Name the two sources of light.​

Answers

Answer 1

Answer:

Natural light (Sun, stars.etc.) and Artificial light (Lamposts, televisions.etc.)

Hope that helps. x


Related Questions

Which of the following best describes reverberation?
A.The wave fronts become mixed and broken up due to contact with a rough or
irregular surface.
B. A change in the sound wave velocity causes the wave to bend in a different
direction.
C. A fraction of the sound waves are absorbed by an object and converted to heat
energy.
D. A single sound undergoes several reflections due to multiple reflecting surfaces.

Answers

Reverberation, in psychoacoustics and acoustics, is a persistence of sound after the sound is produced

Explanation:

I think it is right hope its helps

Answer:

D. A single sound undergoes several reflections due to multiple reflecting surfaces.

Explanation:

Sometimes, the source of a sound is surrounded by multiple reflecting surfaces. The waves traveling from the source strike these different surfaces, causing multiple reflections. For example, a single clap of thunder reflects on several clouds and the earth's surface, causing you to hear a rolling rumble instead of a single sound.

D=12000 m
T= 30min
V=?
Ayudenme en este ejercicio xfa
En m/min y en Km/h

Answers

Ugh can u say it in English im sorry i don’t read Spanish

find the expression for the displacement covered in nth or in last one second​

Answers

Answer:

Snth = u + a/2 ( 2n - 1)

Explanation:

Do you need explanation based on graph, integration or other method?

A plane mirror produces a _____.

virtual image
refracted image
real image

Answers

A plan Murrow produces a virtual image

Answer:

Explanation:

A plane mirror is the kind you look into when you look into a "regular" mirror. The image you see is right-side-up. These images are virtual. Real images are always upside down and are made by mirrors that are "parabolic" in shape. Virtual images are always right-side-up.

A railroad car (with a mass of 3250 kg) moves at 8.1 m/s . It collides with and couples to another car that was initially at rest. After the collision, the two cars move together at a speed of 4.50 m/s. What is the mass of the second car?

Answers

Answer:

m_2 = 2600kg

Explanation:

P_1 = P_2

P = (m_1)*(v_1)+(m_2)(v_2)

P_1 = (3250kg)(8.1m/s)+(m_2)(0m/s)

P_1 = 26,325 kg*m/s

P_2 = (3250kg)(4.5m/s)+(m_2)(4.5m/s)

P_2 = 14,625kg*m/s+(4.5m/s)m_2

26,325 kg*m/s = 14,625kg*m/s+(4.5m/s)m_2

11,700kg*m/s = (4.5m/s)m_2

m_2 = 2600kg

If the velocity of a motorcycle increases from 30 mis too 50m/s in 10 seconds what will be the acceleration of motorcycle?​

Answers

Answer:

2 m/s^2

Explanation:

a = ∆v / ∆t

a = (50 - 30) / 10

a = 20 / 10

a = 2 m/s^2

acceleration = u-v/t

50-30/10

=2m/s

the atomic number of phosphorus is

Answers

Answer:

The atomic number of phosphorus is 15.

Explanation:

It’s found after Si(Silicon) and before S(sulphur)

to all the physicians please help this is for my assignment

Answers

Answer:

Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers

Mathematically, we have;

[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]

Where;

m₁, and m₂ are the masses of the bodies

R = The distance between their centers

G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²

The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers

Q. 2. Newton's law of gravitation in vector form is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]

The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1

The force of gravity of body 2 on 1 is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]

The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)

[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]

Explanation:

Discuss the role of globalization in the development of sI unit​

Answers

Answer:

Sharing of information

Explanation:

The development of SI unit has helped in the sharing of scientific as well as techical information internationally.

Answer:

It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.

2- A copper wire of 3mm diameter with conductivity of 6.7 10' (0.M), and electron mobility of 0.0064 m2 /V sec. Is subjected to an electric field of 30 mV/m. Find (a) the charge density of free electrons, (b) the current density, (c) current flowing in the wire, (d) the electron draft velocity.

Answers

Answer:

a)  [tex]n=5.98*10^{26}/m^3[/tex]

b)  [tex]i=2010000A/m^2[/tex]

c)  [tex]I_w=14.207A[/tex]

d)  [tex]V_e=1.92*10^{-4}m/s[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=3mm=>3*10^{-3}[/tex]

Conductivity [tex]\sigma= 6.7 10^7 (0.M),[/tex]

Electron mobility [tex]\phi= 0.0064 m2 /V sec[/tex]

Electric field [tex]E= 30 mV/m[/tex]

a)

Generally the equation for Charge Density is mathematically given by

[tex]\phi=\frac{\sigma}{n e}[/tex]

Therefore

[tex]n=\frac{\sigma}{\phi e}[/tex]

[tex]n=\frac{6.7 10^7}{1.6*10^{-19} *0.0064}[/tex]

[tex]n=5.98*10^{26}/m^3[/tex]

b)

Generally the equation for current density is mathematically given by

[tex]i=\sigma*E[/tex]

[tex]i= 30*10^{-3] *6.7 10^7[/tex]

[tex]i=2010000A/m^2[/tex]

c)

Generally the equation for current in wire is mathematically given by

[tex]I_w=iA[/tex]

[tex]I_w=i*\pi r^2[/tex]

[tex]I_w=(2010000)*\pi( 1.5*10^{-3})^2[/tex]

[tex]I_w=14.207A[/tex]

d)

Generally the equation for electron draft velocity. is mathematically given by

[tex]V_e=\phi E[/tex]

[tex]V_e=(0.0064)*(30*10^{-3})[/tex]

[tex]V_e=1.92*10^{-4}m/s[/tex]

how is one standard kilogram defined in SI system?​

Answers

Answer:

One standard kilogram is define as the mass of platinium iridium cylinder having equal diameter and height kept at the particular condition of international bureo of weigh and measure sevre near paris.

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.

Answers

Answer:

8392

Explanation:

d=s/t

A tray containing 0.20kg of water at 20degree celsius is placed in a freezer. The temperature of the water drops to 0degree celsius in 10 minutes. Calculate


a) The energy lost by the water asit cools to 0 degree celsius.

b) The average rate at which the water is losing energy in J/s.

c) Estimate the time taken for the water at 0 degree celsius to turn completely into ice.

d) state any assumptios you make​

Answers

Answer:

a. Energy lost, Q = 16,800 Joules.

b. Power = 28 J/s

c. Time, t = 2357.14 seconds

d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

Explanation:

Given the following data;

Mass = 0.20 kgInitial temperature, T1 = 20°CFinal temperature = 0°CTime = 10 minutes

a. To find the energy lost by the water as it cools to 0 degree celsius;

Mathematically, heat capacity is given by the formula;

[tex] Q = mcdt [/tex]

Where;

Q represents the heat capacity or quantity of heat.M represents the mass of an object.C represents the specific heat capacity of water.dt represents the change in temperature.

dt = T2 - T1

dt = 20 - 0

dt = 20°C

We know that the specific heat capacity of water is equal to  4200 J/kg°C

Substituting the values into the formula, we have;

[tex] Q = 0.20 * 4200 * 20 [/tex]

Energy lost, Q = 16,800 Joules.

b. To find the average rate at which the water is losing energy in J/s by using the following formula;

[tex] Power = \frac {energy}{time} [/tex]

First of all, we would have to convert the value of time in minutes to seconds.

Conversion:

1 minute = 60 seconds

10 minutes = X seconds

Cross-multiplying, we have;

X = 60 * 10

X = 600 seconds

Substituting the values into the formula, we have;

[tex] Power = \frac {16800}{600} [/tex]

Power = 28 J/s

c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;

We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.

Mathematically, the latent heat of fusion is calculated by using the formula;

Energy, Q = ml = pt

Substituting the values into the formula, we have;

0.20 * 3.3 * 10⁵ = 28 * t

0.20 * 330000 = 28t

66000 = 28t

[tex] t = \frac {66000}{28} [/tex]

Time, t = 2357.14 seconds.

d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

You observe that you see more mockingbirds in small trees and more hawks in large trees. Which of the following is an appropriate scientific question based on this observation?
How does the size of a tree affect the bird species that prefer to live in it?
How do birds fly?
What type of food do birds eat?
What time of year to birds mate?

Answers

Hi! I think the answer is How does the size of a tree defect the bird species that prefer to live in it or option A. I would go with this option because it is the most logical answer. Option B doesn’t really have anything to do with why the two different types are picking different size of trees to live in. Option C. seems like the second most logical answer. Option D. Wouldn’t make the different types of birds pick different trees It shouldn’t matter when they mate if they are staying in the same types of trees. I hope this helped, Goodluck :)

Answer:

A) How the size of a tree affect the bird species that prefer to live in it

Explanation:

I took the quiz

The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?


A)2.10x10^5J

B)7.78x10^-21J

C)6.30x10^13J

D)2.10x10^61J

Answers

Answer:

[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]

300 ml of a gas at 27°C is Cooled at -3°c at Constant pressure the final volume is plzz answer fast i will mark brainliest ​

Answers

Answer:V₁=300ml

T₁=27°C

V₂=?

T₂= -3°C

as we know

V₁T₁=V₂T₂

By putting values in formula

300ml×27°C=V₂×(-3°C)

300ml×27°C/-3°C=V₂

8100ml/-3=V₂

-2700ml=V₂

or V₂=  -2700ml

A tightrope walker is walking between two buildings holding a pole with length L=14.0 m, and mass mp=17.5 kg. The daredevil grips the pole with each hand a distance d=0.595 m from the center of the pole. A bird of mass mb=560 g lands on the very end of the left‑hand side of the pole. Assuming the daredevil applies upward forces with the left and right hands in a direction perpendicular to the pole, what magnitude of force Fleft and Fright must the left and right hand exert to counteract the torque of the bird?

Answers

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

          Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

          W_bird  L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

          F_left = F_right = F

          W_bird L / 2 - 2 F 0.595 = 0

          F = [tex]\frac{m_{bird} \ g L} { 4 \ 0.595}[/tex]

   

we calculate

         F = 0.560 9.8 14.0 /2.38

         F = 32.28 N

A cricketer throws a ball sideways with an initial velocity of 30 m/s. She releases the ball from a height of 1.3m. Calculate how far the ball travels before hitting the ground.

Answers

Answer:

78.34

Explanation:

1.3/30=78.3m

please answer these diagrammatic questions ASAP and please no spam answers​

Answers

Answer:

i. The pressure of due to the water, P, is given according to the following equation;

P = ρ·g·h

Where;

ρ = The density of the water (a constant) = 997 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the water (minimum h = h₁, maximum h = h₂)

The pressure is directly proportional to the water height, and we have;

The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC

ii. The thrust = The force acting on the body = Pressure × Area

The maximum areas exposed to the water are on side AB and DC

However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC

Explanation:

In a lunar experiment, a 950-g aluminum (920 J/(°Ckg)) sphere is dropped from the space probe while is 75 m above the Lunar ground. If the sphere’s temperature increased by 0.11°C when it hits the ground, what percentage of the initial mechanical energy was absorbed as thermal energy by the aluminum sphere?

Answers

Answer:

13.759 % of the initial mechanical energy is lost as thermal energy.

Explanation:

By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:

[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)

Where [tex]x[/tex] is the percentage of the energy loss, no unit.

By definition of the gravitational potential energy and internal energy, we expand this equation:

[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)

Where:

[tex]m[/tex] - Mass of the object, in kilograms.

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

[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.

[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.

[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.

If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:

[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]

[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]

[tex]x = 13.759\,\%[/tex]

13.759 % of the initial mechanical energy is lost as thermal energy.

The moon Phobos orbits Mars
(mass = 6.42 x 1023 kg) at
a distance
of 9.38 x 106 m. What is its period of
orbit?
[?]s

Answers

Answer:

Explanation:

We are basically needing to solve for the time in the equation d = rt, where d is the distance around Mars (aka the circumference), r is the velocity, and t is time. We need to find the circumference and the velocity. We will begin with the velocity.

Because the gravitational attraction between Phobos and Mars provides the centripetal acceleration necessary to keep Phobos in its (sort of) circular path, the equation we use for this is:

[tex]F_g=F_c[/tex] which says that Force supplied by gravity is equal to the centripetal force. Expanding that:

[tex]\frac{Gm_{Phobos}m_{Mars}}{r^2}=\frac{m_{Phobos}v^2}{r}[/tex]

When we move that around mathematically to solve for the velocity value, what we end up with is:

[tex]v=\sqrt{\frac{Gm_{Mars}}{r}[/tex] and filling in:

[tex]v=\sqrt{\frac{(6.67*10^{-11})(6.42*10^{23})}{9.38*10^6} }[/tex] and we get that

v = 2100 m/s

Now for the circumference:

C = 2πr and

C = 2(3.1415)(9.38 × 10⁶) so

C = 5.9 × 10⁷

Putting that all together in the C = vT equation:

5.9 × 10⁷ = 2100T so

T = 2.8 × 10⁴ sec or 7.8 hours

Indoor pollution experts conducted an analysis of the paint used in many office buildings which revealed that the paint contains traces of lead. The lead seems to rise to the surface of the paint and escape into the building's air supply. Tests show that because of the chemical drying process, the lead is not discernible on the surface until the paint has been on the walls for at least six months. To meet safety standards, owners should repaint walls at least every six months, or cover the walls with a different material.Which of the following, if true, would most weaken the conclusion above?A) The indoor pollution experts had no clear understanding of why it took six months for the lead to become discernible on the paint's surface.B) The indoor pollution experts neglected to examine the paint for traces of other toxic substances such as cadmium and mercury.C) The amount of lead found on the surface of the paint after six months remained constant for the next two years.D) The indoor pollution experts found that even in those offices painted with a different brand of paint, traces of lead were still found in the air workers breathed.E) The indoor pollution experts' research shows that the amounts of lead that come into contact with the air people breathe, even in the office buildings that used the greatest amount of paint, are too low to affect workers.

