How does the suns energy affect the climate of an area

Answers

Answer 1

There wouldn't be any life on Earth and no climate if it weren't for the Sun.

The spherical form of the Earth results in changes in the amount of sunlight that reaches the surface at various latitudes, giving rise to various climatic zones. Seasons are caused by the tilt of the Earth's axis of rotation in relation to the ecliptic plane.

The surface of the sun is around 6000 C hot; at this temperature, the solar mostly produces visible light. The Earth radiates infrared energy back into space to maintain its energy balance after being warmed by incoming solar radiation. The Earth's surface has an average temperature of 15 C, and at this temperature, the planet emits infrared light.

Carbon dioxide, methane, nitrous oxide, and other atmospheric greenhouse gases let solar energy to pass through but absorb infrared radiation that the planet emits back into space. This feature is not shared by other atmospheric elements like oxygen and nitrogen. These greenhouse gases re-emit infrared radiation in all directions to maintain energy balance, some of which is aimed at the Earth's surface and warms it. More greenhouse gases will increase infrared absorption and raise surface temperatures. This is comparable to how much blanket you cover yourself with at night. The blanket's thickness affects how warm you feel. The result of heat brought on by rising levels of greenhouse gases in the atmosphere is climate change.

In the course of billions of years, the Sun will continue to get hotter and brighter, rendering the Earth uninhabitable.

Additionally, all fossil fuels (coal, petroleum, and natural gas) and renewable energy are directly and indirectly derived from the sun (wind, solar, biomass, and hydroelectric).

Thank you,

Eddie

Answer 2
The Sun is the only source of energy for the Earth's climate system. Important to earth's atmosphere is the warming of the atmosphere caused by solar radiation, which also causes global wind patterns and aids in the development of clouds, storms, and rainfall.

Related Questions

(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C

If a charge of −4.72 µC is placed at this point, what are the magnitude and direction of the force on it? Magnitude _______N

Direction?

- toward the left
- upward
-downward
- toward the right

Answers

The electric field strength at a point 1.00 cm to the left of the middle is  -2.0 x 10⁷ N/C.

The magnitude of the force is 94.4 N and direction of the force on it towards the right.

Electric field strength

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

Electric field due to first charge

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

Electric field due to second charge

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

Electric field due to third charge

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

Net electric field

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C

E = -2.0 x 10⁷ N/C

Force on the charge −4.72 µC

F = Eq

F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = 94.4 N

Thus, the direction of the force will be towards the right.

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The object that ends up with less electrons has a __________ charge.
A. Positive
B. Not enough info
C. Neutral
D. Negative
Reset Selection

Answers

The object that ends up with less electrons has a positive charge; option A.

What are electrons?

Electrons are negatively charged particles which form a part of the three fundamental particles in an atom.

Electrons can easily be removed from the vicinity of atoms making the atom to become positively charged. On the other hand, when atoms gain extra electrons, they become negatively-charged.

On way of charging objects is by rubbing them against other objects in order to gain extra electrons or to lose electrons.

When two objects are rubbed together, the object that ends up with less electrons has a positive charge, whereas the object that ends up with more electrons has a negative charge.

In conclusion, objects can become charged when they are rubbed against each other, and they can either become negatively or positively-charged by gain or loss of electrons.

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The object indicated here that ends up with fewer electrons has an unknown charge (not enough info, Option B).

What does atomic charge mean?

The atomic charge makes reference to the overall charge of an atom, which can be evidenced by knowing its net charge (positive or negative charge).

The net charge of a molecule and/or atom is obtained by calculating the amount of electrically negative electrons (e-) and the number of positive protons.

Neutrons are another type of subatomic particle that have not net charge (neither positive nor negative).

In conclusion, the object indicated here that ends up with fewer electrons has an unknown charge (not enough info, Option B).

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. An aeroplane takes 30 min to travel from Biratnagar to Kathmandu. If the air distance between
the Biratnagar and Kathmandu is 450 km. Find the average velocity of the plane.

