Answer:
[tex]v = \sqrt{\frac{2GM}{r} }[/tex]
The formula for escape velocity where:
G - Gravitational constant (9.81 etc.)
M - the mass of the object the escape should be made from
r - distance to the centre of that mass
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on Earth, and her friend Darien is stationed on the star where the race ends. Bob is in a rocket that can travel at 0.7c; whereas Charlie's rocket can reach a speed of 0.866c. Bob and Charlie start at the same time. Draw space- diagrams from each perspective.
Required:
Estimate how long it takes Bob and Charlie to finish the race from each perspective.
Solution :
The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years
But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.
For Bob,
Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]
So the distance appeared to Bob due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]
[tex]$=7.1 \ Ly$[/tex]
Therefore, the time required to finish the race by Bob is
[tex]$t_b = \frac{L_b}{V_b}$[/tex]
[tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]
= 10.143 year
For Charlie,
Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]
So the distance appeared to Charlie due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]
[tex]$=5 \ Ly$[/tex]
The time required to finish the race by Charlie is
[tex]$t_b = \frac{L_c}{V_c}$[/tex]
[tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]
= 5.77 year
Three resistors (16 ohm, 16 ohm and 8 ohm) are connected in parallel. The equivalent
resistance (Re)
Answer:
4 ohm
Explanation:
The equivalent resistance (Re) of three resistors in parallel is given by;
1/Re = 1/R1 + 1/R2 + 1/R3
Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm
1/Re= 1/16 + 1/16 + 1/8
1/Re= (0.0625) + (0.0625) + (0.125)
Re= 4 ohm
Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very few collisions with other molecules. Express your answer using two significant figures.
Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒ [tex]v=\frac{502.1}{\sqrt{3} }[/tex]
[tex]=289.9 \ m/s[/tex]
The time will be:
⇒ [tex]t=\frac{d}{v}[/tex]
[tex]=\frac{2\times 6}{289.9}[/tex]
[tex]=\frac{12}{289.9}[/tex]
[tex]=0.041 \ sec[/tex]
hence,
⇒ [tex]N=\frac{1}{t}[/tex]
[tex]=\frac{1}{0.041}[/tex]
[tex]=24.39 \ per \ sec[/tex]
Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
How far from the wall is she? What is the period of her up-and-down motion?
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
What is used to represent the magnitude of the force in an FBD?
Answer:
See explanation below
Explanation:
The length of the arrow represents the magnitude/size of the force. The longer it is, the higher the force's magnitude is.
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 830 kg car to rest from an initial speed of 1.4 m/s.
Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W =[tex]F^>[/tex] × [tex]x^>[/tex]
W = [tex]Fxcos\theta[/tex] ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = [tex]Fxcos([/tex] 0° [tex])[/tex]
W = [tex]Fx[/tex] ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
[tex]W_{net[/tex] = ΔKE
= [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu²
we make F, the subject of formula
F = [tex]\frac{m}{2x}[/tex]( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Answer:
[tex]V_2=3.3m/s[/tex]
Explanation:
From the question we are told that:
Distance [tex]d_1=1.4m[/tex]
Tangential speed [tex]V=2.2m/s[/tex]
Distance 2 [tex]d_2=2.1m[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]w=\frac{v}{r}[/tex]
Therefore
[tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]
[tex]V_2=\frac{2.2*2.1}{1.4}[/tex]
[tex]V_2=3.3m/s[/tex]
When an AC source is connected across a 17.0 Ω resistor, the output voltage is given by Δv = (190 V)sin(50πt). Determine the following quantities. (a) maximum voltage V (b) rms voltage V (c) rms current A (d) peak current A (e) Find the current when t = 0.0045 s.
Explanation:
the answer is in the above image
(a) The maximum voltage V is 190 Volts.
(b) The rms voltage V is 95√2 Volts.
(c) The rms current in Amperes is 7.9 A.
(d) The peak current Amperes is 11.18 A.
(e)The current when t = 0.0045 s is 7.26 A.
What is current?The current is the stream of charges which flow inside the conductors when connected across the end of voltage.
Given is an AC source is connected across a 17.0 Ω resistor, the output voltage is given by Δv = (190 V)sin(50πt).
(a) From the given voltage equation, maximum voltage V is 190 Volts.
(b) rms voltage =Vmax/√2
Put the values, we get
Vrms = 190/√2 = 95√2 Volts
(c) rms current = Vrms/Resistance
Put the values, we get
Irms = 95√2 /17
Irms = 7.9 Amperes.
