In the period 6 inner transition element series, the 4f orbitals begin to be filled after the element _____.

Answers

Answer 1

In the period 6 inner transition element series, the 4f orbitals begin to be filled after the element Lanthanum(La) and Cerium(Ce).

The d and f Block elements in the groups of 3 to 11 are also called transition elements and inner transition elements respectively. 4f and 5f orbitals of f-block elements are steadily in the latter of two long periods.

Inner transition element:

Inner transition elements are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements. These includes lanthanoids( Z = 58 - 71) and actinoids (Z = 90 - 103).

Therefore period 6 inner transition elements begin with Lanthanum and Cerium.

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Related Questions

Use the information from the graph to answer the
What is the total displacement of the object?
question.
m
Velocity vs. Time
Velocity (m/s)
40
30
20
10
0
Time (s)
0
5
10
15
20
25

Answers

Answer:

The total displacement 560 m

Explanation:

 Given:

t₀ = 0 s

t = 25 s

V₀ = 10 m/s

V = 35 m/s

___________

D - ?

Acceleration:

a =  (V - V₀) / (t - t₀) = (35 - 10) / (25 - 0) = 25 / 25 = 1.0 m/s²

The total displacement:

D = V₀·t + a·t² / 2

D = 10·25 + 1.0·25² / 2 ≈ 560 m

Light passes straight through two vertical polarizers. The axes of the filters are aligned parallel to each other. How does the light that passes through the second filter compare to the light that passed through the first filter?.

Answers

The direction of the polarized light after passing the first and the second filter are the same, however the intensity of the light after passing the first and second polarizer are half and fourth of its original intensity, respectively.

Light polarization is a filter of electromagnetic waves such that it propagates into one transmission axis only.

When light passes through a polarizer, its intensity will decrease by half.

Suppose the intensity of unpolarized light is  I₀,  after passing the first polarizer, its intensity becomes:

I₁ = 1/2 . I₀

After passing the second polarizer, the intensity will further decrease by half, or:

I₂ = 1/2 . I₁

I₂ = 1/2 .1/2 . I₀ = 1/4. I₀

Since both polarizers are vertical filters, then the direction of the polarized light after passing the first and the second filter are the same.

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The temperature, pressure, and precipitation conditions of the atmosphere for a specific place on a given day is called , whereas a longer term view of these same factors, typically taken over a period of many years, is called.

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The temperature, pressure and precipitation condition of the atmosphere for a specific place on a given day is called weather, whereas are long term view of the same factors, typically taken over a period of many years, is called climate.

The basic difference between climate and weather is that climate exist for a very long period of time while weather can change in just a few hours.

Weather is just a way to way information about the temperature, pressure and precipitation condition of a atmosphere on a day today basis.

While climate is a kind weather information which provides information about the weather of a reason for a long period of time.

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if more massive stars have more hydrogen in their cores than less massive stars, why do they have shorter lifetimes?

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Thermonuclear fusion occurs faster in massive stars so larger stars use up all their fuel in less time.

The more mass a star has the faster it will exhaust its fuel supply and the shorter its lifespan. The most massive stars could burn up and explode in a supernova after just a few million years of nuclear fusion. Massive stars are the largest, hottest, and brightest main-sequence stars and are blue, blue-white, or white in color.

Massive stars run out of hydrogen fuel very quickly and therefore have short lives. This is because the more massive the star the greater the fuel consumption. Even if a high-mass star has more fuel it uses it up very quickly so it does not live as long as a low-mass star. A main-sequence star's mass determines the fundamental properties of its luminosity surface temperature radius and lifetime.

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a cylindrical space colony 13 km in diameter and 30 km long has been proposed as living quarters for future space explorers. such a habitat would have cities, land, and lakes on the inside surface and air and clouds in the center. all this would be held in place by the rotation of the cylinder about the long axis. how fast would such a cylinder have to rotate at its perimeter to produce a 1-g gravitational field at the walls of the cylinder?

Answers

0.00048 rad/s should be the angular velocity of the cylinder to produce a 1-g gravitational field at the walls of the cylinder.

