Which of these observations of galileo refuted ptolemy's epicycles? the craters on the moon the rotation of sunspots across the sun's surface the revolution of jupiter's moons around it the complete cycle of venus' phases the visibility of many more stars with the telescope?
The answer is : the complete cycle of venus' phases. This is the observation of Galileo that refuted ptolemy's epicycles. . Using his telescope, Galileo found that Venus went through phases, just like our Moon. But, the nature of these phases could only be explained by Venus going around the Sun, not the Earth.
Galileo's observations that refuted Ptolemy's epicycles were the complete cycle of Venus' phases and the revolution of Jupiter's moons. These observations showed that these bodies were orbiting the Sun and Jupiter respectively, not the Earth as Ptolemy's model suggested.
Explanation:The observation Galileo made that refuted Ptolemy's epicycles was the complete cycle of Venus' phases. Ptolemy's model suggested that all planets revolved in small circles, or 'epicycles', around a point that in turn orbited Earth. However, Galileo's observation showed that Venus goes through a set of phases similar to the Moon's, which could only occur if Venus orbits the Sun, not Earth. This discovery was incompatible with Ptolemy's model which largely placed Earth as the center of the universe. Galileo's discovery of the revolution of Jupiter's moons around it, instead of Earth, also offered further proof against Ptolemy's model.
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A hit-and-run incident is one of the more common cases for paint evidence.
true or false
Calculate δe, if the system absorbs 7.24 kj of heat from the surroundings while its volume remains constant (assume that only p−v work can be done). express your answer using three significant figures.
Answer:
[tex]\delta Q = \delta E = 7.24 kJ[/tex]
Explanation:
Heat absorbed by the system is given as
[tex]\delta Q = 7.24 kJ[/tex]
now from first law of thermodynamics we know that
[tex]\delta Q = \delta E + W[/tex]
here
W = work done by the system
[tex]\delta E[/tex] = change in internal energy
also we know that when volume of the system remains same then work done by the system must be zero
[tex]W = 0[/tex]
so from above equation
[tex]\delta Q = \delta E = 7.24 kJ[/tex]
What will happen to the brightness of a string of bulbs hooked up in series when you increase the number of bulbs in the string?
A. Each light in the string will get brighter.
B. Each light in the string will get dimmer.
C. There will be no change in the brightness of each light.
D. You cannot predict what will happen.,
As you add or remove neutrons from the nucleus, isotopes are formed
A force of 75 N is applied to a spring, causing it to stretch 0.3 m. What is the spring constant of the spring?
0.004 N/m
22.5 N/m
75.3 N/m
250 N/m
Answer:
D 250
Explanation:
EDGE
When red light and green light shine on the same place on a piece of white paper, the spot appears t obe?
Answer:
yellow
Explanation:
red + green light makes yellow
Engine oil (η = 0.20 Pa⋅s) passes through a fine 1.90-mm-diameter tube that is 9.40cm long.
What pressure difference is needed to maintain a flow rate of 6.4mL/min ?,
Here we have to calculate the pressure difference between two ends of the tube.
The diametre of tube [d] =1.90 mm
we know that 1 mm=[tex]10^{-3}[/tex]
The radius of tube [r]=0.95 mm i.e 0.00095 m
The length of the tube [l]=9.40 cm i.e 0.0940 m
The coefficient of visocity of the oil [tex][\eta][/tex] =0.20 pa.s
The rate of flow of oil [tex][\frac{dv}{dt} ]=6.4mL/min[/tex]
we know that 1 mL=[tex]10^{-3} L=10^{-6}m^{3}[/tex] [1L=[tex]10^{-3} m^{3} ][/tex]
Hence 6.4mL/m^3= [tex]1.067*10^{-7}[/tex] m^3/s
we know that [tex]\frac{dv}{dt} =\frac{\pi pr^4}{8\eta l}[/tex]
Hence [tex]p=\frac{dv}{dt} *8\eta l*\frac{1}{\pi r^4}[/tex]
⇒ [tex]p=1.067*10^{-7} *8*0.20*0.0940 *\frac{1}{ 3.14*[0.00095]^4}[/tex] pa
⇒ P=6.274630973*[tex]10^{3}[/tex] pa
An object is places 20.0 cm from the front of a converging lens of focal length 10.0 cm. What is the image distance and magnification?
The image distance and magnification is mathematically given as
i2= - 24.6 cmM= -2.69What are the image distance and magnification?
Question Parameter(s):
An object is placed 20.0 cm from the front of
a converging lens of focal length 10.0 cm.
Generally, the equation for the lens equation is mathematically given as
1 / i + 1 / o = 1 / f
Therefore
i = o f / (o - f)
i1 = 30 * 15.2 / (30 - 15.2)
i1= 30.8 cm
Therefore
i2 = 9.4 * 15.2 / (9.4 - 15.2)
i2= - 24.6 cm
In conclusion, magnification
M = m1 * m2
M= (-30.8 / 30) * (24.6 / 9.4)
M= -2.69
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a body of mass 1.5kg, traveling along the positive x axis with speed 4.5m/s,collides with another body B of mass 3.2kg which,initially is at rest. A is deflected and moves with a speed of 2.1m/s in a direction which is 30 degrees below the x axis. B Is set in motion at angle b above the x axis. calculate the velocity of B after collision.,
What is the velocity of the object at 20.0s
3. The following can assist in positive communication?
A. having another person present.
B. making sure your point is made no matter what.
C. trying to find something to focus your attention on.,
Which of the following describes the role of a refrigerant in an air conditioner? to absorb heat flow from air molecules circulating around the refrigerant tubing cools the air molecules by adding cool energy to the air as it is circulated around the refrigerant tubing increases the specific heat of the air molecules and allows them to release more heat once cooled decreases the specific heat of the air molecules and allows them to absorb more heat before cooling
Answer:
to absorb heat flow from air molecules circuling around the refrigerant tubing
Explanation:
I know this is correct because I just had this question and this was the correct answer.
Two very quick questions!!
1) If a pulley system with an actual mechanical advantage of 2 is used in lifting a 2,000 lb. car, what effort would be needed to move the car?
If a pulley system with an actual mechanical advantage of 2,000,000 is used in lifting a 2,000 lb. car, what effort would be needed to move the car?,
Explanation:
1. Mechanical advantage of a pulley system, m = 2
Mechanical advantage of a machine is defined as the ratio of load to the effort force i.e.
[tex]m=\dfrac{F_L}{F_E}[/tex]
Here, [tex]F_L=2000\ lb[/tex]
[tex]F_E=\dfrac{F_L}{m}[/tex]
[tex]F_E=\dfrac{2000\ lb}{2}[/tex]
[tex]F_E=1000\ lb[/tex]
Hence, 1000 lb effort will be needed to move the car.
2. Mechanical advantage of a pulley system, m = 2,000,000
Value of load, [tex]F_L=2000\ lb[/tex]
Mechanical advantage, [tex]m=\dfrac{F_L}{F_E}[/tex]
[tex]F_E=\dfrac{F_L}{m}[/tex]
[tex]F_E=\dfrac{2000\ lb}{2000000}[/tex]
[tex]F_E=0.001\ lb[/tex]
Hence, 0.001 lb effort would be needed to move the car.
A laser beam of wavelength 439.4 nm is inci- dent on two slits 0.306 mm apart.
How far apart are the bright interference fringes on a screen 6.08 m away from the double slits?
Answer in units of cm.
The cochlea _____. i) amplifies sound vibrations ii) collects sound pressure waves iii) detects the frequency of sounds
Answer:
The correct answer is option (i), amplifies sound vibrations
Explanation:
Cochlea is part of the inner ear that is responsible for receiving the sound vibrations. The received sound vibration is then converted into nerve impulses which are then send to the brain for interpretation. Cochlea amplifies the sound so that they can be easily detected even by a damaged ear and it is also able to stimulate the hearing nerve directly.
The cochlea amplifies sound vibrations.
What is Cochlea ?The cochlea is a spiral, hollow, conical chamber inside the human ears which is made up of a bone.
Cochlea is present inside the inner ear its function is to receive the sound vibrations.
The received sound vibration is then converted into nerve impulses which are then send to the brain for interpretation.
Cochlea amplifies the sound so that they can be easily detected even by a damaged ear and it is also able to stimulate the hearing nerve directly.
Hence The cochlea amplifies sound vibrations.
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If a material is a good conductor of electricity, it is a ______
A. Metal
B. Nonmetal
C. Metalloid
Answer:
Metal
Explanation:
If a material is a good conductor of electricity, it is a metal, therefore the correct answer is option A.
What are metalloids?The elements of the periodic tables that behave as metal, as well as the nonmetal in some chemical or physical aspects, are known as metalloids. Some examples of metalloids are Boron, silicon, germanium, arsenic, antimony, tellurium, etc.
Metals are very good conductors of heat and electricity while on the other hand nonmetals are very poor conductors of heat and electricity.
Metalloids are semiconducting materials that behave as the intermediate behavior between metal and nonmetal.
Thus, If a material is a good conductor of electricity, it is a metal, therefore the correct answer is option A.
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What is the relationship between atomic radius, ionization energy, and electronegativity? What is the relationship between atomic radius, ionization energy, and electronegativity? @Physics,
The relationship between electronegativity, ionization energy, and atomic radius is that larger atoms are less attracted by nuclear force, which impacts their ability to retain and attract electrons.
Atomic Radius:
It is generally defined as the distance from the nucleus to the outermost electron cloud. Atomic radius tends to increase as you move down a group on the periodic table due to the addition of electron shells.Conversely, it decreases as you move across a period from left to right because of the increasing positive charge in the nucleus, which pulls electrons closer.Ionization Energy:
The energy needed to extract an electron from an atom in the gas phase is known as ionization energy.The first ionization energy refers to the energy needed to remove one electron. This energy increases across a period and decreases down a group. Higher ionization energy indicates that an atom holds its electrons more tightly.Electronegativity:
An atom's capacity to draw in and form bonds with electrons is measured by its electronegativity.It goes higher periodically and down within a group.When an atom has a high electronegativity, it draws electrons toward itself in a chemical connection.Relationship Between the Three Properties:
As the atomic radius increases, ionization energy and electronegativity decrease. This is because larger atoms have more electron shells, resulting in weaker attraction between the nucleus and the outermost electrons.Conversely, as the atomic radius decreases, ionization energy and electronegativity increase. This is due to stronger nuclear attraction in smaller atoms, making it harder to remove electrons and easier to attract electrons from other atoms.a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the tension in the rope and the angle the rope makes with the horizontal
Two Jupiter-size planets are released from rest 1.40×10^11m apart. What is their speed as they crash?,
To solve this problem, we derive Newton’s Law of Universal Gravitation as the basis of computation
Where: M₁ = mass of planet #1
M₂ = mass of planet #2
M = total mass
R₁ = radius of planet #1
R₂ = radius of planet #2
d₁ = initial distance between planet centers
d₂ = final distance between planet centers
a = semimajor axis of plunge orbit
v₁ = relative speed of approach at distance d₁
v₂ = relative speed of approach at distance d₂
To determine velocity during the impact of two heavenly bodies, the solution is as follows:
M₁ = M₂ = 1.8986e27 kilograms
M = M₁ + M₂ = 3.7972e27 kg
G = 6.6742e-11 m³ kg⁻¹ sec⁻²
GM = 2.5343e17 m³ sec⁻²
d₁ = 1.4e11 meters
a = d₁/2 = 7e10 meters
R₁ = R₂ = 7.1492e7 meters
d₂ = R₁ + R₂ = 1.42984e8 meters
v₁ = 0
v₂ = √[GM(2/d₂−1/a)]
v₂ = 59508.4 m/s
The final answer is 59508.4 m/s.
Answer:
this verified kid is way too smart for his own good
Explanation:
Why do electromagnetic waves not require a medium for travel?
a.because electromagnetic waves travel too fast for stationary particles to move with them
b.because electromagnetic waves transmit energy without compressing the particles of the medium
c.because electromagnetic waves generate their own particles for compression and use these for movement
d.because electromagnetic waves move in two-dimensional space, the particles of mediums exist in 3-d space?
Unit 2 Lesson 7 Portfolio for physics
According to Coulomb's law, what will happen to the force between two charges if the distance is divided by 2?
A. 1/2 as strong
B. 2 times as strong
C. 1/4 as strong
D. 4 times as strong
The correct answer is D. The force between the two charges will be 4 times as strong if the distance is divided by 2.
According to Coulomb's Law, the electric force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, this is expressed as:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
where [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( k = 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex]).
If the distance [tex]\( r \)[/tex] is divided by 2, the new distance becomes [tex]\( r' = \frac{r}{2} \)[/tex]. The new force [tex]\( F' \)[/tex] can be calculated by substituting [tex]\( r' \)[/tex] into Coulomb's Law:
[tex]\[ F' = k \frac{|q_1 q_2|}{(r')^2} = k \frac{|q_1 q_2|}{\left(\frac{r}{2}\right)^2} = k \frac{|q_1 q_2|}{\frac{r^2}{4}} = k \frac{4 |q_1 q_2|}{r^2} = 4 \left(k \frac{|q_1 q_2|}{r^2}\right) \][/tex]
[tex]\[ F' = 4F \][/tex]
Thus, the force between the two charges is 4 times stronger when the distance between them is halved. This is because the force is inversely proportional to the square of the distance, so reducing the distance by a factor of 2 results in a force that is [tex]\( 2^2 = 4 \)[/tex] times stronger.