(d) peak current =√2 Irms
Substitute the values, we get
Peak current = 7.9 √2 = 11.18 A
(e) The current when t = 0.0045 s is written as
I = (190 V)sin(50πt)/R
Substitute the values, we have
I = (190 )sin(50π x0.0045)/17
I = 7.26 A.
Learn more about current.
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Consider a 92.0 kg ice skater who is spinning on the ice. What is the moment of inertia of the skater, if the skater is approximated to be a solid cylinder that has a 0.140 m radius and is rotating about the center axis of the cylinder.
Answer:
[tex]I=0.902kg*m^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=92.0kg[/tex]
Radius [tex]r=0.140m[/tex]
Generally the equation for moment of Inertia is mathematically given by
[tex]I = 0.5*m*r^2[/tex]
[tex]I=0.5*(92)*0.14^2[/tex]
[tex]I=0.902kg*m^2[/tex]
In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.
a. True
b. False
If it were to snow in Phoenix in July, which type of Earth scientist would be most
surprised?
Answer:
Economists, I guess.
Explanation:
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm.
a) What is the wave?
b) Wrote the wave function, y(x,t)
c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
Explanation:wave=wavelength×frequency,
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
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. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content of the system.
Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
Short-term memory is active, while long-term memory is:
A dynamic
B
reflective.
c) passive
D
recessive.
Answer:
the answer is b. reflective
Which type of balance is key to sitting?
dynamic
static
bosu
level
Explanation:
bosu
here is your answer
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down? Assume the speed of sound is 343 m/s.
Answer:
The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
Explanation:
We can calculate the speed of the train using the Doppler equation:
[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]
Where:
f₀: is the emitted frequency
f: is the frequency heard by the observer
v: is the speed of the sound = 343 m/s
[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)
[tex] v_{s}[/tex]: is the speed of the source =?
The frequency of the train before slowing down is given by:
[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex] (1)
Now, the frequency of the train after slowing down is:
[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex] (2)
Dividing equation (1) by (2) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex] (3)
Also, we know that the speed of the train when it is slowing down is half the initial speed so:
[tex] v_{s_{b}} = 2v_{s_{a}} [/tex] (4)
Now, by entering equation (4) into (3) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]
[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]
By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:
[tex] v_{s_{a}} = 11.06 m/s [/tex]
Finally, the speed of the train before slowing down is:
[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]
Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
I hope it helps you!
Match the atmospheric energy transfer process that best fits each of the following scenarios:
Warming of the Earth's surface on a sunny day
[ Choose ] convection conduction radiation advection
On a sunny afternoon, you watch cumulus clouds forming
[ Choose ] convection conduction radiation advection
A very shallow layer of air in contact with the ground is warmed
[ Choose ] convection conduction radiation advection
A south wind carries warm air into the central United States
[ Choose ] convection conduction radiation advection
Answer:
a) RADIATION, b) CONVECTION, c) CONDUCTION, d) CONVECTION
Explanation:
In the heating processes there can be three types: conduction, convention and radiation.
The conduction process occurs when the movement of atoms or thermal agitation of molecules creates the transfer of thermal energy.
The convention process occurs when there is a movement of matter creating the transfer of energy
The process of Radiation an electromagnetic wave indexes on a material and is absorbed, creating the process of energy transfer.
Now let's examine each situation>
a)Warming of the Earth's surface on a sunny day
in this case the sunlight heats the earth as it is absorbed, which is why it is a RADIATION process
b) On a sunny afternoon, you watch cumulus clouds forming
in this case the nines rise from the surface, since it is a moving mass, the process is CONVECTION
c) A very shallow layer of air in contact with the ground is warmed
in this case the thermal movement of the layer molecules heat the earth, for which the process of CONDUCTION
d) A south wind carries warm air into the central United States
As we have a movement of a mass of matter, the process is CONVECTION.
Water from a fire hose knocks over a wooden shed. Compared with the pressure within
the water, the pressure exerted against the shed is
a) less.
b) the same.
c) more.
d) nonexistent
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 24 N, and F3 = 13 N. The radial distances are R2 = 0.22 m and R3 = 0.10 m. The forces F2 and F3 are perpendicular to the radial lines labeled R2 and R3. The moment of inertia of the cylinder is 1/2 MR 2/2. Find the magnitude of the angular acceleration of the cylinder about the axis of rotation.
