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lesya692 [45]
2 years ago
15

Chose the word that correctly completes the following sentence: Kirchoff's junction law expresses the conservation of ____. a) m

ass charge b) volume c) energy quilet
Physics
1 answer:
Radda [10]2 years ago
5 0

Answer:A- mass charge.

This can also be called current.

Explanation:

This is Kirchhoff’s 2nd law.

Kirchhoff’s junction law states that the sum of current(mass charge) flowing in and out of the junction must be equal to zero. This law emphasizes conservation of charge and energy. Charge is also a form of energy and it can neither be created nor destroyed.

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Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why d
nydimaria [60]

<u>Answer:</u>

<h3>During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.</h3>

<u>Sources:</u>

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

6 0
2 years ago
A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
Y_Kistochka [10]

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The speed with which the liquid sulfur left the volcano is 529.15 m/s

7 0
2 years ago
A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if
saul85 [17]

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

F = 0.15×10×2×sin0

F = 0

Goodluck

7 0
2 years ago
Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass
zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value into the formula

8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v

v=\dfrac{-19.375}{14.25}

v=-1.35\ m/s

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2

K.E_{i}=239.96\ J

We need to calculate the final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2

K.E_{f}=12.98\ J

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

\Delta K.E=K.E_{f}-K.E_{i}

Put the value into the formula

\Delta K.E=12.98-239.96

\Delta K.E=-226.98\ J

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0
2 years ago
Read 2 more answers
A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char
Aleksandr [31]

Answer:

Ek =  8,79 [J]

Explanation:

We are going to solve this problem, using  the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁  = Ek  + U₁

As charge are at rest Ek = 0

And  U₁ has two components

U₁₂   = K * Q₁*Q₂ / 0,4          and    U₃₂  = K*Q₃*Q₂ / 0,6

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4  ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6  ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J]   ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then   E₁ = E₁₂ + E₃₂    

E₁ = 10,5 [J]

At  the moment of Q₂ passing x = 40 cm  or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂  =  U₁₂  + U₃₂

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7   ⇒  U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7       U₁₂ = 7,71 [J]

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3  ⇒ U₃₂ = -  9*20*10*10⁻³/0,3

U₃₂ = -  9*20*10⁻²/0,3  

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as  

E₂  = Ek + U₂  and  E₂ = E₁

Then

Ek + U₂ = E₁

Ek =  10,5 - U₂    

Ek  = 10,5 - 1,71

Ek =  8,79 [J]

5 0
1 year ago
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