The following problem requires understanding angular velocity and centripetal acceleration to obtain the result. The angular velocity of a body is defined by its virtue of velocity at a particular angle on an axis.

Centripetal acceleration is acceleration when the body is rotating and that acts towards the midpoint of the circular movement.

centripetal acceleration = rω² (derived from v²/r)

here

ω = angular velocity of the cylinder

r = radius of the space colony = 13km / 2 ⇒ 6.5km or 6500m

Substituting the values,

a = g ⇒ 9.8m/s²

ω = √9.8 / 6500 = 0.00048 rad/s

∴The cylinder should rotate at 0.00048 rad/s for producing a 1-g gravitational field.

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At rest, a car's horn has a frequency of
395 Hz. Car A passes car B on the street
in the same direction. If car A is traveling
at 22.0 m/s and car B is traveling
at 19.5 m/s, what frequency does
car B hear when car A honks?

(Speed of sound = 343 m/s)
(Unit = Hz)

Answers

The frequency heard by car A is determined as 398.4 Hz.

What is the frequency heard by car A?

The frequency heard by car A is determined by applying the following equation.

f = fs(v - v₀) / (v - vs)

where;

v is the speed of sound = 343 m/sv₀ is the speed car B = 19.2 m/svs is the speed of car A = 22 m/sfs is the frequency of car A = 395 Hzf is the frequency of car B = ?

f = 395(343 - 19.2) / (343 - 22)

f = 395(1.0087)

f = 398.4 Hz

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flux i - suppose just a north pole of a permanent magnet is placed inside a gaussian surface. what is a true statement about the magnetic flux through the gaussian surface in this situation?

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The true statement about magnetic flux through the Gaussian surface when a north pole of a permanent magnet is placed inside a Gaussian surface in this situation is "The net flux is zero," option A.

What is the magnetic flux through any surface?

Magnet flux through a surface, in physics, is a surface that is integral of a normal component with a magnetic field, B over the given surface. This means that if a section of a magnet is enclosed in a chosen Gaussian surface, the net magnetic flux moving through the surface will be zero.

This magnetic flux continues inside the magnet and it leaves the Gaussian surface where the magnet is cut.

The full question is:

flux i - suppose just a north pole of a permanent magnet is placed inside a Gaussian surface. what is a true statement about the magnetic flux through the Gaussian surface in this situation?

a. The net flux is zero

b. The net flux is positive

c. The net flux is negative

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falling stars can only be observed by scanning the sky at night. given your knowledge of falling stars, what specific schedule of reinforcement maintains scanning the skies in search of falling stars?

Answers

The specific schedule of reinforcement maintains scanning the skies in search of falling stars is VI (Variable-interval schedule).

Which reinforcement schedule demands the end of a specific?

Before reinforcement is given, a ratio plan demands the completion of a certain number of responses, whereas an interval schedule calls for the occurrence of at least one right response after a predetermined amount of time.

Opportunistic conditioning uses a schedule of reinforcement as a strategy. By setting up a gap between reinforcers and replies, the goal is to determine how and when a desired behavior occurs.

This is an illustration of a schedule with variable intervals, where the reinforcement is based on the passage of time but the intervals change in a random manner.

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A spherical object has a certain density. If it is compressed under high pressure to half of its original diameter, its density will now be.

Answers

To calculate the density, known the volume and density formula.

What is density?

How tightly a material is packed together is determined by its density. The definition of it is the mass per unit volume. Density Formula: D = m/V, where D stands for density, m for mass, and V for volume, denotes the mass and volume of the object, respectively.

What is volume?

Any three-dimensional solid's volume is simply the amount of space it takes up. A cube, cuboid, cone, cylinder, or sphere can be one of these solids.

Let mass of the sphere be m and radius be r, then,

m= v × p ∴ v= volume

or

m= 4/3πr³× p⇒1

now new diameter, d¹= 1/4d and d=2r

so, d¹= 1/4 × 2r⇒ d¹=1/2r also,

d¹=2r¹ so, 2r¹ = 1/2r⇒  r¹= 1/4r

logs is same, there is no change in it then,

m=v¹×p¹

m=4/3πr¹^3ₓp¹

or m=4/3π(1/4r)^3×p¹⇒2

(1) is equal to (2) so,

4/3π(1/4r)^3×p¹=4/3πr^3×p

1/64r^3×p¹ = r^3×p

p¹=64p

Therefore denisty will be 64.