Answers

Answer:

average velocity = [tex]\bf 900 \space\ km/h[/tex]

Explanation:

We can find the average velocity using the following equation:

[tex]\boxed{average \space\ velocity = \frac{total \space\ distance \space\ travelled}{ \space\ time \space\ taken}}[/tex] .

In this case:

• total distance travelled = 450 km

• time taken = 30 min = 0.5 h

Substituting these values into the equation:

[tex]average \space\ velocity = \frac{450 \space\ km }{ 0.5 \space\ h}[/tex]

                          ⇒ [tex]\bf 900 \space\ km/h[/tex]

What is the equation for the enthalpy of vaporization?

A. Q = m*Lv
B. Q = m/Lv
C. Q = 1/m*Lv
D. Q = Lv/m

Answers

Answer:

A: Q = m*Lv

Explanation:

The equation formula for the enthalpy of vaporization is;

Q = m*ΔHv

Where;

q = heat energy

m = mass

ΔHv = heat of vaporization

A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?

Answers

A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then

(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s

(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.

c) The speed of the 0.950- kg puck after the collision is 2.82 m/s

d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.

Given:

Mass of ice puck, m₁ = 0.350 kg

Mass of another puck, m₂ = 0.950 kg

Velocity of ice puck, v₁ = 5.22 m/s

Velocity of another puck, v₂ = 0 m/s

[tex]v^{'}_1= ?[/tex]

[tex]v^{'}_2= ?[/tex]

[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]

[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]

[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]

Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.

[tex]v^{'}_2 = v^{'} + v_1[/tex]

    [tex]= -2.40+5.22[/tex]

    [tex]= 2.82 m/s[/tex]

Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.

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An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How will the final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected?.

Answers

The kinetic energy of the ejected electron (photoelectron) is equal to the energy of the photon minus the work function (E required to eject the photoelectron).

The  process of final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected -

The electron has a lot of potential energy and very little kinetic energy when it is liberated from the cathode (as it is not moving yet or moving very slowly). We are currently on the left side of the picture above. The electron accelerates and speeds up as it goes away from the cathode and toward the anode. As a result, the kinetic energy rises.The electron is likely to have the highest kinetic energy and the lowest potential energy when it reaches the anode.Therefore, the kinetic energy at the end of the electron's journey is approximately equal as the potential energy at the beginning (when the electron was released).

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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?

Answers

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: [tex]900\; {\rm J}[/tex].

Work done when firing this projectile: [tex]2500\; {\rm J}[/tex].

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity [tex]v_{x}[/tex] of this projectile would stay the same (at [tex]30\; {\rm m\cdot s^{-1}}[/tex]) throughout the flight.

The vertical velocity [tex]v_{y}[/tex] of this projectile would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] at the vertex (highest point) of its trajectory. (Otherwise, if [tex]v_{y} > 0[/tex], this projectile would continue moving up and reach an even higher point. If [tex]v_{y} < 0[/tex], the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be [tex]v = 30\; {\rm m\cdot s^{-1}}\![/tex] when it is at the top of the trajectory. The kinetic energy [tex]\text{KE}[/tex] of this projectile (mass [tex]m = 2.0\; {\rm kg}[/tex]) at the vertex of its trajectory would be:

[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}[/tex].

Apply the Pythagorean Theorem to find the initial speed of this projectile:

[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Hence, the initial kinetic energy [tex]\text{KE}[/tex] of this projectile would be:

[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}[/tex].

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be [tex]2500\; {\rm J}[/tex].

If a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)

a) How far and at what angle is the Aster's final position from her initial position?

b) In what direction would she has to head to return to her initial position

Answers

Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position

What is Displacement ?

Displacement is the distance travelled in a specific direction. It is a vector quantity.

Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.

a) Let P be the Aster's final position from her initial position?

We can use bearing by using Cosine formula to solve this question.

