Supposing that 9 six-sided dice are rolled. Then, the entropy of rolling a sum of 10 would be 3.03e-23 J/K.
Calculating entropy of the dice:
The entropy of rolling a sum of 10 can be calculated using the formula for entropy:
S = k ln(W)
Where k is the Boltzmann constant and W is the number of microstates corresponding to the given macrostate. In this case, the macrostate is the sum of the outcomes of 9 dice, which can range from 9 to 54. To calculate W, we need to count the number of ways in which we can get a sum of 10 with 9 dice.
One way to approach this is to use a generating function, which is a polynomial that encodes the probabilities of different outcomes. The generating function for a single die is:
(\(x^{1}\) + \(x^{2}\) + \(x^{3}\) + \(x^{4}\) + \(x^{5}\) + \(x^{6}\))
(where x represents the variable that tracks the outcome of the die)
The generating function for the sum of 9 dice is obtained by multiplying the generating function of a single die by itself 9 times:
(\(x^{1}\) + \(x^{2}\) + \(x^{3}\) + \(x^{4}\) + \(x^{5}\) + \(x^{6}\))^9
Expanding this polynomial would give us all the possible ways to obtain a sum of 10 with 9 dice, but that would be a tedious task.
The expression in parentheses is the generating function for the sum of 3 dice (multiplied by x to shift the outcomes by 6), raised to the power of 3. We can use the binomial theorem to expand this expression and get the coefficients of the polynomial, which represent the number of ways to get each sum. All the other coefficients are zero.
Therefore, the number of microstates corresponding to a sum of 10 is W = 27. Plugging this into the formula for entropy, we get:
S = k ln(W) = k ln(27) ≈ 3.03e-23 J/K
So the answer is (b).
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Suppose that, through the malicious act of an eight-dimensional alien being, the strong force was suddenly turned off throughout the universe. What would happen almost immediately to atoms
If the strong force was suddenly turned off throughout the universe, atoms would almost immediately disintegrate as the strong force is responsible for holding the nucleus of an atom together.
This would result in a release of energy as the protons and neutrons in the nucleus repel each other due to the electromagnetic force. The energy released would be so great that it would cause a massive explosion, similar to a nuclear explosion. Furthermore, the absence of the strong force would also affect the stability of neutron stars and supernovae, which rely on the strong force to maintain their structure. Overall, the absence of the strong force would result in a catastrophic and potentially apocalyptic scenario for the universe.
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(b) 32g of dry ice was added to 200g of water at 25°C in a beaker of negligible heat
capacity. When all ice had melted the temperature of water was found to be 10°C. 9 (Take specific
heat capacity of water to be 4.0J/gk)
(i) Calculate the heat lost by water
The heat lost by water is equal to the heat gained by ice here. The heat lost from water for a temperature change of 25 to 10 degree Celsius is 12300 J.
What is calorimetric ?Calorimetry is an analytical technique used to determine the heat energy absorbed or evolved by a system. The calorimetric equation relating the heat energy q with the mass m, specific heat c and the temperature difference ΔT is :
q = m c ΔT
Here, the heat energy gained by the dry ice is equal to the heat lost from water.
temperature difference for water = 25- 10 °C = 15°C
thus, 15°C is lost from water.
mass of water = 200 g
q =200 g × 4.12 J/°C g × 10°C = 12300 J
Therefore, the heat energy lost from water is 12300 J.
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In the heat equation what does Q represent
The heat equation is Q = mcT
Q is the total amount of heat transferred to or from something.
If the velocity v of a particle moving along a straight line decreases linearly with its displacement from 20m/s to a value approaching zero at s = 30m , determine the acceleration of the particle when s =15 m
The acceleration of the particle when s = 15 m is -2/3 m/s².
To determine the acceleration of the particle when the displacement is 15 m, we need to consider the given information that the velocity of the particle decreases linearly with its displacement. Let's break down the problem into steps:
1. Identify the initial and final velocities: The problem states that the velocity decreases from 20 m/s to a value approaching zero at s = 30 m.
2. Calculate the rate of change of velocity: Since the velocity decreases linearly with displacement, we can determine the rate of change of velocity using the formula:
Acceleration = (Change in velocity) / (Change in displacement)
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Initial displacement (s₁) = 0 m (assuming the particle starts at s = 0)
Final displacement (s₂) = 30 m
Change in velocity = v - u = 0 - 20 = -20 m/s
Change in displacement = s₂ - s₁ = 30 - 0 = 30 m
Acceleration = (-20 m/s) / (30 m) = -2/3 m/s²
3. Determine the acceleration at s = 15 m: Since the velocity is decreasing linearly with displacement, we can assume that the acceleration is constant throughout. Therefore, the acceleration of the particle when the displacement is 15 m is also -2/3 m/s².
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For the force field F ( p ) = - ( k / p² ) e , find the angular frequency of polar vibrations around stable circular motion . Show that this frequency is equal to the rotational angular frequency of circular motion.
The force field F(p) = -(k/p^2)e represents an inverse square force law, where k is a constant and p is the radial distance from the origin. To find the angular frequency of polar vibrations around stable circular motion, we can consider the equation of motion for a particle in a circular orbit.
For circular motion, the centripetal force is given by Fc = mω^2p, where m is the mass of the particle, ω is the angular frequency, and p is the radial distance.Equating the centripetal force to the force field, we have:-
(k/p^2)e = mω^2p
Simplifying, we get:
k/p^3 = mω^2
Taking the square root and rearranging, we find:
ω = √(k/(mp^3))
This is the angular frequency of polar vibrations around stable circular motion.To show that this frequency is equal to the rotational angular frequency of circular motion, we can recall that in circular motion, the rotational angular frequency ω_rot = v/p, where v is the tangential velocity.Since the tangential velocity v = ω_rot * p, we can substitute this into the equation for ω:
ω = √(k/(m * p^3)) = √(k/(m * p^2)) * (1/p)
Using the relation v = ω_rot * p, we can rewrite the expression as:
ω = √(k/(m * p^2)) * (1/p) = √(k/(m * p^2)) * (1/(v/p))
Simplifying, we get:
ω = √(k/(m * p^2)) * (p/v) = v/√(m * k)
This shows that the angular frequency ω is equal to the rotational angular frequency ω_rot. Therefore, the angular frequency of polar vibrations around stable circular motion is equal to the rotational angular frequency of circular motion.
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The rate of change of momentum?
Answer:
Answer in below and plz mark me as brainlist plz
Explanation:
The rate of change of momentum =tm(v−u) Rate of change of momentum = force applied. Force∝tm(v−u) Velocity is the rate of change of displacement and acceleration is the rate of change of velocity. Impulse is a change in momentum
The rate of change of momentum =tm(v−u) Rate of change of momentum = force applied. Force∝tm(v−u) Velocity is the rate of change of displacement and acceleration is the rate of change of velocity. Impulse is a change in momentum
Which measurement is a potential difference?
O A. 115 N
O B. 115 C
O C. 115 J
O D. 115 V
Answer:
option d is answer because pd is measured in volt.
A 0. 8-kg ball is thrown into the air with an initial velocity of 10 m/s. How much kinetic energy does the ball have when it is released from the hand?
The kinetic energy of the ball when it is released from the hand is 40 Joules (J).
The kinetic energy of a 0.8-kg ball when released from the hand with an initial velocity of 10 m/s will be 40 Joules (J)
Kinetic energy is the energy possessed by a moving object. It is calculated by multiplying half of the mass of an object by its velocity squared. The formula is as follows: \(K. E = 1/2 * m * v^2\)
The scalar quantity of kinetic energy is expressed in joules (J) units. An object's mass and velocity both affect its kinetic energy. It is a fundamental idea in physics and is crucial to comprehending how energy behaves and changes throughout many systems and occurrences.
where:
K. E is the kinetic energy of the object,m is the mass of the object,v is the velocity of the object.
Using the given values, we can calculate the kinetic energy of the ball.
K. E =\(1/2 * m * v^2*K. E\)= 1/2 * 0.8 kg * \((10 m/s)^2\)
K. E = 1/2 * 0.8 kg * \(100 m^2/s^2K. E\) = 40 J
Therefore, the kinetic energy of the ball when it is released from the hand is 40 Joules (J).
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The work function for chromium metal is 4.37 ev. What wavelenght of radation must be used to eject?
Wavelength of radiation must be used to eject electrons with a velocity of 2300 km/s is 63.9nm.
Equation :Einstein's equation for the photoelectric effect
hv = A + W
Where,
Energy of photon : E = hvx = hxc/λ
The work function for chromium is A
Planck's constant h= 6.626X 10⁻³⁴ Js
speed of light c = 2.998X10⁸ m/s
Velocity of 2300 km/s = 2300 X 10⁶ m/s
The kinetic energy of emitted electron
W = mv²/2 = (9.10939 X 10⁻³¹ kg X (2300X10⁶ m/s)²)/2 = 2.409X10⁻¹⁸
Hence,
λ = ch/A+W
λ = (6.626X10⁻³⁴JsX2.998X10⁸m/s)/(7X10⁻¹⁹J + 2.409X10⁻¹⁸J)
= 19.865X10²⁶Jm/3.109X10⁻¹⁸J = 6.39X10⁻⁸= 63.9 nm
Wavelength of radiation must be used to eject electrons with a velocity of 2300 km/s is 63.9nm
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I understand that the question you are looking for is "The work function for chromium metal is 7.0 x 10-19 J (4.37 eV). What wavelength of radiation must be used to eject electrons with a velocity of 2300 km/s?"
How long does it take to fly 108 km east if you are travelling at 270 km/h?
Answer:
0.4 hour or 24 minutes
Explanation:
Knowing the formula:
\( \displaystyle{ \vec v = \dfrac{ \vec s}{t}}\)
v is velocity, s is displacement and t is time. We know two values which are s = 108 km and v = 270 km/h. Substitute in the formula:
\( \displaystyle{ 270 \ \text{km/h} = \dfrac{108 \ \text{km}}{t}}\)
Solve for t:
\( \displaystyle{ 270 \ \text{km/h} \cdot t= 108 \ \text{km}} \\ \\ \displaystyle{t= \dfrac{108 \ \text{km}}{270 \ \text{km/h}}} \\ \\ \displaystyle{t= 0.4 \ \text{h}}\)
Therefore, it will take around 0.4 hour or converting to minutes. We know that 1 hour = 60 minutes. Multiplying 0.4 both sides, hence:
0.4 hour = 0.4 × 60 minutes = 24 minutes.
Calculate the magnitude of the force required to give a 0.145 kg baseball an acceleration a = 1.00 m/s^2 in the direction of the baseball's initial velocity when this velocity has a magnitude of 10.
The magnitude of the force required is 0.145 N.
To calculate the magnitude of the force required, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a).
In this case, the mass of the baseball is given as 0.145 kg, and the acceleration is given as 1.00 m/s².
So, the equation we can use is: F = m * a
Substituting the given values, we have:
F = 0.145 kg * 1.00 m/s²
Multiplying these values together, we find that the force required is:
F = 0.145 N
Therefore, the magnitude of the force required to give the baseball an acceleration of 1.00 m/s² in the direction of its initial velocity is 0.145 N.
In conclusion, the direct answer to the question is that the magnitude of the force required is 0.145 N.
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Imagine an astronaut in a spaceship so far away from any celestial objects that we can neglect the effects of gravity.
Assume that there is no air resistance. The astronaut attaches a rock to a rope and starts rotating it fast in a circle.
As there is no ground or other normal force to offset the pull of gravity, an astronaut orbiting the Earth does feel weightless.
Why, despite being in a weightless state, an astronaut in an orbiting spaceship is not in zero gravity? As there is no ground or other normal force to offset the pull of gravity, an astronaut orbiting the Earth does feel weightless.This causes the astronaut to drop.But rather than colliding with the Earth, the astronaut is instead constantly falling around it since they are both traveling so quickly.Inverse relationship: the force of gravity decreases with increasing distance between things, and increases with decreasing distance between objects.Not just planets, stars, and moons are affected by this; every object in the universe is.Centripetal force produces no work at all.The amount of effort done by centripetal force will be zero because it is perpendicular to the direction of motion.To learn more about gravity refer
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Which city has the lowest temperature and slowest speed of sound?
a. Baguio
b. Manila
c. Lipa
d. Cagayan
e. Cebu
f. Tagaytay
ITS A a AAA A A a A A A A A A A
a wire can withstand a force of 500 n before breaking. a 30.5 kg pumpkin is tied to the wire and spun in a horizontal circle with a radius of 2 m. what is the maximum speed of the pumpkin when the wire breaks?
The force acting on the pumpkin here is a centripetal force. Hence the speed of the pumpkin weighing 30.5 Kg has after the wire breaks is 5.72 m/s.
What is centripetal force?Centripetal force is the force acting on a body which moves through a curvature. The centripetal force is directed perpendicular to the motion of the object and along the radius of the circular path.
The centripetal force F = mv²/r
Given that the force is 500 N and the radius of the circular path is 2 m, mass of pumpkin is 30.5 Kg. Thus velocity can be calculated as :
v = √(F r/ m)
= √(500 N × 2 m) / 30.5 Kg
= 5.75 m/s.
Therefore, the maximum speed of the pumpkin will be 5.75 m/s.
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Explain the mechanism that is responsible for the formation of snowflakes.
The mechanism that is responsible for the formation of snowflakes is the nucleation of ice crystals.
What is a Snowflake?This is defined as a piece of snow which falls from the sky as a result of an extremely cold climate condition and is common in the arctic regions of the world.
Snow flakes are formed when ice crystals stick together to form the flakes which usually has a dust or pollen being formed around the area being talked about.
This is also regarded as a type of precipitation such as rain etc and is therefore the most appropriate choice.
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The force that pulls falling objects toward Earth is called *
gravity
free fall
acceleration
air resistance
Answer:
Gravity
Explanation:
That's easy because gravity is the only thing that can pull us down that hard at that fast without anything helping it.
Two cars travel along a straight road. When
a stopwatch reads t = 0.00 h, car A is at
XA= 48.0 km moving at a constant speed of
36.0 km/h. Later, when the watch reads
t = 0.50 h, car B is at XB = 0.00 km moving
at 48.0 km/h. Answer the following
questions, first graphically by creating a
position-time graph, and then algebraically
by writing equations for the positions XA
and XB as a function of the stopwatch
cime, t.
The equation for position of car A is \(x_{a}\) ( t ) = 36 t + 48.
The equation for position of car B is \(x_{b}\) ( 0.5 ) = 36 ( 0.5 ) + c.
The position function in one dimension is written as x ( t ) = vt + c. This equation is for a function starting at t = 0.
x = Distance
t = Time
v = Initial Velocity constant
c = Initial position constant
For car A ,
t = 0
c = 48 km
v = 36 km / h
\(x_{a}\) ( 0 ) = 36 ( 0 ) + 48
\(x_{a}\) ( 0 ) = 48 km, which is the distance at which car A starts to move from.
For car B,
t = 0.5 hr
c = 0
v = 48 km / h
\(x_{b}\) ( 0.5 ) = 48 ( 0.5 ) + 0
\(x_{b}\) ( 0.5 ) = 24 km
\(x_{b}\) ( 0 ) = 48 ( 0 ) + c
\(x_{b}\) ( 0 ) = c
Therefore,
The equation for position of car A is \(x_{a}\) ( t ) = 36 t + 48.The equation for position of car B is \(x_{b}\) ( 0.5 ) = 36 ( 0.5 ) + c.To know more about position equation as a function of time
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Light is incident at an angle of 56 degrees on the surface of a diamond. The index of refraction of diamond is 2.42. Find the angle of refraction. Answer in units of degrees.
The angle of refraction, based on the provided information, is 22.02 degrees.
To find the angle of refraction in this scenario, we can use Snell's Law:
n1 * sin(angle1) = n2 * sin(angle2)
where n1 is the index of refraction of the first medium (air, which is approximately 1), angle1 is the angle of incidence (56 degrees), n2 is the index of refraction of the second medium (diamond, 2.42), and angle2 is the angle of refraction we need to find.
1 * sin(56) = 2.42 * sin(angle2)
To find the angle of refraction, angle2, we can rearrange the equation:
sin(angle2) = sin(56) / 2.42
Now, we can calculate the angle:
angle2 = arcsin(sin(56) / 2.42)
angle2 ≈ 22.02 degrees
So, the angle of refraction is approximately 22.02 degrees.
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A string with a natural frequency of 15 Hz will likely show
resonance at which frequencies: (could be more than one)
A) 20 Hz; B) 30 Hz; C) 40 Hz; D) 50 Hz; E) 60 Hz
What other frequencies would show resonance?
Frequencies would show resonance will be 30 Hz,45 Hz and 60 Hz. Option B and E are correct.
What is resonance frequency?Mechanical systems and charged particles all have a frequency that they vibrate at. this frequency is referred to as the resonance frequency or natural frequency.
When the matching vibrations of another item enhance the amplitude of an object's oscillations, this is known as resonance.
15 Hz is the fundamental frequency of the string.
Higher frequencies, such as the first overtone, second overtone, third overtone, and so on, must be used to cause the string to vibrate in a resonant manner.
F₁ = 2F₀ = 2 x 15 = 30 Hz is the initial overtone.
F₂ = 3F₀ = 3 x 15 = 45 Hz is the second overtone.
F₃ = 4F₀ = 4 × 15 = 60 Hz is the third overtone.
Hence,frequencies would show resonance will be 30 Hz,45 Hz and 60 Hz. Option B and E are correct.
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what is definition of viscosity ?
Viscosity is the drag force applied by the layer of liquid on a body.
When a liquid settles down it usually forms a layer by layer structure in it.
When another body of any shape goes between these layer a drag force is applied on the body in the direction opposite to the motion of the body. This drag force is what we call the viscosity of the liquid.
The viscosity depends on the compressive and adhesive forces present in the particle of the liquid. It also depends on the density of the liquid.
If the body has higher density than it is expected to have a higher viscosity in it.
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he mass of a muon is 207 times the electron mass. in one experiment, a muon (of proper lifetime 2.20 μs) is measured to have a lifetime of 6.90 μs in the lab frame. as measured in the lab frame,
For the moving muons in this experiment, a) the speed factor (β) is 0.948, b) the kinetic energy (K) is 227 MeV, and c) the momentum (p) is 315 MeV/c.
(a) For finding the speed factor (β), use the time dilation formula. The time dilation factor (γ) is given by:
\(\gamma = \tau_0/\tau\)
where \(\tau_0\) is the lifetime at rest and τ is the measured lifetime. Plugging in the values:
γ = 2.20 μs / 6.90 μs = 0.3197.
The speed factor β is the square root of \((1 - \gamma^2)\), which gives \(\beta = \sqrt(1 - 0.3197^2) = 0.948.\)
(b) The kinetic energy (K) of a moving muon can be calculated using the relativistic kinetic energy formula:
\(K = (\gamma - 1)mc^2,\)
where γ is the time dilation factor and \(mc^2\) is the rest energy of the muon. Substituting the values:
\(K = (0.3197 - 1) * (207 * electron \;mass) * c^2 = 227 MeV\)
Here, the mass of electron and its value is \(9.109*10^{-31}\)
(c) The momentum (p) of a muon can be determined using the relativistic momentum formula:
p = γmv,
where γ is the time dilation factor, m is the mass of the muon, and v is its velocity. Since β = v/c, rewrite the formula as
p = γmβc.
Plugging in the values:
p = 0.3197 * (207 * electron mass) * 0.948 * c = 315 MeV/c.
Here, the mass of electron and its value is \(9.109*10^{-31}\)
Therefore, for the moving muons in this experiment, the speed factor (β) is 0.948, the kinetic energy (K) is 227 MeV, and the momentum (p) is 315 MeV/c.
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The complete question is:
The mass of a muon is 207 times the electron mass; the average lifetime of muons at rest is \(2.20 \mu s\) . In a certain experiment, muons moving through a laboratory are measured to have an average lifetime of \(6.90 \mu s\). For the moving muons, what are (a) \beta (b) K, and (c) p (in MeV/c)?
A light sensor is based on a photodiode that requires a minimum photon energy of 1.30 eVeV to create mobile electrons.
The wavelength of the light sensor is 15.23 × 10^ -7.
E = h c / λ
\(1.30 * 1.6 * 10^{-19} = 6.6 * 10^{-34} * 3 * 10 ^ {8}/\) λ
λ = 15.23 *\(10^{-7\)
What is a photon?
We are aware that photons have their own energy. The wavelength has no bearing on the amount of energy because the energy is inversely proportional to the electromagnetic frequency of the photon.
The photon's energy will be high if its frequency is high. Therefore, we can conclude that a photon's energy is lower if its wavelength is longer.
We can claim that a more powerful red light than a less intense blue light can deliver more power to a certain location. We are aware that photon energy is measured in joules and electron volts (eV).
The equation E=hf describes the photon's energy.
where E is the energy of a photon, h is Planck's constant, and f is the electromagnetic frequency.
The given equation holds true for a single photon. E = n* h* f is the formula for the number of photons that are emitted when there are more than n photons.
Depending on the unit system being used, energy is measured in Joules or electron volts (eV).
Joules equal 6.24 x 1018 eV.
The large units aid in describing the photon energy with higher energy and frequency for radiations like gamma rays.
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1. Calculate the centripetal force exerted on a car that rounds a radius curve on horizontal ground at . 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
The magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line is 2000 N.
How to calculate the centripetal force and magnitude of the frictional force?To calculate the centripetal force exerted on a car that rounds a radius curve on horizontal ground, we can use the following formula:
\(F = mv^2 / r\)
where F is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the curve.
Without the frictional force, the car would slide off in a straight line due to its inertia, so the frictional force must be equal in magnitude and opposite in direction to the centrifugal force, which is the force that tends to pull the car away from the center of the curve.
The maximum static frictional force that can act between the tires and the road without causing the car to slip is given by:
f = μsN
where f is the frictional force, μs is the coefficient of static friction between the tires and the road, and N is the normal force acting on the car due to the road.
Since the car is traveling on a horizontal surface, the normal force is equal in magnitude to the weight of the car, which can be calculated as:
N = mg
where g is the acceleration due to gravity.
Combining the above equations, we get:
f = μsN = μsmg
The maximum value of the frictional force that allows the car to round the curve without sliding off in a straight line is equal to the centripetal force, which can be equated to mv^2/r, as shown earlier.
Therefore, we can write:
\(f = mv^2 / r\)
Equating this expression to the previous expression for f, we get:
\(mv^2 / r = μsmg\)
Solving for the frictional force, we get:
\(f = μsmg = mv^2 / r\)
\(f = (m/r) v^2\)
Substituting the given values, we get:
\(f = (m/r) v^2 = (1000 kg / 50 m) (10 m/s)^2\)
f = 2000 N
Therefore, the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line is 2000 N.
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how did barnett newman increase the capacity of color to communicate emotion?
Barnett Newman, an American abstract expressionist painter, increased the capacity of color to communicate emotion by using color in an abstract way. Newman is well-known for his "zip" paintings, which are large canvases divided by a vertical line of color. He argued that the viewer's experience of these paintings was not just visual but physical, invoking emotions like awe, transcendence, and mystery.
He believed that his use of color had the capacity to evoke a spiritual experience in viewers. In his work, he made extensive use of large fields of pure color, which he believed had the power to convey deep emotions and spiritual states. He aimed to create an almost mystical experience for the viewer by immersing them in the color and allowing them to feel its intensity and purity.
In conclusion, Barnett Newman increased the capacity of color to communicate emotion by using pure color fields in his abstract paintings, which evoked a sense of awe, transcendence, and mystery, which he believed had the capacity to create a spiritual experience for the viewer.
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On a trip to the store around the block a bike rider goes 75m toward the west and 75m toward the north at a constant speed of 1.5m/s. what is the magnitude of the average velocity?
The bike rider's average velocity is 1.5m/s in the northwest direction. This means that on average, the rider covers a distance of 1.5 meters every second in the northwest direction.
The magnitude of the average velocity can be determined by finding the total displacement and dividing it by the total time taken. In this case, the bike rider travels 75m toward the west and 75m toward the north, resulting in a total displacement of 75m in the northwest direction.
To find the average velocity, we need to calculate the total time taken. Since the distance traveled in each direction is the same and the speed is constant at 1.5m/s, we can divide the total distance by the speed to find the total time. In this case, 75m / 1.5m/s = 50s.
Next, we divide the total displacement (75m) by the total time (50s) to find the average velocity. 75m / 50s = 1.5m/s.
Therefore, the magnitude of the average velocity is 1.5m/s.
In summary, the bike rider's average velocity is 1.5m/s in the northwest direction. This means that on average, the rider covers a distance of 1.5 meters every second in the northwest direction.
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A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the rod when the ball approaches it closely this first time?.
What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
ElectrostaticsI have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
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explain why einstein argued that the constancy of the speed of light (postulate 2) actually follows from the principle of relativity (postulate 1).
Einstein's theory of special relativity is based on two postulates: the principle of relativity and the constancy of the speed of light.
The principle of relativity states that the laws of physics are the same in all inertial frames of reference. The constancy of the speed of light states that the speed of light is the same for all observers, regardless of their motion relative to the light source.
Einstein argued that the constancy of the speed of light actually follows from the principle of relativity. This is because the principle of relativity implies that there is no preferred frame of reference, and therefore no preferred speed.
If there were a preferred speed, then there would be a preferred frame of reference, which would violate the principle of relativity. Therefore, the speed of light must be constant in all frames of reference, in order to be consistent with the principle of relativity.
In other words, the constancy of the speed of light is a consequence of the fact that the laws of physics are the same in all inertial frames of reference. If the speed of light were not constant, then the laws of physics would be different in different frames of reference, which would violate the principle of relativity.
Therefore, Einstein's argument is that the constancy of the speed of light actually follows from the principle of relativity, because the principle of relativity implies that there is no preferred frame of reference, and therefore no preferred speed.
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b. Two vectors with dimensions A = 5i + 3j + k and B = 4i + j + 2k are used for the following calculation. Determine: i. ii. iv. The dot product A.B. [2 marks] [3 Marks] The angle between vectors A and B. The cross product A XB. [2 marks] The area of the parallelogram spanned by the vectors A and B. [3 Marks]
The dot product is 25, the angle is \(\theta = cos^{-1} \frac{25}{\sqrt{35} \times \sqrt{21}}\), the cross product is 1i + (-6)j + (-7)k, and the area of the parallelogram spanned by vectors A and B is \(\sqrt{86}\).
Given,
A = 5i + 3j + k
B = 4i + j + 2k
i. Dot Product (A · B):
The dot product of two vectors A and B is given by the sum of the products of their corresponding components.
