Answers
F = 153 N
[tex]\theta = 11.3°[/tex]
Explanation:
Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:
[tex]F_{x}[/tex] = 350 N - 200 N = 150 N
[tex]F_{y}[/tex] = 180 N - 150 N = 30 N
The magnitude of the resultant force F is given by
[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]
[tex]\:\:\:\:\:\:=153\:N[/tex]
To find the direction [tex]\theta[/tex], we use
[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]
or
[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]
Related Questions
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.
Answers
Answer:
The electric field in the center is [tex]\frac{2k}{\pi R^{2}}\left ( Q_{2}-Q_{1 } \right )[/tex].
Explanation:
Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cladding have the same refractive indices.
a. True
b. False
Answers
Answer:
False
Explanation:
Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.
A 2.7 kg mass is connected to a spring (k=159 N/m) and is sliding on a horizontal frictionless surface. The mass is given an initial displacement of +19 cm and released with an initial velocity of -13 cm/s. Determine the acceleration of the spring at t=3.4 seconds.
Answers
Answer:
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
Explanation:
The mass-spring system experiments a Simple Harmonic Motion, whose kinematic expression is the following:
[tex]x(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t + \phi\right)[/tex] (1)
Where:
[tex]x(t)[/tex] - Position, in centimeters.
[tex]A[/tex] - Amplitude, in centimeters.
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]m[/tex] - Mass, in kilograms.
[tex]\phi[/tex] - Phase, in radians.
By Differential Calculus, we derive expression for the velocity and acceleration of the mass-spring system:
[tex]v(t) = -\sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (2)
[tex]a(t) = -\frac{k\cdot A}{m}\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (3)
Where [tex]v(t)[/tex] and [tex]a(t)[/tex] are the velocity and acceleration of the system, in centimeters per second and centimeters per square second, respectively.
If we know [tex]m = 2.7\,kg[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]t = 0\,s[/tex], [tex]x(t) = 19\,cm[/tex] and [tex]v(t) = -13\,\frac{cm}{s}[/tex], then we have the following system of nonlinear equations:
[tex]A \cdot \cos \phi = 19[/tex] (1)
[tex]-7.674\cdot A \cdot \sin \phi = - 13[/tex] (2)
If we divide (2) by (1):
[tex]-7.674\cdot \tan \phi = -0.684[/tex]
[tex]\tan \phi = 0.089[/tex]
[tex]\phi = \tan^{-1} 0.089[/tex]
[tex]\phi \approx 0.028\,rad[/tex]
By (1), we get the value of the amplitude:
[tex]A = \frac{19}{\cos \phi}[/tex]
[tex]A = 19.075\,cm[/tex]
If we know that [tex]A = 19.075\,cm[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]m = 2.7\,kg[/tex], [tex]t = 3.4\,s[/tex] and [tex]\phi \approx 0.028\,rad[/tex], then the acceleration of the spring is:
[tex]a(t) = -29339.947\,\frac{cm}{s^{2}}[/tex]
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answer to two significant figures.
The spring constant, k, is
StartFraction N over m EndFraction.
Answers
Answer: 4800 N/m
Explanation:
Given
mass of rock [tex]m=2.5\ kg[/tex]
Height of cliff [tex]h=32\ m[/tex]
compression in the spring [tex]x=57\ cm[/tex]
Here, potential energy is converted into kinetic energy which in turn converts to elastic potential energy of the spring
[tex]\Rightarrow mgh=\dfrac{1}{2}kx^2\\\\\Rightarrow k=\dfrac{2mgh}{x^2}\\\\\Rightarrow k=\dfrac{1568}{0.3249}\\\\\Rightarrow k=4826.100\approx 4800\ N/m[/tex]
Answer:
480
Explanation:
2021
A car on the roller coaster begins with 0j of kinetic energy and 12,928j of potential energy and finishes the track with 3,715j of potential energy. How much kinetic energy dose the car finish with????
Answers
Answer:
9213 J
Explanation:
Change in Kinetic energy = Change in Potential energy
= 12,928J - 3715J
=9213 J
For more assistance: +1 (304) 223-3136
In addition to acceleration, what else will be a maximum at the amplitude for SHM?
A. Potential energy
B. Kinetic energy
C. Nuclear energy
D. Chemical energy
Answers
It is Potential energy's
Answer the following questions
1. Heat in liquid travels from
a) bottom to top
b) top to bottom
c) left to right
d) right to left
2. The direction of flow of heat is
a) always from a cooler body to a hotter body
b) always from a hotter body to cooler body
c) always from a body at a lower temperature to a body at a higher temperature
d) all the above
3. A cold steel spoon is dipped in a cup of hot milk. The steel spoon transfer the heat to its other end by the process of
a) convection
b) conduction
c) radiation
d) none of the above
Answers
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal speed be right before it hits the ground?
A. 15 m/s
B. 0 m/s
C. 30 m/s
D. 40 m/s
Answers
Answer:
C
Explanation:
horizintal speed stays same
only vertical speed changes
so H speed will stay 30 m/s
The more bonds an atom can make, the more likely it is to combine with other aton
Which element is most likely able to make the greatest variety of bonds?
nitrogen
hydrogen
oxygen
carbon
Answers
Answer:
carbon has tetra valency while oxygen can make 2 bonds while nitrogen can make 3 bonds and hydrogen can make 1 bond so i think answer is definatly carbon
Explanation:
calculate the work function that requires a 455 nm photon to eject an electron of a value 0.73 eV
Answers
Answer:
W = 2 eV
Explanation:
Given that,
The wavelength of a photon = 455 nm
The kinetic energy of a photon, K = 0.73 eV
We need to find the work function of the electron. It can be solved using Einstein's equation such that,
[tex]W=E-K[/tex]
E is the energy of the photon
So,
[tex]W=\dfrac{hc}{\lambda}-K\\\\W=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{455\times 10^{-9}\times 1.6\times 10^{-19}}-0.73\\\\W= 2.73\ eV-0.73\ eV\\\\W=2\ eV[/tex]
So, the work function of the metal is 2 eV.
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and speed (in m/s) when t = 5. f(t) = 18 + 48/t + 1
Answers
Answer:
The velocity of the particle = -1.92 m/s
The speed of the particle = 5.72 m/s
Explanation:
Given equation of motion;
[tex]f(t) = 18 \ + \ \frac{48}{t} \ + \ 1[/tex]
Velocity is defined as the change in displacement with time.
[tex]V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s[/tex]
The distance traveled by the particle in 5 s:
[tex]s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m[/tex]
The speed of the particle when t = 5s
[tex]Speed = \frac{28.6}{5} = 5.72 \ m/s[/tex]
what is the resistance of a bulb of 40w connected in a line of 70v
Answers
Answer:
122.5 ohm
Explanation:
Given :
P=40 w
V= 70 V
R=?
Resistance can be calculated as :
[tex]P=\frac{V^{2} }{R} \\40=\frac{(70)^{2} }{R}\\40=\frac{4900}{R} \\R=\frac{4900}{40} \\R=122.5 ohm[/tex]
Therefore, resistance of the bulb will be 122.5 ohm
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1. Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.
2. The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinite
Answers
Answer:pp
Explanation:
ii
Which type of balance is key to sitting?
dynamic
static
bosu
level
Answers
Explanation:
bosu
here is your answer
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answers
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
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.
Answers
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
You sit on ice and shove a heavy box with your feet with a given force. What will you and the box share? *
A) Same acceleration
B) equal and opposite acceleration
C) the equal and opposite force
D) same force
explain please
Answers
Answer:
the correct answer is C
Explanation:
In this exercise we will analyze the situation.
When the person is on the ice, the friction coefficient is very small, if the box is in a place where there is no ice, the coefficient is different, so the friction force on each body is different.
Therefore the acceleration that each body acquires is different.
If we apply the conservation of momentum, each body moves in the opposite direction, but with different speeds.
If we use Newton's third law, the force applied to each body has the same magnitude and opposite direction, which is why the force is of the action and reaction type
Consequently the correct answer is C
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force. Ultimately the board breaks at an applied force of 870 N and a deflection of 1.4 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a distance of 1.2 cm, what is the average force on the hand?
Answers
Answer:
A. The hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand = 1025 N
Explanation:
A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.
Workdone = force × distance
Minimum force required = 870 N;
Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M
Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J
This work done = Kinetic energy of the hand
Kinetic energy = mv²/2 ; where m is mass and v is velocity
Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J
v² = 2 KE/m
v = √2KE/m
v = √(2 × 12.18/0.50)
v = 6.98 m/s
Therefore, the hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand
This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration
From the equation of motion, a = v² - u²/2s
At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m
a = 0² - 6.98²/ 2 × 0.012
a = -2030 m/s²
Force = 2030 m/s² × 0. 50 kg = 1015 N
Therefore, Average force on the hand = 1025 N
Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter cup containing 844 g of water at 14.6°C?
Answers
Answer:
T = 20.84°C
Explanation:
From the law of conservation of energy:
Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water
[tex]m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a[/tex]
where,
[tex]m_c[/tex] = mass of copper = 227 g
[tex]m_w[/tex] = mass of water = 844 g
[tex]m_a[/tex] = mass of aluminum = 155 g
[tex]C_c[/tex] = specific heat capacity of calorimeter = 385 J/kg.°C
[tex]C_w[/tex] = specific heat capacity of water = 4200 J/kg.°C
[tex]C_a[/tex] = specific heat capacity of aluminum = 890 J/kg.°C
[tex]\Delta T_c[/tex] = change in temperature of copper = 283°C - T
[tex]\Delta T_w[/tex] = change in temperature of water = T - 14.6°C
[tex]\Delta T_a[/tex] = change in temperature of aluminum = T - 14.6°C
T = equilibrium temperature = ?
Therefore,
[tex](227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}[/tex]
T = 20.84°C
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 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:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?
Answers
Answer:
a) [tex]d=11m[/tex]
b) [tex]T=4.68s[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=22m[/tex]
Velocity [tex]v=4.6m/s[/tex]
a)
Generally the equation for distance between her and the wall d is mathematically given by
Since
The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]
Therefore
[tex]d= \frac{\22}{2}[/tex]
[tex]d=11m[/tex]
b)
Generally the equation for her up-and-down motion is mathematically given by
[tex]T=\frac{22}{4.7}[/tex]
[tex]T=4.68s[/tex]
what is effort arm
don't say the answer of gogle
Answers
Answer:
effort arm mean the use of any work by using your hand force motion or by hand power
If distance between two charges increased by 2 times then force
Answers
Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.64 of her regular weight. Calculate the magnitude of the acceleration of the elevator.
Answers
Answer:
The downwards acceleration is 3.53 m/s2.
Explanation:
Let the true weight is m g.
The reading of the balance, R = 0.64 mg
Let the acceleration is a.
As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.
Use Newton's second law
m g - R = m a
m g - 0.64 m g = m a
0.36 g = a
a = 3.53 m/s2
help plzzzzzzzzzzzz ?
Answers
Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
who won the battle of Buxar
Answers
Answer:
British East India Company
The battle was fought at Buxar, a "small fortified town" within the territory of Bihar, located on the banks of the Ganga river about 130 kilometres (81 mi) west of Patna; it was a decisive victory for the British East India Company.
To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direction of rain...
Answers
Answer:
The actual speed of the rain is 5 m/s and its direction is -53.13°
Explanation:
The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.
V = v + v'
We add these vectorially.
Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j
So, V = v + v'
V = 3i + (-4j)
V = 3i - 4j
So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s
The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°
So, the actual speed of the rain is 5 m/s and its direction is -53.13°
HELP ME PLEASE!!!
Which 2 statements are true about this chemical reaction that forms acid rain?
Answers
Answer:
B.
Explanation:
HNO2 is less stable thus dissociates easily to HNO3 + NO + H2O while HNO3 is a strong acid. Thus when they react with H2O they form acid rain
Answer:
B
Explanation:
dont have one just trust me
Which of the following is a good example of a contact force?
ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up
Answers
Answer:
A bridge suspended by cables
Explanation:
Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.
An empty parallel plate capacitor is connected between the terminals of a 8.85-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answers
Answer:
The new voltage is 17.7 V.
Explanation:
Voltage, V = 8.85 V
The spacing is doubled.
When it is disconnected, the charge remains same,
q = C V ..... (1)
where, C is the capacitance, V is the voltage.
The capacitance is inversely proportional to the distance between the two plates.
So, when the spacing is doubled, the capacitance is halved.
Let the new voltage is V'.
C V = C' V'
C x 8.85 = C/2 x V'
V' = 17.7 V
When you release some of the paint from a spray paint can and the can remains at the same temperature, which gas law does this represent?
Answers
Answer:
Boyle's law.
Explanation:
Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
[tex] PV = K[/tex]
[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]
Where;
P1 is the original pressure.
P2 is the final pressure.
V1 is the original volume.
V2 is the final volume.
Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.