P² = 70² + 82² - 2 × 70 × 82 cos 73

P² = 4900 + 6724 - 11480 cos 73

P² = 11624 - 3356.43

P² = 8267.57

P = √8267.57

P = 90.9 m

P = 90.9 - 28

P = 62.9 m

We can get the angle by using Sine rule

82/ sin Ф = 90.9 / sin 73

sin Ф = 0.8627

Ф = [tex]Sin^{-1}[/tex] (0.8627)

Ф = 59.6°

Ф = 60°

b)  She will head north west direction to return to her initial position

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An object with an initial velocity of
0.12rad accelerates at 0.11rad over a
distance of 0.25 radians. What is the
final angular velocity of the object?
rad
S

Answers

Answer:

100rad because it angular velocity

Maya made this picture to represent a chemical reaction:

Which of the following statements best explains the type of chemical reaction represented by Maya's picture?

A. It is neither a synthesis reaction nor a decomposition reaction because the total mass of the products is less than the total mass of the reactants.

B. It is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

C.It represents a synthesis reaction because the same atoms are present in the reactants and products.

D. It represents a decomposition reaction because two reactants break apart and form two products.

Answers

The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.

What is a chemical reaction?

A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.

A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.

A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.

Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

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Physics occurs all the time but often goes unnoticed. Here is your chance to reflect on physics in action. Other than the examples used in this lesson, think of a time where you witnessed the conservation of angular momentum. Describe the objects that had angular momentum and how angular momentum was conserved. You may also create an example if you cannot recall one in your personal experience.

Answers

Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

What is principle of conservation of angular momentum?

The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no  external torque acting on the object.

Angular momentum of a system is conserved as long as there is no net external torque acting on the system.

Examples of conservation of angular momentumthe spinning ice skatersomeone spinning in an office chaira child spinning on roller coaster  

Thus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

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Friction typically __________ objects. A. Speeds up B. Slows down C. Doesn't affect D. Destroys

Answers

Answer:

B

Explanation:

friction opposes motion

Friction typically Slows down objects

Option "B"

How does friction affect speed?Friction and Speed

While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.

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A jet of water squirts out horizontally from a hole near the bottom of the tank, as seen in the figure.
If the hole has a diameter of 3.13 mm, what is the height, h, of the water level in the tank? Assume that x = 1.33 m and y = 1.72 m.

Answers

The height of the water level in the water tank is 0.011 m.

What is the time taken by a fluid particle to reach the ground?From Newtown's equation of motion, y=U×t+(1/2)gt²As, initial velocity (U) is zero, t= 2y/g

t = (2× 1.72)/9.8 = 0.35 s

What's the velocity along x direction?

Velocity along x direction= x × t

= 1.33 × 0.35

= 0.46 m/s

What's the height of water level in the tank?

As per Bernoulli's theorem, 1/2 × ρ × v² = ρ × g× h

=> h = v²/2g = 0.46²/(2×9.8)

= 0.011 m

Thus, we can conclude that the height of water level is 0.011m.

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Complete the following:

When light is incident through the focal point, and then strikes a mirror, ___

the light will reflect parallel to the principal axis
the light will reflect back through the focal point
the light will not reflect at all

Answers

Answer:

the light will reflect parallel to the principal axis

which element has more than 5 valency​

Answers

Answer:

Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.

Explanation:

The electrons found in an element's outermost atomic shell are known as valence electrons.

Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.

However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

Thank you ,

Eddie

A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
a)What is the velocity of the space capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut.
b)If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other?
Express your answer to three significant figures and include the appropriate units.
c)What is the kinetic energy of the astronaut after the push in the reference frame?
Express your answer to three significant figures and include the appropriate units.
d)What is the kinetic energy of the capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units.