Answer:
α = 13.7 rad / s²
Explanation:
Let's use Newton's second law for rotational motion
∑ τ = I α
we will assume that the counterclockwise turns are positive
F₁ 0 + F₂ R₂ - F₃ R₃ = I α
give us the cylinder moment of inertia
I = ½ M R₂²
α = (F₂ R₂ - F₃ R₃) [tex]\frac{2}{M R_2^2}[/tex]
let's calculate
α = (24 0.22 - 13 0.10) [tex]\frac{2}{12 \ 0.22^2}[/tex]2/12 0.22²
α = 13.7 rad / s²
You are helping to design a new electron microscope to investigate the structure of the HIV virus. A new device to position the electron beam consists of a charged circle of conductor. This circle is divided into two half circles separated by a thin insulator so that half of the circle can be charged positively and the other half can be charged negatively. The electron beam will go through the center of the circle. To complete the design your job is to calculate the electric field in the center of the circle as a function of the amount of positive charge on the half circle, the amount of negative charge on the other half circle, and the radius of the circle.
Answer:
The electric field in the center is [tex]\frac{2k}{\pi R^{2}}\left ( Q_{2}-Q_{1 } \right )[/tex].
Explanation:
Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that's 4.0 m high. On the slope, Monica's bike pulls on the trailer with a constant force of 9.0 N. They start out at the bottom of the slope with a speed of 5.0 m/s.
What is their speed at the top of the slope?
Answer:
Explanation:
From the given information:
total mass = 25 kg
distance d = 100 m
height = 4.0 m
Force F = 9.0 N
The speed at (bottom) u = 5.0 m/s
Using the concept of energy conservation;
[tex]\dfrac{1}{2}mu^2 + W = \dfrac{1}{2}mv^2 + mgh[/tex]
divide both sides by m
[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]
multiply both sides by 2
[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]
[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]
Recall that:
W = Fd
∴
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]
[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]
[tex]v^2 =(25.0 ) - 78.48 +72[/tex]
[tex]v^2 = 18.52 \ m^2/s^2[/tex]
[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]
v = 4.30 m/s
Match each planet to an accurate characteristic
Answer:
venus - 2
earth - 3
mars - 4
mercury - 1
Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?
Answer:
Rahul's weight
Explanation:
In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.
Using formulas, Rahul's weight is equal to
W=mg
where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).
A car has a mass of 900 kg is accelerated from rest at a rate of 1.2 m/s calculate the time taken to reach 30/s
Answer:
12+2=24+30+2=66
Explanation:
Please help, only answer if your 1000% correct im in summer school and need to pass this class
Action reaction forces never cancel each other out because .............................? *
1 point
they do not act on objects.
they act on the same objects
they act on different objects.
none of the above
A force can be described as? *
1 point
push or pull action
pulling direction
direction of change
push action only
Using EquatIO calculate the resultant force for the horizontal (x-axis) component. Show your calculation step. *
2 points
Captionless Image
A player hits a baseball with a bat. The action force is the impact of the bat against the ball. The reaction force is .......................? *
1 point
Captionless Image
the grip of the player's hands on the ball
the weight of the ball
the air resistance of the ball
the force of the ball against the bat
Which law is this? "An object at rest tends to stay at rest and an object in motion tends to stay in motion unless acted upon by an unbalanced force." *
1 point
Newtons third law
Newtons second law
Newtons first law
None of the above
Which law is this? "If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A." *
1 point
None of the above
Newtons second law
Newtons first law
Newtons third law
What can be said about a force on an object that is not stationary? *
1 point
Captionless Image
there are no forces acting on the object
there are unequal forces acting on the object
there is only one force acting on the object
there are equal and opposite acting on the object
Using EquatIO calculate the resultant force for the vertical (y-axis) component. Show your calculation step. *
2 points
Captionless Image
Identify the different force acting on a moving vehicle shown in the mage below. *
5 points
Captionless Image
A B C D E
Reaction force
Weight
Friction
Air resistance
Thrust
Reaction force
Weight
Friction
Air resistance
Thrust
Which law is this? "The acceleration of an object is directly related to the net force and inversely related to its mass." *
1 point
None of the above
Newtons third law
Newtons first law
Newtons second law
If forces acting on an object are unbalanced, what can occur to the object? *
1 point
acceleration
all of the above
change of shape
deceleration
change in direction
If a player hits a baseball with a force of 870N Calculate the acceleration (state the units) of the ball . Show your calculation step using only EquatIO. *
2 points
Captionless Image
Action reaction forces never cancel each other out because .............................? *
1 point
they act on different objects.
none of the above
they act on the same objects
they do not act on objects.
because they work on the same object
Explanation:
A push and pull factor
A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.
Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation of charge the law of conservation of energy. the law of conservation of angular momentum. Newton's second law
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.
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm