How many moles of oxygen atoms are in 2 moles of water?

Answer:

What we see in the sky

Where are they in the general context of the overview of the Universe?

With the naked eye, the main astronomical objects we can see are the Sun, the Moon, the stars, and some of the planets

All of the stars we see are in the Milky Way galaxy, most of them relatively close to the Sun.

The stars we see in the sky come in a range of brightnesses, partly because stars come in different intrinsic brightnesses, and partly because some are closer than others.

When we look at an astronomical object ``by eye'', we can't tell just by looking how far away it is (because not all objects have the same intrinsic brightness). All we can see is what direction it is in.

As a result, when looking ``by eye'', the positions of stars on the sky are described by their direction only; you can imagine that the sky is a big sphere with astronomical objects located at different positions on it. This is called the celestial sphere

The positions of the stars can be described with a sort of astronomical longitude and latitude, called right ascension and declination.

Constellations are patterns of stars seen in the sky. However, although the stars in any constellation are all in the same general direction in the sky, the different stars in a constellation may be at very different distances from Earth, hence constellations may not be real associations of stars in space, just stars in the same general direction as seen from Earth.

Only the nearest galaxies are visible to the naked eye, as relatively faint smudges of light, although two of the very nearest galaxies, the Magellanic Clouds, are visible are moderately large clouds from the Southern hemisphere.

Motions in the sky

Hope this helps! :)

Answer:

Salt dissolving

Explanation:

Dis solving salt in water doesn't change it's chemical composition

Answer:

Iron Rusting

6th grade stuff or 5th i think

Explanation:

Answer:

The original length of the iron rod is approximately 572.189 meters.

Explanation:

This is a case of linear expansion, which is defined by the following differential equation:

[tex]\alpha = \frac{1}{L}\cdot \frac{dL}{dT}[/tex] (1)

Where:

[tex]\alpha[/tex] - Linear expansion coefficient, measured in [tex]\frac{1}{^{\circ}C}[/tex].

[tex]L[/tex] - Length of the element, measured in centimeters.

[tex]\frac{dL}{dT}[/tex] - First derivative of the length of the element with respect to temperature, measured in centimeters per degree Celsius.

If we assume that thermal deformation are small regarding the length of the element, then we simplify (1) in the following form:

[tex]\alpha \approx \frac{\Delta L_{o}}{L\cdot \Delta T}[/tex]

[tex]L_{f} -L_{o} = \alpha \cdot L_{o} \cdot \Delta T[/tex]

[tex]L_{f} = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})][/tex] (2)

Where:

[tex]L_{o}[/tex], [tex]L_{f}[/tex] - Initial and final lengths of the element, measured in centimeters.

[tex]T_{o}, T_{f}[/tex] - Initial and final temperatures of the element, measured in degrees Celsius.

Given that brass has a higher coefficient of linear expansion, it is suppose that initial length is less than the initial length of the iron element. Then, we have the following system of linear equations:

Brass

[tex]L = L_{o}\cdot [1+\alpha_{B}\cdot (T_{f}-T_{o})][/tex] (3)

Iron

[tex]L = (L_{o}+28)\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})][/tex] (4)

Where [tex]\alpha_{B}[/tex], [tex]\alpha_{I}[/tex] are coefficients of linear expansion of brass and iron, measured in [tex]\frac{1}{^{\circ}C}[/tex].

By equalizing (3) and (4), we have the following formula:

[tex]L_{o} \cdot [1+\alpha_{B}\cdot (T_{f}-T_{o})] = (L_{o}+28)\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})][/tex]

[tex]L_{o} \cdot (\alpha_{B}-\alpha_{I})\cdot (T_{f}-T_{o}) = 28\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})][/tex]

[tex]L_{o} = \frac{28\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})]}{(\alpha_{B}-\alpha_{I})\cdot (T_{f}-T_{o} )}[/tex]

If we know that [tex]\alpha_{B} = 1.9\times 10^{-5}\,\frac{1}{^{\circ}C}[/tex], [tex]\alpha_{I} = 1.2\times 10^{-5}\,\frac{1}{^{\circ}C}[/tex], [tex]T_{o} = 20\,^{\circ}C[/tex] and [tex]T_{f} = 90\,^{\circ}C[/tex], then the initial length of the iron rod is:

[tex]L_{o} = \frac{28\cdot [1+\left(1.2\times 10^{-5}\,\frac{1}{^{\circ}C} \right)\cdot (90\,^{\circ}C-20\,^{\circ}C)]}{\left(1.9\times 10^{-5}\,\frac{1}{^{\circ}C}-1.2\times 10^{-5}\,\frac{1}{^{\circ}C} \right)\cdot (90\,^{\circ}C-20\,^{\circ}C)}[/tex]

[tex]L_{o} = 57190.857\,cm[/tex]

[tex]L_{o, I} = 57218.857\,cm[/tex]

[tex]L_{o,I} = 572.189\,m[/tex]

The original length of the iron rod is approximately 572.189 meters.

Answer:

The speed of the car is 18.75 m/s

Explanation:

Impulse-Momentum Change Equation

The impulse received by an object is equal to the change in momentum:

Impulse = Change in momentum

The change in momentum is:

[tex]\Delta p=m\Delta v=m(v_2-v_1)[/tex]

Where m is the mass of the object, v2 is the final speed, and v1 is the initial speed. This means the impulse J is:

[tex]J=m(v_2-v_1)[/tex]

It's given an m=1,600 kg car is moving at v1=25 m/s North and then an impulse of J=-10,000 N.s is applied South. The minus sign indicates the impulse is opposite to the original direction of the velocity.

Solving for v2:

[tex]\displaystyle v_2=\frac{J}{m}+v_1[/tex]

Substituting:

[tex]\displaystyle v_2=\frac{-10,000}{1,600}+25[/tex]

[tex]\displaystyle v_2=-6.25+25=18.75[/tex]

[tex]v_2=18.75\ m/s[/tex]

The speed of the car is 18.75 m/s

Complete Question:

A force of 5.0 N moves a 6.0 kg object along a rough floor at a constant speed of 2.5 m/s.

(a) How much work is done in 25 s?

(b) What power is being used?

(c) What force of friction is acting on the object?

Answer:

a. Workdone = 312.5 Joules.

b. Power = 12.5 Watts.

c. Force of friction = 5 Newton.

Explanation:

Given the following data;

Force = 5N

Speed = 2.5m/s

Mass = 6kg

Time = 25 secs

a. To find the work done.

We would first solve for the distance covered by the object.

Distance = speed * time

Distance = 2.5 * 25

Distance = 62.5 meters.

Workdone = force * distance

Workdone = 5 * 62.5

Workdone = 312.5 Joules.

b. To find the amount of power being used.

Power = workdone/time

Power = 312.5/25

Power = 12.5 Watts.

c. To find the force of friction is acting on the object.

Since we know that there is no net force acting on the body. Therefore, the force of friction is equal to 5 Newton.

a) Power is being used will be 12.5 Watts.

b)  Force of friction is acting on the object will be 5 Newton.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

m is the mass of object =6.0 kg

u is the speed =2.5 m/s

a) Power is being used will be 12.5 Watts.

The distance traveled by the object is;

[tex]\rm d = v \times t \\\\ \rm d = 2.5 \times 25 \\\\ \rm d = 62.5 m[/tex]

The work done is the product of force and displacement;

[tex]\rm W = F \times d \\\\ \rm W = 5 \times 62.5 \\\\ \rm W = 312.5 \ J[/tex]

The power is the ratio of work done to the  time;

[tex]\rm P= \frac{W}{t} \\\\ \rm P= \frac{312.5 }{25 } \\\\\ \rm P=12.5 \ Watt[/tex]

Hence Power is being used will be 12.5 Watts.

b) Force of friction is acting on the object will be 5 Newton.

If the all unbalanced force is zero the net force is equal to the force of friction;

The force of friction= Net force

Force of friction= 5 Newton.

Hence the force of friction is acting on the object will be 5 Newton.

To learn more about the friction force refer to the link;

https://brainly.com/question/1714663

Answer:

1 or 2

Explanation:

Answer:

HI!

Here is ur answer

Magnetic PE depends on both distance and orientation in the magnetic field. ... The magnetic fields are aligned. This makes an attractive force between the two magnets. The farther apart they are, the higher their potential energy.