Answers

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Given:

Mass of the astronaut, [tex]m_a[/tex] = 126 kg

Speed he acquires, [tex]v_{a}[/tex]  = 2.70 m/s

Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

[tex]P_f = m_av_a + m_cv_c[/tex]

[tex]P_I[/tex] = 0

[tex]m_av_a + m_cv_c = 0[/tex]

[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]

   [tex]= \frac{126* 2.70}{1800}[/tex]

   [tex]= - 0.189[/tex] m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     [tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     = [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

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a 30kg crate is pulled up a ramp 15m long and 2m high by a constant force of 100n. the crate starts from resta nd has a velocity of 2m/s when it reaches the ramp. what is the frictional force between the crate and the ramp. use the principal of conservation energy.​

Answers

The frictional force between the crate and the ramp is determined as 35.2 N.

Energy lost to frictional force

Apply the principle of conservation energy to calculate the change in the energy of the crate.

Change in energy of the crate = energy loss to friction

P.Ei - K.Ef = E

mgh - ¹/₂mv² = E

where;

m is mass of the crateh is vertical height traveled by the cratev is the final velocity of the crate

(30)(9.8)(2) - (0.5)(30)(2²) = E

528 J = E

Frictional force between the crate and the ramp

E = Fd

where;

F is the frictional forced is the distance traveled by the crate

F = E/d

F = (528)/(15)

F = 35.2 N

Thus, the frictional force between the crate and the ramp is determined as 35.2 N.

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A visitor at a Michigan lighthouse is trying to measure the height of the lighthouse. She has a big spool of string, but she doesn't have a measuring tape. She has a wrist watch with a timer function on it. She attaches a weight to the end of the string and she makes a simple pendulum. She hangs the pendulum down from the top of the spiral staircase and she measures the period of the pendulum. The period of the pendulum is 8.48 s. What is the height of the light house?

Answers

The height of the light house is 17.86m.

To find the answer, we have to know about the simple pendulum.

How to find the height of the light house?We have the expression for time period of the pendulum as,

                       [tex]T=2\pi \sqrt{\frac{l}{g} } \\[/tex]

where, g is acceleration due to gravity, and l is the length of the pendulum.

For the pendulum with time period T kept at some height under the influence of gravity, then the height will be equal to,

              [tex]h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m[/tex]

Thus, we can conclude that, the height of the light house is 17.86m.

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A fighter plane is flying overhead at mach 1.20. What angle does the wave front of the shock wave produced make relative to the plane's direction of motion (in degrees)?

Answers

Answer: 56.44°

Explanation:

Given:

Let u represent the current speed of the plane, 1.2 Mach

Converting to SI Units (m/s):

= (1.2 mach)(340 ms^-1 / 1 Mach)

u  = 408 m/s

Speed of sound in air, v = 340 m/s

Find:

Angle the wave front of the shock wave relative to the plane's direction of motion, θ

We have, sinθ = speed of sound / speed of object

               sinθ = v / u

                   θ = sin^-1 (v / u)  

                      = sin^-1 (340 / 408)

                   θ = 56.44°

iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weight of iron ball in air?​

Answers

Answer:

453 gm

Explanation:

Immersed objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm  in air

A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N. What is the speed of transverse waves on the wire?

Answers

The speed of transverse waves on the wire is 1.555.

Given: Length= 1.150 m

Mass= 4.80 g

Speed =mass/length

U=M/L formula

U=1.150 *10^(-3)/4.80

v=√t/u

v=√580*4.80/1.150*10^(-3)=1.555

1.555

The derivative of the displacement with respect to time gives the transverse velocity in the y direction, as velocity is the rate of change of position: Where k = 2/ is referred to as the wave number and = 2f as before, v(x,t) = y(x,t)/t = -A cos (k x - t + ).The linear density and tension v=FT can be used to determine the wave's speed. v = F T μ . According to the equation v=FT, v = F T, the tension would need to be increased by a factor of 20 if the linear density were to be doubled by almost as much.

