The transfer of energy as heat caused by the collision of molecules is called _________. radiation convection conduction contact

The examples of kinetic energy is ceiling fan and washing machine.

The examples of potential energy is stretched rubber band and electric battery.

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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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:

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]

(-6C₆) / x⁷ newton; it is an attractive force.  

Given:

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:

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.

Answer:

yellow

Explanation:

red + green light makes yellow

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

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.

Answer:

D 250

Explanation:

EDGE

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.

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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C. it is the most powerful explanation scientists have to offer at this time. I am 100% right, I just finish the test and got it right.

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.

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 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:

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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Answer:

Metal

Explanation:

If a material is a good conductor of electricity, it is a metal, therefore the correct answer is option A.

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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The work done in pulling a bucket full of water with a weight of 15 kg that lost 6 kg of it from the hole from a well 15 m deep is 1350 Joules

Thus; Work done is measured in Joules, J, or Nm

In this case;

Mass of the bucket is 15 Kg

However, 6 kg of water flew out due to the whole, therefore the remaining mass is 9 kg.

Force or Weight = Mass (kg) × 10 N/kg

Therefore;

Force = 90 Newtons

Distance = 15 m

Therefore;

Work done = 90 N × 15 m

                  = 1350 Joules

Hence; Work done pulling out the bucket from the well is 1350 Joules.

Key words: Work done, Force, Distance

Level: High school

Subject: Physics

Topic: Work, power and simple machines

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:

Ionization Energy:

Electronegativity:

Relationship Between the Three Properties:

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.

Answer:

the answer is B, 10-kg object

Explanation:

Because of Newton we know that if a force is applied to an object, it will have an acceleration on the same direction of the force applied.

[tex]F=m*a\\\\where:\\m=mass\\a=acceleration[/tex]

in order to obtain the acceleration we have to reordenate the formula:

[tex]a=\frac{F}{m}[/tex]

The acceleration of the 10 kg object is:

[tex]a=\frac{20N}{10kg}=2\frac{m}{s^2}[/tex]

The acceleration of the object of 18 kg is:

[tex]a=\frac{30N}{18kg}=1.67\frac{m}{s^2}[/tex]

The biggest acceleration was for the object with less mass.

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.

The image distance and magnification is mathematically given as

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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