Answer:

Check attachment for your answer

Good luck

Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

By Newton's second law,

• the net vertical force on the box is

∑ F = n - mg = 0

where n = magnitude of the normal force, m = mass of the box, and g = acceleration due to gravity; then

n = mg = (20 kg) (9.8 m/s²) = 196 N

• the net horizontal force on the box is

∑ F = -f = -ma

where f = mag. of kinetic friction and a = acceleration of the box. Since

f = µn

where m is the coefficient of kinetic friction, so that

f = 0.5 (196 N) = 98 N

Then the box's acceleration is

-98 N = - (20 kg) a   →   a = (98 N)/(20 kg) = 4.9 m/s²

Answer:

7 hours.

Explanation:

This is because you are traveling 5 kilometers per hour so 5 multiplied by 7= 35.

Answer:

1.5 km/s²

Explanation:

Given that:

a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s

after time (t) = 20 seconds

the final velocity = 108 km/hr = 30 m/s

The acceleration (a) of the car can be determined by using the formula:

[tex]a = \dfrac{v-u}{t}[/tex]

[tex]a = \dfrac{30\ m/s -0 \ m/s}{20 \ s}[/tex]

[tex]a = \dfrac{30 \ m/s}{20 \ s}[/tex]

a = 1.5 km/s²

Answer:

The velocity of the second ball is 15.88 m/s East

Explanation:

The given parameters are;

The type of collision = Perfectly elastic collision

The mass of the ball moving East, m₁ = 0.400 kg

The velocity of the ball moving East, v₁ = 3.7 m/s

The mass of the ball at rest, m₂ = 0.200 kg

The initial velocity of the ball at rest, v₂ = 0 m/s

The velocity of the first ball, 'm₁', after the collision, v₃ = 4.24 m/s (The direction is assumed as being towards the West)

The velocity, 'v₄', of the second ball, 'm₂', after the collision is given by the law of conservation of linear momentum as follows;

m₁·v₁ + m₂·v₂ = m₁·v₃ + m₂·v₄

Substituting the known values gives;

0.400 kg × 3.7 m/s + 0.200 kg × 0 m/s = 0.400 kg × (-4.24 m/s) + 0.200 kg × v₄

∴ 0.200 kg × v₄ = 0.400 kg × 3.7 m/s - 0.400 kg × (-4.24 m/s)

Therefore;

0.200 kg × v₄ = 0.400 kg × 3.7 m/s + 0.400 kg × 4.24 m/s = 3.176 kg·m/s

v₄ = 3.176 kg·m/s/(0.200 kg) = 15.88 m/s

The velocity of the second ball = v₄ = 15.88 m/s East.

Explanation:

Parameters

Number of oscillation = 40 oscillations

Time of oscillation = 128s

Formular

Period (T) = No of oscillation

Time taken

T = 128

40

T=3.2s

Answer:

The baseball was falling during 1.733 seconds.

Explanation:

The baseball experimented a free fall, which is a particular case of uniformly accelerated motion, in which object is accelerated by gravity. In this case, we need to determine the time spent by the ball before hitting the ground ([tex]t[/tex]), measured in seconds, which is determined by the following kinematic formula:

[tex]t = \frac{v-v_{o}}{g}[/tex] (1)

Where:

[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the baseball, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 17\,\frac{m}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the time spent by the baseball is:

[tex]t = \frac{17\,\frac{m}{s}-0\,\frac{m}{s} }{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]t = 1.733\,s[/tex]

The baseball was falling during 1.733 seconds.

M = mass of the ball A = 5.0 kg

m = mass of the ball B = ?

V = initial velocity of the ball A before collision = 20 m/s

v = initial velocity of the ball B  before collision = 10 m/s

V' = final velocity of the ball A after collision = 10 m/s

v' = final velocity of the ball B after collision = 15 m/s

using conservation of momentum

M V + m v = M V' + m v'

(5.0) (20) + m (10) = (5.0) (10) + m (15)

100 + 10 m= 50 + 15 m

5 m = 50

m= 10 m/s

Answer: 5

Explanation:

Answer:

Explanation:

If a cat walk 100m East, this means that it is walking in the positive x direction, the distance will therefore be +100m

If it turns around and walks 25 m west, the direction of movement is in the negative x direction i.e -25m

Taking the sum;

Displacement = +100m - 25m

Displacement of the cat = 75m

hence the cats displacement is 75m

Answer:

The greater the luminosity of a star, the longer its period of oscillation.

Answer:

the angle of reflection also increases

Explanation:

this is because the law of reflection states that the angle of incidence equals to the angle of reflection, thus, if the angle of incidence increases, angle of reflection should also increase, basically , think as if the angle of incidence is the same thing as the angle of reflection

hope this helps, if not please report it someone else can try it

Answer:

300Nm

Explanation:

work = force x distance = 60x5=300

The work required to pull a sled by 5 meters using 60 N of force is 300 joules.

Work done in physics is defined as the dot product of force and displacement caused by it. Mathematically -

W = F.x

W = |F| |x| cos Ф

where Ф is the angle between force and displacement

Given is to calculate the work required to pull a sled by 5 meters by using 60 N of force.

We can write the following data -

Applied Force [F] = 60 Newtons

Displacement needed [x] = 5 meters

Angle between force and displacement [Ф] = 0°

Using the formula for work done, we can write -

W = |F| |x| cos Ф

W = 60 x 5 x cos(0)

W = 300 x 1

W = 300 Joules

Therefore, the work required to pull a sled by 5 meters using 60 N of force is 300 joules.

To solve more questions on Work, energy and power, visit the link below-

brainly.com/question/3893991

#SPJ6

Explanation:

A machine with a velocity ratio of 30 moves a load of 3000N when an effort of 200N is applied.the efficiency of the machine is what?

Answer: small intestine

Explanation:

I took a test

Answer:

A) A wedge

B) The efficiency of the machine is 50%

Explanation:

The type of simple machine Scott used is the wedge

A) A wedge is a simple machine that consists of a long thick metal or wood stem that tapers to a thin (Relatively sharp) end. It is used for a splitting, tightening,  and lifting.

An example of a wedge is an axe

B) The efficiency of a machine is given approximately as the ratio of the work output to the work input as follows;

The amount of input energy the axe uses = 200 Joules

The amount of output work Scott receives from the axe  = 100 Joules

Therefore, the efficiency of the machine, η, is given as follows;

[tex]\eta = \dfrac{The \ out.put \ work \ Scott \ receives \ from \ the \ axe}{The \ amount \ of \ in.put \ energy \ the \ axe \ uses} \times 100[/tex]

[tex]\therefore \eta = = \dfrac{100 \ J}{200 \ J} \times 100 = 50\%[/tex]

The efficiency of the machine, η = 50%

Answer:

Forms over water, warm humid air mass, it's a polar air mass

Explanation: I think that's right sorry if it's not..

GL! :)

Answer:

The magnitude of the gravitational force of attraction between the two objects when they are 416 meters apart is;

D. 4.5 × 10⁻¹⁰ Newtons

Explanation:

The given parameters are;

The distance between the center of mass of the cow and the tractor, r = 208 m

The gravitational attraction between the cow and the tractor = 1.8 × 10⁻⁹ N

The formula for finding the gravitational force, 'F', between the cow and the tractor is given as follows;

[tex]F =G\cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}[/tex]

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹m³·kg⁻¹·s⁻²

m₁·m₂ = The product of the mass of the cow and the tractor

Therefore, we have;

[tex]F = 1.8 \times 10^{-9} =G\cdot \dfrac{m_{1} \cdot m_{2}}{208^{2}}[/tex]

1.8 × 10⁻⁹ × 208² ≈ 7.78752 × 10⁻⁵ = G × m₁ × m₂

Therefore, when the distance between the two objects are 416 meters apart, we have;

[tex]F = \dfrac{G \cdot m_{1} \cdot m_{2}}{r^{2}} = \dfrac{7.78752 \times 10^{-5}}{416^2} = 4.5 \times 10^{-10}[/tex]

The magnitude of the gravitational force of attraction between the cow and the tractor when they are 416 meters apart, F = 4.5 × 10⁻¹⁰ N.

Answer:

a = F/m

a = 9.80÷ 10.0

a 0.98 m/s²