The greater the surface area of a piece of fission material, the

Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation

hammer [34]

Answer:

the law of conservation of energy.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.

The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.

Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.

One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.

Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.

8 0

3 years ago

What temperature resistivity of tungsten is 4 times resistivity of silver?

mihalych1998 [28]

Answer:

Slightly above 20°c

Explanation:

The resistivity of silver is 1.59 x 10-8Ωm{ as discovered by experiment from scientist}

The resistivity of tungsten at 20°c is 5.6x10-8Ωm

We see the value is just an approximation of 4 times not exactly.

6 0

4 years ago

Precipitation clouds are observed at a distance of about 720 miles to the west of a city. The clouds are seen moving westwards a

tankabanditka [31]

This question may require more information for an accurate answer. If the clouds are observed west of the city and they are moving in a westerly direction it means that the clouds are moving away. Precipitation refers to rainfall, sleet, hail or snow. Since no reference was made to the temperature, season or location of the city it could be rain or snow clouds. The clouds are moving away from the city so the possible answers could be b. or d. Furthermore, by dividing the distance by the speed one would get an answer of twenty four hours. So even if the clouds were moving towards the city it would take a day to reach and precipitation would not be likely in 24 hrs.

3 0

4 years ago

Read 2 more answers

How much force does it take to accelerate a 50.8 kg person at 3.50 m/s^2?

Vikki [24]

Apply Newton's second law to the person's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

m = 50.8kg, a = 3.50m/s²

Plug in and solve for F:

F = 50.8(3.50)

F = 178N

8 0

3 years ago

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

3 years ago