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5. In an emergency, the driver of a 1.3 x 10³-kg car slams on the brakes, causing the car to skid forward on the road. The coefficient of kinetic friction between the tires and the road is 0.97, and the car comes to a stop after travelling 27 m horizontally. Determine the work done by the force of friction during the skidding.​

Answers

The work done by the force of friction during the skidding, given that the car comes to a stop after travelling 27 m horizontally is 333660.6 J

How do I determine the work done?

We'll begin by obtaining the frictional force. This can be obtained as follow:

Mass (m) = 1.3 x 10³ KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 1.3 x 10³ × 9.8 = 12740 NCoefficient of kinetic friction (μK) = 0.97Frictional force (F) = ?

F = μKN

F = 0.97 × 12740

Frictional force = 12357.8 N

Finally, we shall determine the work done by the force of friction. This is illustrated below:

Frictional force = 12357.8 NDistance (d) = 27 mWorkdone (Wd) =?

Work done (Wd) = force (F) × distance (d)

Wd = 12357.8 × 27

Wd = 333660.6 J

Thus, we can conclude that the work done by the force of friction is 333660.6 J

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the top of a ladder slides down a vertical wall at a rate of 0.15 mys. at the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 mys. how long is the ladder?

Answers

The length of the ladder which is placed 3 meters from the wall is           5 meters

The rate of the change of ladder decreasing in vertical wall = - 0.15 m/s

The distance between the bottom of the ladder and the wall = 3 meter

The rate of change of the bottom of the ladder away from the wall = 0.2 m/s.

The length of the ladder can be found using the Pythagoras theorem,

               x² + y² = L²

where x is the distance between the bottom of the ladder and the wall

           y is the distance from the top of the ladder to the bottom of the wall.

           L is the length of the ladder

Let us differentiate in terms of the rate of change in the above equation,

              2x dx/dt + 2y dy/dt = 0

Now let us substitute the known values,

                 2(3)(0.2) + 2y(-0.15) = 0

                           1.2  - 0.3y = 0

                                 0.3y = 1.2

                                      y = 1.2 / 0.3

                                        = 4

Then, the length of the ladder is

               3² + 4² = L²

                 9 + 16 = L²

                        L = √25

                        L = 5

Therefore, the length of the ladder is 5 meters

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if the ladder is just on the verge of slipping when the firefighter is 9.10 m from the bottom, what is the coefficient of static friction between ladder and ground?

Answers

Static friction between the ladder and the ground has a coefficient of

[tex]\mu_{s}[/tex] = 0.303.

A measure of static friction.

The maximum static friction force (F) that can exist between two surfaces before movement starts is divided by the normal force (N) to determine the coefficient of static friction.

x₁ = 8

x₂ = a₁

x₃ = x₁ cos θ

x₃ = 8 cos61 = 3.87m

x₄ = x₂cosθ = 9.1cos 61

x₄ = Lsin 60 = 16sin 60

[tex]x_{s}[/tex] = 13.856m

w₁ = mg = 840

w₂ = mg = 490

N₁ = mg + mg = 490+840 =1330N

N₂ = f = [tex]\mu_{s}[/tex]N₁

Torque about point is zero

[tex]\sum[/tex]Z = 0

mgx₃ +mgx₁ = N₂[tex]x_{s}[/tex]

N₂ = 490×387+840×4.41/13856

N₂ = 404.268

[tex]\mu_{s}[/tex] = N₂/N₁ = 404.268/1330 = 0.303

[tex]\mu_{s}[/tex] = 0.303

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when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about:

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When an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about  the amount of energy the star gives off each second.

Luminosity and apparent brightness are about brightness, but from a different point of view. The difference between luminosity and apparent brightness is that luminosity tells us exactly how bright a star really is while apparent brightness only tells us its brightness seen from Earth.