What examples can you find in your home that are examples of kinetic and potential energy name two for each
The examples of kinetic energy is ceiling fan and washing machine.
The examples of potential energy is stretched rubber band and electric battery.
What are kinetic energy and potential energy?The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes.
Potential energy in physics is the energy that an item retains as a result of its position in relation to other objects, internal tensions, electric charge, or other elements.
The gravitational potential energy of an item, the elastic potential energy of a stretched spring, and the electric potential energy of an electric charge in an electric field are examples of common types of potential energy.
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Calculate the total force on the Earth due to Venus, Jupiter, and Saturn, assuming all four planets are in a line. The masses are Mv=0.815ME, MJ=318ME, Msat=95.1ME, Msun=1.99x10^30kg, ME=5.98x10^24kg and the mean distances of the four planets from the Sun are 108, 150, 778, and 1430 million km. Apparently the answer is 9.56x10^17 N but I'm not sure how to get to that .-.,
Answer: Total Force = [tex]9.56*10^{17}[/tex]
Explanation:
Line points are: Sun - Venus - Earth - Jupiter - Saturn
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This means:
[tex]F=G\frac{m_{1} m_{2}}{r^{2} }[/tex]
Where,
G is the gravitational constant,
m1 and m2 are the masses of the objects,
and r is the distance between the centers of their masses.
So, if G value is [tex]6.674*10^{-11} [\frac{m^{3}}{kg*s^{2}}][/tex], then we replace the equation with the corresponding values:
[tex]F=6.674*10^{-11} (-\frac{0.815ME^{2}}{(4.2*10^{10})^{2}} + \frac{318ME^{2}}{(6.28*10^{11})^{2}} + \frac{95.1ME^{2}}{(1.28*10^{12})^{2}})[/tex]
To get the distances we subtract the distances between the sun and earth and the distances between the other planets and the sun.
in a lab investigation. one group of students (group A) measures the speed of a 0.1-kilogram car at 2.5 m/sec at the bottom of a hill. Another group of students (group B) measures the speed of the car at 3 m/sec at the bottom of the hill. the car's starting position at the top of the hill is one-meter high. what is the potential energy of the car at the beginning of the experiment before its speed is measured?
The potential energy of the car at the beginning of the experiment, before its speed is measured, is approximately 0.98 Joules.
To calculate the potential energy of the car at the beginning of the experiment, we can use the formula for gravitational potential energy:
Potential energy (PE) = mass (m) × gravitational acceleration (g) × height (h)
Given:
Mass of the car (m) = 0.1 kg
Height of the hill (h) = 1 meter
Gravitational acceleration (g) = 9.8 m/s² (approximately)
The potential energy (PE) is as follows:
PE = m × g × h
= 0.1 × 9.8 × 1
= 0.98 Joules
Therefore, the potential energy of the car at the beginning of the experiment, before its speed is measured, is approximately 0.98 Joules.
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A hexagonal aquarium is 14.5 cm on each side of its base and 26.5 cm high. the lateral sides are molded from one piece of plastic. what is the area of this piece of plastic? 3458.25 cm2 2305.5 cm2 1152.75 cm2 384.25 cm2
True or false A physical change occurs when matter changes state as from a liquid to a gas Becuase
A positive charge of 6.0 × 10-4 C is in an electric field that exerts a force of 4.5 × 10-4 N on it. What is the strength of the electric field
The magnitude of the electric force exerted on a charge in an electric field is given by
[tex] F=qE [/tex]
where
q is the charge
E is the magnitude of the electric field
In this problem, we have a charge of [tex] q=6.0 \cdot 10^{-4} C [/tex], while the force exerted on it is [tex] F=4.5 \cdot 10^{-4}N [/tex], so we can rearrange the previous formula to calculate the magnitude of the electric field:
[tex] E=\frac{F}{q}=\frac{4.5 \cdot 10^{-4} N}{6.0 \cdot 10^{-4} C}=0.75 N/C [/tex]
Answer:
0.75
Explanation:
What is the frequency of a microwave that has a wavelength of 0.050 m?
Answer:
[tex]6x10^9Hz[/tex]
Explanation:
we use a formula that relates the frequency and the wavelength:
[tex]f=\frac{c}{\lambda}[/tex]
where [tex]f[/tex] is the frequency of the wave, [tex]c[/tex] is the speed of light [tex]c=3x10^8m/s[/tex], and [tex]\lambda[/tex] is the wavelength.
We know that the wavelength is: [tex]\lambda=0.050m[/tex], so substituting the known values in the equation for the frequency we get:
[tex]f=\frac{3x10^8m/s}{0.05m} \\f=6x10^9s^{-1}=6x10^9Hz[/tex]
The frequency is:
[tex]6x10^9Hz[/tex]
Final answer:
The frequency of a microwave with a wavelength of 0.050 m is [tex]6*10^9[/tex] Hz, or 6000 MHz, calculated using the speed of light and the wavelength-to-frequency formula.
Explanation:
To calculate the frequency of a microwave that has a wavelength of 0.050 m, you can use the formula
c = λf, where c is the speed of light in a vacuum (approximately [tex]3*10^8[/tex] m/s), λ is the wavelength, and f is the frequency. Solving for f, the equation becomes
f = c / λ.
Let’s plug in the values:
f = ([tex]3*10^8[/tex] m/s) / (0.050 m)
After calculating, the frequency f is [tex]6*10^9[/tex] Hz, or 6000 MHz.
The potential energy of a pair of hydrogen atoms separated by a large distance x is given by u(x)=−c6/x6, where c6 is a positive constant. is this force attractive or repulsive?
(-6C₆) / x⁷ newton; it is an attractive force.
Further explanationGiven:
The potential energy of a pair of hydrogen atoms separated by a large distance x is given by [tex]\boxed{ \ U(x) = - \frac{C_6}{x^6} \ }[/tex], where C₆ is a positive constant.
Question:
What is the force that one atom exerts on the other? Is this force attractive or repulsive?The Process:
Let us rewrite the potential energy of the pair of the hydrogen atom:
[tex]\boxed{ \ U(x) = - \frac{C_6}{x^6} \ }[/tex]
The component of a conservative force in a particular direction, equals the negative of the derivative of the corresponding potential energy, properly concerning a displacement in that direction.
For one-dimensional motion, say along the x-axis, the relationship between force and potential energy is as follows:
[tex]\boxed{ \ \overrightarrow{F} = F_x \hat{i} = - \frac{\delta U}{\delta x} \hat{i} \ }[/tex]
Let us determine the force.
[tex]\boxed{ \ F_x \hat{i} = - \frac{\delta}{\delta x} \Big( - \frac{C_6}{x^6} \Big) \ \hat{i} \ }[/tex]
[tex]\boxed{ \ F_x \hat{i} = - (- C_6) \frac{\delta}{\delta x} \Big( \frac{1}{x^6} \Big) \ \hat{i} \ }[/tex]
[tex]\boxed{ \ F_x \hat{i} = C_6 \frac{\delta}{\delta x} (x^{-6}) \ \hat{i}} \ }[/tex]
[tex]\boxed{ \ F_x \hat{i} = (C_6)(-6)x^{-7} \ \hat{i} \ }[/tex]
Thus, we get: [tex]\boxed{\boxed{ \ F_x = - \frac{6C_6}{x^7} \ \hat{i} \ }}[/tex]
From the data problem above, we know that C₆ is a positive constant. Hence, Fₓ is positive when x is negative and vice versa. Thus Fₓ is always directed toward the origin. Besides, a negative sign means that an attractive force occurs.
In essence, the directions are toward the origin, since this is the potential energy for a restoring force.
_ _ _ _ _ _ _ _ _ _
Notes:
We know that the difference of potential energy from point 1 to point 2 as the negative of the work done:
[tex]\boxed{ \ \Delta U_{12} = U_2 - U_1 = - W_{12} \ }[/tex]
The work done by the given force as the particle moves from coordinate x to (x + d x) in one dimension is
[tex]\boxed{ \ dW = \overrightarrow{F} \cdot d \vec{r} \ } \rightarrow \boxed{ \ \overrightarrow{F} = \frac{dW}{d \vec{r}} \ }[/tex]
Therefore, for motion along a straight line, a conservative force is the negative derivative of its associated potential energy function, i.e.,
[tex]\boxed{ \ \overrightarrow{F} = F_x \hat{i} = - \frac{\delta U}{\delta x} \hat{i} \ }[/tex]
_ _ _ _ _ _ _ _ _ _
Conservative force : force that does work independent of path.
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