\(A.B = (A_x \times B_x) + (A_y \times B_y) + (A_z \times B_z)\\A.B = (5 \times 4) + (3 \times 1) + (1 \times 2) \\= 20 + 3 + 2 \\= 25\)
ii. Angle between vectors A and B:
The angle between two vectors A and B can be calculated using the dot product and the magnitudes of the vectors.
\(cos\theta = (A.B) / (|A| \times |B|)\\\theta = \frac{1}{cos} ((A.B) / (|A| \times |B|))\\A = \sqrt{(5^2 + 3^2 + 1^2)} =\\ \sqrt{35}\\B = \sqrt{(4^2 + 1^2 + 2^2)} \\= \sqrt{21}cos\theta = \frac{(A.B) / (|A| \times |B|)\\\theta = \frac{1}{cos} \frac{25}{\sqrt{35} \times \sqrt{21}}}\)
iv. Cross Product (A × B):
The cross product of two vectors A and B is a vector that is perpendicular to both A and B and its magnitude is equal to the area of the parallelogram spanned by A and B.
\(A\times B = (A_y \timesB_z - A_z \timesB_y)i + (A_z \timesB_x - A_x \timesB_z)j + (A_x \times B_y - A_y \times B_x)k\\A\times B = ((3 \times 2) - (1 \times 1))i + ((1 \times 4) - (5 \times 2))j + ((5 \times 1) - (3 \times 4))k\\= 1i + (-6)j + (-7)k\)
Area of the parallelogram spanned by vectors A and B:
The magnitude of the cross product A × B gives us the area of the parallelogram spanned by A and B.
Area = |A × B|
Area of the parallelogram spanned by vectors A and B:
Area = |A × B| =
\(\sqrt{(1^2 + (-6)^2 + (-7)^2}\\\sqrt{1+36+49\\\\\sqrt{86}\)
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The gas law for an ideal at absolute temperature (in kelvins), pressure Pin atmospheres)and volume Vinters PV = ART, Where is the number of males of the - 0.0671 gal constant. Suppose that, at a certain instant, Postm and is increasing at a rate of 0.11 atm/min and verzand it decreasing at a rate of 0.27 min. Find the rate of change of with resped To time (in/min) at that instantin = 10 mo [Round your answer to four decimal places) K/min mit A
The rate οf change οf temperature with respect tο time is apprοximately -0.4223 K/min.
How to find the rate οf change οf temperature ?Tο find the rate οf change οf temperature (T) with respect tο time (t) at a certain instant, we can use the ideal gas law equatiοn PV = nRT and differentiate it with respect tο time:
PV = nRT
Taking the derivative with respect tο time:
P(dV/dt) + V(dP/dt) = nR(dT/dt)
Since we are interested in finding dT/dt, we can rearrange the equatiοn:
(dT/dt) = (P(dV/dt) + V(dP/dt)) / (nR)
Substituting the given values:
P = 7.0 atm
dV/dt = -0.17 L/min (negative sign indicates a decrease in vοlume)
dP/dt = 0.11 atm/min
n = 10 mοl
R = 0.0621 L·atm/(mοl·K)
(dT/dt) = (7.0 atm * (-0.17 L/min) + 12 L * 0.11 atm/min) / (10 mοl * 0.0621 L·atm/(mοl·K))
Calculating the rate οf change οf temperature:
(dT/dt) ≈ -0.4223 K/min
Therefοre, at that instant, the rate οf change οf temperature with respect tο time is apprοximately -0.4223 K/min.
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The force of fraction between a 10kg mass and a leveled ground is found to be 6N.what applied force is required to accelerate the mass at 1.2 meter per second square?
The net force acting on the mass parallel to the surface is
∑ F[horizontal] = F[applied] - F[friction] = ma
where m = 10 kg. Given that F[friction] = 6 N and a = 1.2 m/s², we have
F[applied] - 6 N = (10 kg) (1.2 m/s²)
F[applied] = 12 N + 6 N
F[applied] = 18 N