L / Lsun = ( d / dsun )² = b / bsun

L = Luminosity of a star

Lsun = Luminosity of sun

d = Distance of star

dsun = Distance of sun

b = Brightness of star

bsun = Brightness of sun

Therefore, when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about  the amount of energy the star gives off each second.

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a gyroscope flywheel of radius 2.83 cm is accelerated from rest at 14.2 rad/s2 until its angular speed is 2760 rev/min. (a) what is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) what is the radial acceleration of this point when the flywheel is spinning at full speed? (c) through what distance does a point on the rim move during the spin-up?

Answers

The tangential acceleration of the gyroscope flywheel is 40.186 x [tex]10^{-2[/tex] m/[tex]s^2[/tex], Radial acceleration of  gyroscope flywheel is  2364.08 [tex]m/s^2[/tex] and During the spin up process the distance moved by a point on rim  is 83.24 m.

What is meant by tangential acceleration?The centripetal force causes an acceleration known as radial acceleration, which is measured in radians per square second. Radial acceleration is directed toward the center.Tangential acceleration occurs when a body or object moves at a speed that is not uniform.

a gyroscope flywheel of radius, R = 2.83 cm =2.83 x [tex]10^{-2}[/tex] m

[tex]\omega_0 = 0[/tex]

[tex]\omega = 2760 X \frac{2\pi}{60} rad/sec[/tex]

[tex]\alpha = 14.2 rad/ sec^2[/tex]

The tangential acceleration :

[tex]a_t = R\alpha[/tex] =2.83 x [tex]10^{-2}[/tex] m x 14.2 [tex]rad/s^2[/tex] =40.186 x [tex]10^{-2[/tex] m/[tex]s^2[/tex]

The radial Acceleration :

[tex]a_r = R\omega^2\\=(2.83 X 10^-2) (2760 X \frac{\pi}{30} )^2[/tex] =2364.08 [tex]m/s^2[/tex]

During the spin up process the distance moved by a point on rim

d = [tex]R\theta[/tex]

where [tex]\theta =\frac{ \omega^2 - \omega_0^{2}}{2\alpha}[/tex]

= [tex]\frac{(2760 X \pi/30)^2}{2 X 14.2}[/tex] =2941.42 rad

d = 2.83 x [tex]10^{-2}[/tex] x 2941.42 =83.24.m

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two cars, car x and car y, are moving with the same velocity, and slam on the brakes coming to a stop. car x is more massive than car y. for which car is the amount of work done by friction in stopping it the highest?

Answers

The amount of work done by friction in stopping the car is given as

Wx > Wy.

There are two cars moving with the same velocity

car X is more massive than car Y

mass of car X = Mx

mass of car Y = My

here given,  Mx >  My

The distance travelled will be the same as the initial velocity, and it is independent of mass.

Let, μ be the coefficient of kinetic friction

work done by the frictional force is given as:

F = μMg

Fx = μMxg

Fy = μMyg

W = F.s

Wx = μMxg.s

Wy = μMyg.s

Wx > Wy

The amount of work done by friction in stopping the car is given as

Wx > Wy.  

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if someone asked you whether their house could be heated with solar heat, what characteristics would you look for?

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Solar house would have linear east-west floor designs with huge south-facing windows & roof overhangs to protect against the intense summer sun.

South-facing glass & thermal mass are the two main components of passive solar heat since they both collect, store, and disperse heat. The methods for putting such aspects into practice vary. A solar collector, thermal absorber, and distribution system make up the real living area.

This collection includes sizable circular or linear seating components that come with adjustable backrests and corner backrests.

The ideal windows for daylighting are those facing south. They provide all-day illumination and are less affected by heat and glare than windows with an east or west orientation.

The eaves of the house or the rafters on your roof are two names for this overhang. When this overhang is done, the bottom is referred to as the soffit, meaning literally translates to "something attached beneath".

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suppose we compare to something that simply slides with zero friction (but doesn't roll at all). should our rolling cylinder be released from the same height, or from a larger height, or from a smaller height if we would like our cylinder to (barely) completely make it through the circle?

Answers

The cylinder must be released from a greater height for it to complete the circle.

From the concept of work-energy theorem we know that for the sliding object, the beginning work done/ energy of the object is the potential energy due to gravity at a height.

P.E = m.g.h

Here h=height

At the top of the circle, total energy will include both Kinetic energy and potential energy

From conservation of energy, it’s inferred

m.g.h= (½ mv²) + (mg(2R))

Here R=radius of circular motion

Taking m common

m.g.h= m(v²/2 + 2gR)

Cancelling m

We get height for releasing the object as

h= (v²/2g) + (2R)

The similar formulas can be applied for the rolling object as well

but along with rotational energy

m.g.h = (½ mv²) + (½ Iω²) + (mg(2R))

Here I = moment of inertia of the rolling object

And angular velocity is denoted by ω

h = ½ (v²/g) + ½ (Iω²/mg)+ 2R

From the extra term in the above equation i.e. ½ (Iω²/mg) it’s deduced that the release height is more in case of a rolling object.

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after the switch is closed for a little while, the current through r1 is 1.4 a. what is the voltage across the capacitor?

Answers

To calculate the voltage across the capacitor is (1.44)R.

What is speed?

The speed at which electrons move past a particular location in an electrical circuit is known as the "current." In the most basic terms, current = flow. The international unit for measuring current is the ampere, pronounced "amp" (AM-pir).

What is voltage?

Voltage is the "pressure" under which electricity is pushed. Higher voltages result in more electricity flowing to an electronic device. The amount of voltage is measured in units known as volts (V).

I=1.4A

V=?

V=(1.44)R

R⇒ resistance

Therefore, the voltage across the capacitor is (1.44)R.

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a car speeds up from 22 m/s to 26 m/s in 2 seconds. what is the average acceleration of the car

Answers

Answer:

2 m/s^2

Explanation:

Acceleration is change in velocity over change in time

(26-22) m/s / 2 s =  2 m/s^2

a skater spins with an angular speed of 12.0rad/s with his arms outstretched. he lowers his arms, decreasing his moment of inertia from 41kgm2 to 36kgm2. a. calculate his initial and final rotational kinetic energy. b. does his rotational kinetic energy increase or decrease? c. what causes this change in kinetic energy?

Answers

It has a 2952 J initial rotational energy.

492 kg-m2/s represents its moment of momentum.

13.67 rad/s of angular momentum is conserved.3364J

A) is the final rotational kinetic energy. The rotational energy of it at first is

KE=[tex]\frac{1}{2}\\[/tex]Iω²

=[tex]\frac{1}{2}\\[/tex]

=[tex]2952J[/tex]

B) Her angular momentum is

L=Iω

=(41)(12)

= 492 kg-m²/s

C) Angle momentum is kept constant. Therefore, even if he draws his hands in, nothing will change. She still has 492 kg-m2/s of angular momentum after pulling her arms back. Her moment of inertia has altered, though. L must therefore change in order for its angular velocity to remain constant.

L=Iω

ω=L/I

=492/36

= 13.67 rad/s

D) So her final rotational kinetic energy is

KE=[tex]\frac{1}{2}\\[/tex]Iω²

 =[tex]\frac{1}{2}\\[/tex][tex](36) (13.67)[/tex]

=[tex]3364J[/tex]

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3) mary looked in her science book at a picture of the solar system. the planets were large and colorful but she knew it was not an accurate model. why?

Answers

mary looked in her science book at a picture of the solar system. the planets were large and colorful but she knew it was not an accurate model. The planets should have been much farther apart.

The gravitationally bound system of the Sun and the satellites in its orbit is referred to as the Solar System. The gravitational collapse of a massive interstellar molecular cloud gave it birth 4.6 billion years ago. The Sun is the system's primary mass, accounting for 99.86% of its total mass, with Jupiter making up the majority of the remaining mass. Mercury, Venus, Earth, and Mars are the four planets in the inner solar system, and they are all terrestrial planets with rocky and metallic cores. In comparison to the terrestrial planets, the four giant planets of the outer solar system are significantly bigger and more massive. The two biggest, Jupiter and Saturn, are gas giants made primarily of hydrogen and helium; the next two, Uranus and Neptune, are ice giants made primarily of highly volatile substances.

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a ball is thrown straight up. it passes a 2.45 m high window 7.45 m off the ground on its path up and takes 1.45 s to go past the window. what was the ball's initial velocity (in m/s)?

Answers

The ball's initial velocity  is  14 m/s

Since the equation of motion we referring to for  calculating  the initial velocity are :

s=ut+1/2at²  ....1

v²= u²+2as .... 2

where u and v are initial and final velocity, t is the time taken, a is the acceleration due to gravity and s is the distance or we can also  say the height .As we are given h₁ and  h₂ which are 2.45 m and  7.45 m and the time is taken, t which is  1.45 s, and at last we are having a = -9.8ms^-2

using the 2 equations, we get  

v₁= √(u²−2gh₁),

from equation 1, we get

h₂ =√(u₂−2gh₁t)−1/2gt₂

there the initial velocity will be:

u= √(((h₂+1/2gt²)/t)²+2gh1

 =√(((7.35+1/2*9.80*(1.45)^2)/1.45)^2+2×9.80×2.45)

 = 14

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Calculate the average speed of a complete round trip in which the outgoing 300 km is covered at 97 km/h , followed by a 1.0- h lunch break, and the return 300 km is covered at 55 km/h .

Answers

The average speed of the complete round trip is 62.85 km/h.

The average speed is nothing but the total distance travelled per unit time.

or in other words,

Average speed = [tex]\frac{Total distance travelled}{total time taken}[/tex]

Now, Total distance travelled = 300km + 300 km = 600 km

and the time taken = t1 + t2 + t3

t1 = distance/speed

t1 = 300km/(97km/h)

t1 = 300/97 Hour = 3.0927 Hour

t2 = 1 Hour

and t3 = distance/speed

t3 = 300km/(55km/h)

t3 = 300/55 Hour = 5.4545 Hour

Total time = t1+t2+t3

T = 3.0927+1+5.4545 = 9.547 Hour

So, average speed = total distance / total time

Average speed = 600km/9.547 Hour

Average speed = 62.85 km/h

Hence the average speed is 62.85 km/h.

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A gas receives from an external thermal source an amount of heat equal to 1000 J. This energy, in addition to producing heating in the gas, causes its expansion, with the consequent performance of work equivalent to 600 J. What was the change in the internal energy of the gas? gas?

help someone help me​

Answers

Hello..!

Subject: Thermodynamics

The first law of thermodynamics relates work and transferred heat exchanged in a system through a new thermodynamic variable, internal energy. This energy is neither created nor destroyed, only transformed.

We can think of gas as a thermodynamic system, all because gases can work and absorb heat, and then they can turn all that into energy.

The formula for the change in energy is given by the first law of thermodynamics expressed as:

[tex] \: \: \: \: \: \: \: \: \: \: {\boxed{\boxed{ \sf\large \rm \Delta U = Q - W }}}[/tex]

Being:

ΔU = change in energyQ = added heatW = Work done

Problem:

A gas receives from an external thermal source an amount of heat equal to 1000 J. This energy, in addition to producing heating in the gas, causes its expansion, with consequent performance of work equivalent to 600 J. What was the change in the internal energy of the gas? gas?

Data:

ΔU = ¿? (Meet)Q = 1000 JW = 600 J

Now adding the data in the formula to find the energy change:

[tex] \sf\large \rm \Delta U = 1000J - 600J[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{\boxed{\boxed{\large \rm \Delta U = 400 J}}}[/tex]

[tex]\begin{gathered}\rule{7cm}{0.01mm}\\\texttt{Good studies! :D}\\\rule{7cm}{0.01mm}\end{gathered}[/tex]

kinetic energy problem an 82-kg stunt man falls off the top of a large building. after falling for 65 meters he has a speed of 22 m/s. assuming he was initially at rest, what was the average force of air resistance exerted on the stunt man during his fall?

Answers

The average force of air resistance exerted on the stunt man during his fall is 305.29 N

How do I determine the average force?

We'll beging by obtaining the kinetic energy of the stunt man. This can be obtained as follow:

Mass (m) = 82 KgVelocity (v) = 22 m/sKinetic energy (KE) =?

