A red light, green light, and blue light differ in A. amplitude. B. complexity. C. wavelength. D. purity.

spray him in the eyes and you have until he washes it put

Answer: the expanding universe

Explanation:

Hope that helps!

There are around one billion light years across galaxies on ordinary. The observable cosmos has around 100 billion galaxies.

In research, observation is vital. Scientists gather and record data through observation, which allows them to create and subsequently test concepts and hypotheses. Scientists can observe in many different ways, including using their own senses or instruments like telescopes, thermometers, satellites, or stethoscopes.

In addition, Hubble proved a basic fact now known as Hubble's Law: galaxies distant from us are retreating faster than those closer to us. The Big Bang Theory's central tenet is the notion of an expanding cosmos. The first understanding of the beginnings of our cosmos came from Hubble's observations.

There are around one billion light years between galaxies on normal. The observable cosmos has around 100 billion galaxies.

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Answer:

a. 104.5 C b. 7.94 × 10⁶ A/m² c. 5.83 × 10⁻⁴ m/s

Explanation:

a. How much electric charge passes through the starter motor?

Using Q = It where Q = electric charge passing through the starter motor, I = current = 110 A and t = time = 0.95 s

So, Q = It = 110 A × 0.95 s = 104.5 C

b. What is the current density in the wire?

The current density, J = I/A where I = current = 110 A and A = cross-sectional area = πd²/4 where d = diameter of copper wire = 4.20 mm = 4.20 × 10⁻³ m

So, J = I/A

= I/πd²/4

= 4I/πd²

= 4 × 110 A/π(4.20 × 10⁻³ m)²

= 440 A/55.42 × 10⁻⁶ m²

= 7.94 × 10⁶ A/m²

c. How far does an electron travel along the wire while the starter motor is on? (The density of conduction electrons in copper is n = 8.50×1028 1/m3.)

To find how far the electron travels, we need to find the electron drift velocity from

J = nev where J = current density = 7.94 × 10⁶ A/m², n = electron density = 8.50 × 10²⁸ m⁻³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of copper wire = 4.20 mm = 4.20 × 10⁻³ m

So, v = J/ne

Substituting the values of the variables into the equation, we have

v = 7.94 × 10⁶ A/m² ÷ (8.50 × 10²⁸ m⁻³ × 1.602 × 10⁻¹⁹ C)

v = 7.94 × 10⁶ A/m² ÷ (13.617 × 10⁹ Cm⁻³)

v = 0.583 × 10⁻³ m/s

v = 5.83 × 10⁻⁴ m/s

Explanation:

Physics is the branch of science concerned with the study of the properties and interactions of space, time, matter and energy.

Answer:

if you are very excited to know why ball doesn't go up

Answer:

chinese

Explanation:

Answer:

B chinese

hope this helps you

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = [tex]\frac{1}{2}[/tex] mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  [tex]\frac{1}{2}[/tex] at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²

mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{5}[/tex]mR²) ω²

v = √( [tex]\frac{10}{7}[/tex]gh₁  )

so we substitute √( [tex]\frac{10}{7}[/tex]gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( [tex]\frac{10}{7}[/tex]gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²

mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{3}[/tex]mR²) ω²

v = √( [tex]\frac{6}{5}[/tex]gh₁ )

so we substitute √( [tex]\frac{6}{5}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( [tex]\frac{6}{5}[/tex]gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²

mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{5}[/tex]mR²) ω²

v = √( [tex]\frac{10}{7}[/tex]gh₁  )

so we substitute √( [tex]\frac{10}{7}[/tex]gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( [tex]\frac{10}{7}[/tex]gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

We have that for the Question it can be said that

From the question we are told

Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R= 15mm in each case; assume that the density of steel is 7.8 g/cm3; and assume that the density of aluminum is 2.7 g/cm3.

Generally the equation for sliding without friction is mathematically given as

[tex]V = \sqrt{4Hh}[/tex]

the equation for sliding without slipping is mathematically given as

[tex]X = \sqrt{\frac{4Hh}{1+I/mR^2}}[/tex]

A) A solid steel sphere sliding down the ramp without friction.[tex]V = \sqrt{4*1.5*0.4}\\\\= 1.55m[/tex]

B) .A solid steel sphere rollingdown the ramp without slipping.[tex]I = 2/5 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m[/tex]

C) A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping.

[tex]I = 2/3 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/3}}\\\\= 1.2m[/tex]

D) A solid aluminum sphere rolling down the ramp without slipping.

[tex]X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m[/tex]

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Answer:

the mouse

Explanation:

the mouse has a momentum of 33 m kg/s

while the elephant has a momentum of 27 m kg/s

i found this out using p=mv

Answer:

B) μ = 0.23

Explanation:

The coefficient of friction is equal to μ = F / N where μ (mu) is the coefficient of friction, F is the friction force, and N is the normal force (the force of an object being applied onto the earth by gravity).

F, the friction force, is given as 67 N

Since the weight of the crate is 289 N, that means the normal force is 289 N.

This means that the coefficient of friction is  μ = F / N = 67 N / 289 N = 0.23183391 = 0.23

Therefore, B is the correct answer

Yes it can, an object can be moving a certain direction while the ACCELERATION is in the opposite direction.

Lets say your riding a bike... if your squeezing your handle bar breaks, the acceleration of the bike would be pushing in the opposite direction of the direction the bike is moving.

Hope this helped!

Answer:

The net force will be:

[tex] F_{net} = 142.53\: N[/tex]

Explanation:

The net force is given by:

[tex] F_{net} = W_{b}+F[/tex]

[tex] F_{net} = m_{b}g+15[/tex]

[tex] F_{net} = 142.53\: N[/tex]

I hope it helps you!

Answer:

Explanation:

emf   due to movement of a rod of length L in a perpendicular to magnetic field B with velocity v is given as

emf = BLv

Putting in the given values ,

E = 2.45 x 12.5 x 10⁻² x 2.4

= .735 V

This emf produces current in resistance . Power consumed by resistance

V² / R where R is resistance , V is emf induced .

Power = .735² / 1.2

= .45 W .

Answer:

momentum = mass × velocity = 6× 20 =120 kg.ms-1

Explanation:

not sure if this is right

Answer:

T₁ = 15.66 N

Explanation:

From the given information:

Using the free body system in the chain.

T - mg = ma

47 - 3.12 × 9.8 = 3.12a

47 - 30.576  = 3.12a

16.424 = 3.12a

a = 16.424/3.12

a = 5.26 m/s²

Now, by the free body diagram of the lowest link; the tension (T₁ ) in the chain is:

T₁ - (3.12/3)  × 9.8 = (3.12/3) × 5.26

T₁ - 1.04 × 9.8 = 1.04  × 5.26

T₁ - 10.192 = 5.4704

T₁ = 5.4704 + 10.192

T₁ = 15.6624

T₁ = 15.66 N

Answer:

yes it is a great question. Thank you .

Answer:

umm it says he but I'm a she :p

Answer:

I don't understand the question

Explanation:

sorry I cant help because I am just a first former

Answer:

house away far from

Explanation:

I dont know tbh

Answer:

Yes.

Explanation:

Answer:

I do believe it is. (more characters for character limit)

Answer:

1)  LIQUID, 2) LIQUID, 3) GASEOU, 4)  GASEOUS,    v = 0.76 m³ / kg

Explanation:

In this exercise it is asked to determine the state of the water, that is, if it is solid, liquid or gas. For this we must use a phase diagram of water which is a graph of Pressure versus Temperature. Let's describe the water diagram

* Below 0ºC and normal pressure the water is in a solid state

* Below this temperature and at low pressure it becomes a gaseous state

* Above 0.01ºC and normal pressure is in liquid state

* Above 0.01ºC and low pressure is in a gaseous state

there are two important points

* The triple point at t + 0.01ºC and P = 0.006 atm where the three states coexist

* The critical point T = 374ºC and P = 218 atm where water decomposes into hydrogen and oxygen

Specific volume is related to density

        v = 1 / ρ

        ρ = 1 / v

the density of water is approximately 1000 kg / m³ in the liquid state at t = 4ºC and decreases with increasing pressure 960 kg / m³ at T = 100ºC (but without changing to the gaseous state

With the above considerations we can answer the questions

1) P = 300 kPa = 3 105 Pa

   v = 0.5 m³ / kg

atmospheric pressure is Patm = 1.01 105 Pa

     P = 3 105 Pa (1 atm / 1.01 105 Pa) = 3 atm

From the phase diagram described, the water can be in two liquid or gaseous states, depending on the temperature, as indicated by the speed of the same state, the water must be LIQUID

the parameter v that you indicate is the

with the other parameter we can calculate the density

     rho = 1 / 0.5

     rho = 2.0 kg / m³

in a thermodynamic system the three basic properties are: pressure, volume and temperature,

we can calculate the body temperature

The body must be at a temperature between 0 <T <100ºC

2) P = 28 Mpa = 28 106 Pa

   P = 28 106 Mpa (1 atm / 1.01 105 Pa) = 280 atm

   T = 200ºC

When examining the diagram it can be seen that the water is in the range of the LIQUID state

3) P = 1MPa

   P = 1 106 Pa (1atm / 1.01 105Pa) = 10 atm

   T = 405ºC

In this case, the only accessible state is the GASEOUS

4) T = 100ºC

     x = 60%

Examining the phase diagram at this temperature depending on the pressure the possible states are Vapor and liquid, for pressures below 1 atm the state is GASEOUS

for the gaseous state we can use the ideal gas equation

            PV = nR T

let us perform the calculation for a mole of gas n = 1, the ideal gas constant is R = 8,206 10-2 atm / mol K

          V = nRT / p

           V = 1 8,206 10-2 100/1

           V = 8,206 m³

          v = V / m

The mass of the water is 18 Kg / mol, which indicates that 60% is in the gaseous state, so the mass in this state is

         m = 0.60 18

         m = 10.8 kg

          v = 8.206 / 10.8

          v = 0.76 m³ / kg

Explanation:

It depends upon the magnitude of the charge, the velocity of the particle and its strength .

Answer:

volume = 8 cm^3 + 32 g / 0.8 g/ cm^3 = 48 cm^3

mass = 32 + 8 = 40 g

40 g / 48 cm^3

Explanation:

The density of a mixture can be calculated from the individual densities. The average density of the mixture of 32 g kerosene and 8 g of water is 0.9 g/cm³.

Density of a substance is the measure of its mass per unit volume. Thus, mathematically it is the ratio of mass to the volume of the substance. Density depends on the mass, volume, temperature,bond type and pressure.

The ratio of density of a substance to the density of water is called specific gravity. If the specific gravity is greater than 1 then the object will sink on water and if it is less than 1 then it will float on water.

If two liquids of densities ρ₁ and ρ₂ then after mixing their densities will be:

[tex]\rho =\frac{ \rho_{1} + \rho_{2} }{2}[/tex]

Hence the density of mixture of water (1 g/cm³) and kerosene (0.8 g/cm³) is :

= (0.8 + 1) /2

=0.9 g/cm³.

Thus the density of the resulting mixture be 0.9 g/cm³

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Answer:

–I6 N

Explanation:

From the question given above, the following data were obtained:

Spring constant (K) = 80 N/m

Initial position = 0.6 m

Final position = 0.4 m

Force (F) =?

Next, we shall determine the compression of the spring. This can be obtained as follow:

Initial position = 0.6 m

Final position = 0.4 m

Compression (x) =?

Compression = Final position – Initial position

Compression (x) = 0.4 – 0.6

Compression (x) = – 0.2 m

Finally, we shall determine the force. This can be obtained as follow:

Spring constant (K) = 80 N/m

Compression (x) = – 0.2 m

Force (F) =?

F = Kx

F = 80 × –0.2

F = –I6 N

Answer:

for the first paragraph introduce the definition. for the second paragraph write about the first law, for the third paragraph write about the 2nd for the fourth paragraph write about the 3rd law. for the last paragraph do a brief summary of what you wrote and a conclusion about the laws.

i hope this helps a bit

Answer:

You can start off with the first law of motion (newtons first law), talk about what it is or what it does, give examples.

Answer: False

Explanation:

Answer:

Explanation:

From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.

So, in light of that, to determine the electric charges values & signs on each of them, we have:

on a = +3 C

on b = -7 C

on c = -7 C

on d = +3 C

on e = -7 C

Answer:

True.

Explanation:

The density of an object is given by its mass divided by its volume. It can be given as follows :

[tex]d=\dfrac{m}{V}[/tex]

It can be seen that the density of an object is directly proportional to its mass. It means if the mass of an object increase, its density will also increase. Hence, the given statement is true.