:

The answer is below

Explanation:

The resistance of a wire is directly proportional to the length of the wire and inversely proportional to its area. The resistance (R) is given by:

[tex]R=\frac{\rho L}{A}\\\\where\ L=length \ of\ wire,A=cross\ sectional\ area, \rho=resistivity\ of\ wire.[/tex]

Let us assume that all the wires have the same resistivity.

a) Wire of Length L and area A

[tex]R_1=\frac{\rho L}{A}[/tex]

b) Wire of Length 2L and area A

[tex]R_2=\frac{\rho *2L}{A}=2R_1[/tex]

C) Wire of Length L and area 2A

[tex]R_3=\frac{\rho L}{2A}=\frac{1}{2}R_1[/tex]

Therefore the wire of least resistance is R3 and R2 has the highest resistivity.

R₃ < R₁ < R₂

Therefore, the ranking of the wires from most current (least resistance) to least current (most resistance) is:

R₃ < R₁ < R₂

Describe Kinetic Energy and Potential Energy (in your own words please!!)

Answers

Answer:

Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.

Explanation:

Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide details about each piece in her model. In order for her model to be realistic, which of the following should she include in her representation of the Sun?

Answers

Answer:

she should write about how big is it and what the sun looks and how far away is it from earth.

Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation

Answers

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

how much heat energy is needed to raise the temperature of 2.0 kg of concrete from 10c to 30c

Answers

50 because read step by step explanation

Help me !!!

What is the velocity of a ball with
a momentum of -4.5 kg*m/s and a
mass of 0.45 kg?

Answers

Answer:−4.05

Explanation:

Define the following soil conservation technique. Make sure to include pro’s and con’s of this method.
(Terraces)

Answers

Answer: Terraces on moderate to steep irregular slopes pro- ... sure of infertile or toxic soils. ... Following are terms used to define distances mea- ... the soil in the entire field will be disturbed to con-.

Explanation:

A material through which electricity cannot flow is called:
a conductor
an insulator
an electrode
a wet cell

Answers

Answer:

el conductor

Explanation:

gracias por los puntitoss

Answer:

conductor

Explanation:

Collisions question plzz help mee

Answers

ban yarichin b club pls

Do you believe you can create a Controlled experiment without an Observational Study? Why or Why not. Include scientific evidence to support your response. PLEASE HELP I BEG YOU.

Answers

Answer:

No, it is very unlikely to perform a controlled experiment, because you need to observe the amount or anything from something. Consider someone on the busy street of a New York neighborhood asking random people that pass by how many pets they have, then taking this data and using it to decide if there should be more pet food stores in that area.

what is the difference between mass and weight

Answers

Answer:

The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.

Answer:

Explanation:

The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).

A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg and the mountain's height was 1000 m)​

Answers

Answer:

no kinetic energy

hope this helps! :-D

Explanation:

the monk is not moving

A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?

Answers

Answer:

:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)

To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)

If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is


364N

185N

173N

73N



Answer and I will give you brainiliest

Answers

73 Newton is the correct answer

The distance to the other planets in the Solar System from the Earth depends on where they are in their orbit. The closest that Mars gets to the Earth is 0.5 AU. At this time it is observed to have an angular diameter of 18 arcseconds. The furthest it gets from Earth is 2.5 AU. What would its angular diameter be for this observation

Answers

Answer:

3.6 arcsec

Explanation:

 angular diameter = diameter / distance

diameter is constant

so angular diameter ∝ 1 / distance

angular diameter = k / distance

For first case ,

18 = k / .5

for second case let angular diameter be D .

D = k / 2.5

dividing ,

D / 18 = .5 / 2.5 = 1 / 5

D = 18 / 5 = 3.6 arcsec

3.6 arcsec is the answer .

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M and positive charge Q hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk.

Required:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
b. At what height h does the sphere hover?

Answers

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Connecting math to physics

Answers

Interesting I guess not much you put

Answer:

wat

Explanation:

A Car is moving at a speed of 20 m/s. How Much Distance it will cover in 1 min? Express the answer in km.

Answers

Answer:

d=20m/sx60s=1200m=1200/1000Km=1.2km

Explanation:

A student pulls on a cart and applies a 20N force at an angle of 25 degrees above the horizontal to
move a crate a distance of 10m across the floor. How much work does the student do?

Answers

Answer:

W = 181.26 J

Explanation:

Given that,

The force acting on the cart, F = 20 N

It is at an angle of 25 degrees above the horizontal to  move a crate a distance of 10m across the floor.