KE = ½mv²

KE = ½ × 82 × 22²

KE = 41 × 484

KE = 19844 J

Finally, we shall determine the average force of air resistance exerted on the stunt. This can be obtained as follow:

Distance = 65 metersEnergy = 19844 JAverage force =?

Energy = work

Work (W) = force (F) × distance (d)

W = Fd

Thus,

E = W = Fd

19844 = F × 65

Divide both sides 65

F = 19844 / 65

F = 305.29 N

Thus, from the above calculation, we can conclude that the average force is 305.29 N

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two balls of mass 4kg and 2kg are moving with speed 10 m/s and 8m/s respectively with the ball of heavier mass behind the lighter ball

Answers

Answer:

Explanation:

And what to find in the problem?

The oxidation of copper(I) oxide, Cu 2 O ( s ) , to copper(II) oxide, CuO ( s ) , is an exothermic process. 2 Cu 2 O ( s ) + O 2 ( g ) ⟶ 4 CuO ( s ) The change in enthalpy upon reaction of 98. 93 g Cu 2 O ( s ) is − 100. 9 kJ. Calculate the work, w , and energy change, Δ U rxn , when 98. 93 g Cu 2 O ( s ) is oxidized at a constant pressure of 1. 00 bar and a constant temperature of 25 ∘ C. Note that Δ E rxn is sometimes used as the symbol for energy change instead of Δ U rxn

Answers

Copper II oxide is created through the oxidation of copper I oxide, cu2os cu2os. Cuo s An exothermic reaction is Cuo S. The formula is 2cu2o s)+ o2 G 4cuo S.

A simple definition of oxidation is what?

Cu2O(s), a copper(I) oxide, is oxidized to copper(II) oxide, CuO, by exothermic processes (s). Cu2O(s), a copper(I) oxide, is oxidized to copper(II) oxide, CuO, by exothermic processes (s).

An oxidation reaction is a chemical process that takes place when an object comes into contact with oxygen or another oxidizing agent (OK-sih-DAY-shun). Oxidation is characterized by the brown hue of sliced apples and rust. In the past, the term "oxidation" was used to describe any chemical reaction that involved oxygen and an element. When magnesium metal and oxygen react, magnesium oxide is produced as a result, which serves as an illustration of this.

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How long does it take a car to travel 330 m at a speed of 30.0 m/s

Answers

It will take 11 seconds

330/30=11

It will take 11 seconds for a car to travel 330 m at a speed of 30.0 m/s

By going through the formula of calculating the speed of the vehicle

that is, S=D/T

We get the equation T=D/S.........(i)

Where T stands for Time

D stands for Distance

S stands for Speed

Now, as per the given question

Distance(D)= 330 meters

Speed(S)= 30.0 meters per second

Time(T)= ?

Putting the values in equation (i),

                      T=330 meters/30 meters per second

                      T = 11 seconds.

Therefore, A car moving at a speed of 30.0 meters per second will traverse 330 meters in 11 seconds.

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The two prisms shown in the diagram below are made of glass. A ray of red light enters each prism from the air, as shown. The critical angle for red light at the glass-air boundary is 42°.
On the diagram, complete the paths of the rays passing through each prism and out into the air again.

Answers

Answer:

That's all i know

Anyway welcome

(a) the control in the experiment was 0 khz, or no sound. what information is gained by using the control with no sound that could not be obtained if no control were used?

Answers

In order to prevent deer from crossing the highway, the experiment involved placing sound-emitting equipment at the perimeter.

A procedure used to confirm or deny a hypothesis, as well as assess the effectiveness or likelihood of something that has never been tried before, is called an experiment. By showing what happens when a particular factor is manipulated, experiments shed light on cause-and-effect relationships. Experiments have a wide range of objectives and sizes but always rely on a repeatable process and a logical analysis of the outcomes. Additionally, there are experiment conducted in nature.

While teams of scientists may spend years conducting methodical research to advance their understanding of a phenomenon, a child may conduct simple experiment to understanding how things fall to the ground. In the science classroom, students benefit greatly from experiments and other hands-on activities. Experiments can improve test results and increase student engagement.

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