Answers

Answer:

E) The indoor pollution experts' research shows that the amounts of lead that come into contact with the air people breathe, even in the office buildings that used the greatest amount of paint, are too low to affect workers

Explanation:

The conclusion that would weaken the above statement if true was that the amount of lead found in air was not hazardous for the people living in building and the effects of such paint was quite low o affect workers.

What is the decay constant for Oxygen-19 if it has a half-life of 26.5s?


A)0.0262/s

B)18.4

C)0.0377/s

C)38.2/s

Answers

Answer:

Option A.

Explanation:

We define the half time T as the time such that an initial quantity A reduces to its half.

So we can model the quantity as a function of time like:

P(t) = A*e^(-k*t)

Then for the half time, T, we will have:

P(T) = A/2 = A*e^(-k*T)

solving for k, we get:

A/2 = A*e^(-k*T)

1/2 = e^(-k*T)

ln(1/2) = ln( e^(-k*T)) = -k*T

-ln(1/2)/T = k

Here we know that the half time is T = 26.5s

if we input that in the above equation, we get:

-ln(1/2)/26.5s = k = 0.0262 s^-1

Then the correct option is A

1. Find the temperature when the degrees of the Celsius scale will be one fifth of the corresponding degrees of the Fahrenheit scale

2. How much heat is necessary to warm 500g of water from 20°C to 65°C?​

Answers

Answer:

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50  

Check:

C = 5/9 ( F - 32)  = 5/9 (50 - 32) = 10  as requested

Q = c m change in temp

Q = 1 cal/gm-deg C * 500 gm * 45 deg C = 22,500 calories

50 Fahrenheit heat required.

How much heat required?

Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy  between physical systems.

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50

The answer is 50 Fahrenheit.

Learn more about heat transfer

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We reduce friction in machines? why

Answers

Answer:

friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.

Answer:

Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries

Explanation:

Sort the processes based on the type of energy transfer they involve. condensation freezing deposition sublimation evaporation melting thermal energy added thermal energy removed

Answers

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

Explanation:

Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.

Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.

On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse

Answers

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

(Q022) A negative magnetic anomaly a. occurs when the Earth's magnetic field measured in ancient rocks is the same as it is today. b. is created when weak magnetic forces in basalt grains add to the force produced by the Earth's dipole. c. describes the sawtooth pattern of magnetic signal strength measured along the Atlantic Ocean seafloor. d. is indicated when a magnetometer measures intervals of magnetism that are weaker than expecte

Answers

Answer:

d. Is indicated when a magnetometer measures intervals of magnetism that are weaker than expected

Explanation:

Magnetic field intensity is measured with a magnetometer on the surface of the Earth. Areas in which the magnetic field strength is lower or more than average are known as areas with magnetic anomalies, which may be due to  the presence of rocks that have a different magnetic characteristics

Where the magnetic anomaly is negative, it is indicative of a magnetic field strength reading that is lower than average magnetic field which is generally obtainable

Therefore, the correct option is indicated when a magnetometer measures intervals of magnetism that are weaker than expected

84. Three resitors each of value 30 respectively are connected in a parallel

combination across a 10 V battery the current through each resitor is​

Answers

Answer:

each resistor draws 1/3 of an amp or 0.33333 amps

Explanation:

V = I * R

V = 10 volts

R = 30 ohms

10 = I * 30       Divide by 30

10/30 = I

I = 0.33333

10. Explain the principle of electric motor. Write its uses.

Answers

Explanation:

The principle of an electric motor is based on the current carrying conductor which produces magnetic field around it. A current carrying conductor is placed perpendicular to the magnetic field so that it experiences a force.

The largest electric motors are used for ship propulsion, pipeline compression and pumped-storage applications with ratings reaching 100 megawatts. Electric motors are found in industrial fans, blowers and pumps, machine tools, household appliances, power tools and disk drives.

Other Questions
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