We need to find work done by the student. The work done by the student is given by :

[tex]W=Fd\cos\theta\\\\W=20\times 10\times \cos25\\W=181.26\ J[/tex]

So, the required work done is 181.26 J.

An 80-kg firefighter slides down a fire pole. After 1.3 seconds of sliding, the firefighter is sliding at a velocity of 6.5 m/s, straight down the pole. Once this velocity is reached, the firefighter grips the pole so that the force of friction exerted by the firefighter's hands on the pole is equal to the force of gravity. At this point what is the downward acceleration of the firefighter

Answers

Answer:

a= 0

Explanation:

In the vertical direction, if the friction force (directed upward) is equal to the force of gravity (downward) this means that no net force is acting on the firefighter.According to Newton's 2nd Law, if no net force is present, the acceleration in this direction is just zero, as follows:

       [tex]F_{net} = m*a = 0 (1)[/tex]

      ⇒ a = 0

Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide details about each piece in her model. In order for her model to be realistic, which of the following should she include in her representation of the Sun?


She should show that sunspots can ve seen as white areas on the Sun's surface.


She should explain that the Sun is made up of gaseous layers that surround an iron core.


She should show that the Sun revolves around the planets, determining the length of the year.


She should explain that the Sun rotates, even though different parts rotate at different rates.

Answers

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart that is initially at rest. In trial 2, the setup is identical except the carts stick together during the collision. How does the speed of the two-cart system's center of mass change, if at all, during the collision in each trial

Answers

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]

in this case we have two bodies

          x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]

          x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = [tex]\frac{m_1}{m_1+m_2}[/tex]  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         [tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)

final

         [tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v

         v_{cm f} =  [tex]\frac{m_1}{M^2} v_o[/tex]

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]

           

           

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]

             v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]

now let's find the velocity of the center of mass

initial

          [tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀

final

          [tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex]  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = [tex]\frac{1}{M}[/tex] [  [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

1.0 kg clay ball traveling straight down at -10 m/s hits the floor and and sticks on it

Answers

Answer:

What am I suppose to solve

Explanation:

what is the angle between 3i-2j-3k and the negative x axis

Answers

Answer:

Um its the vbuck card on the 3 thrid level

Explanation:

Bc its a vbuck card you know sihdg;aig

PHYSICS QUESTION PLS HELP

Answers

The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is

mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J

The total energy is the same, 970,200 J.

Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.

At point B, the coaster has dropped to a height of 10 m, so it has PE

mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J

which means it must have KE

970,200 J = KE + 294,000 J   →   KE = 676,200 J

which gives the coast a speed v at point B of

1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J   →   v21.2 m/s

At point C, the coaster has a speed of 16.0 m/s, so it has KE

1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J

and hence PE

970,200 J = 384,000 J + PE   →   PE = 586,200 J

This lets us determine the height h at C:

mgh = (3000 kg) (9.80 m/s²) h = 586,200 J   →   h ≈ 19.939 m

which means the loop has diameter h - 10 m ≈ 9.94 m.

At point D, the coaster is 15 m above the ground so its PE at D is

mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J

and so its KE is

970,200 J = KE + 441,000 J   →   KE = 529,200 J

and hence has speed v at D

1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J   →   v18.9 m/s

why would the bulb not light?

Answers

are you a dmbss? the bulb and wire must be connected to both end

HELP ! ILL MARK BRAINLIEST HELP ASAP

Answers

Answer:

A

Explanation:

In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.

Hope this helps!

What is the value of the angle of inclination of the slide?



Answer and I will give you brainiliest

Answers

Answer:

63°

Explanation:

90-27 =63

I am not completely sure

6. A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at
30.00°C). The final temperature of the system is 40.22°C. Find the specific heat of iron.
90​

Answers

Answer:

[tex]c=0.45\ J/g^{\circ} C[/tex]

Explanation:

Given that,

A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at  30.00°C). The final temperature of the system is 40.22°C.

We need to find the specific heat of iron.

It can be calculated as:

Cooler water gains = hot metal loses

mc∆T = - mc∆T

Put all the values,

[tex]200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]

So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]

why no tempature can be lower than 0 kelvin

Answers

Answer:

At zero kelvin (minus 273 degrees Celsius) the particles stop moving and all disorder disappears. Thus, nothing can be colder than absolute zero on the Kelvin scale. Physicists have now created an atomic gas in the laboratory that nonetheless has negative Kelvin values.

Explanation:

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