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Nina is doing a workout where she runs at 50% speed for 3 minutes, 75% speed for 2 minutes, and 100% speed for 30 seconds. This is an example of
a. Specificity training
b. Overload training
c. Circuit training
d. Interval training

Answers

D. Interval training
d. Interval training
Not sure if I’m right

How much energy is needed to move an electron in a hydrogen atom from the ground state (n = 1) to n = 3?

Answers

The energy needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3 will be 1.93 *10^-18J and 12.09 eV.

How to compute the value?

The following can be deduced:

Energy of electron in hydrogen atom is

En = -13.6 /n2 eV

where n is principal quantum number of orbit.

Energy of electron in first orbit = E1 = -13.6 / 12 = - 13.6eV

Energy of electron in third orbit = E3 = -13.6 /32 = -1.51 eV

Energy required to move an electron fromfirst to thirdorbit ΔE = E3- E1

ΔE = -1.51 - ( 13.6) = 12.09 eV

Energy in Joule = 12.09 *l/× 1.6 × 10^-19 = 1.93 × 10^-18 J.

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Complete question:

How much energy is needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3? Give theanswer (a) in joules and (b) in eV.

In the hydrogen atom, which electronic transition corresponds to the largest energy emission?

Answers

In hydrogen atom the electronic transition from n= ∞ to n=1 corresponds to largest energy emission.

What is electronic transition ?

Electronic transition is the jump of an electron from one energy level to another energy level .

Which electronic transition in Hydrogen atom corresponds to highest energy emission?The fromula for energy emission is

E = hc/λ

where E= emited energy

h= plank constant =6.62*10^(-34)

c=speed of light

λ= wavelength of emited electron .

The electronic transition from infinity to ground state corresponds to lowest wavelength .

Thus , we can conclude that the electronic transition from infinity to n=1 (ground state) corresponds to largest energy emission .

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Is India a rich country?

Answers

Explanation:

India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.

Answer:

50505050128128128128

Explanation:

The cricket ball has a mass of 0.16kg and it hits the bat with a speed of 25 m/s. After being in contact with the bat for 0.0013 s, the ball rebounds with a speed of 22 m/s in the direction exactly opposite to its original direction.
A) state the difference between speed and velocity
B) calculate.
I) the change in velocity of the cricket ball
Ii) the average acceleration of the ball whilst it is in contact with tge bat
Iii) the average force exerted on the ball by the bat​

Answers

change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².

Give  some difference between velocity and speed.

1) Speed is scalar quantity but velocity is vector quantity.

2) Speed is distance followed by an individual with respect to  time and velocity is displacement with respect to time .

1) final velocity- intial velocity

 25-22= 3m/s

2) The force is

F= ma

 . 0.16×30= 4.8N

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*please refer to photo*
Use meters and seconds for all calculations. Show the appropriate units for measured and calculated values

Answers

Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.

What is average velocity?

An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.

What is an average?

An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.

How to calculate the average velocity?

Mathematically, the average velocity for this data set would be calculated by using this formula:

Average = [F(v)]/n

Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5

Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:

v₁ = 100 m/sv₂ = 150 m/sv₃ = 200 m/sv₄ = 250 m/sv₅ = 300 m/s

Substituting the parameters into the formula, we have:

Vavg = [300 + 450 + 500 + 250 + 300)/5

Vavg = 1800/5

Vavg = 360 m/s.

Next, we would calculate the percent difference by using this formula:

[tex]Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100[/tex]

Percent difference = [360 - 343]/360 × 100

Percent difference = 17/360 × 100

Percent difference = 0.0472 × 100

Percent difference = 4.72%.

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According to current evidence, what happens when gays or lesbians rear a child?
a. the child is more likely than average to develop depression or anxiety disorders.
O b. the child develops about the same as other children.
O c. the child is more likely than average to become gay or lesbian.
O d. the child does poorly in school.

Answers

Answer: The answer is A.

Explanation:

Just because u have gay parents doesnt mean theyre bad lol XD

NASA wants a satellite to revolve around Earth 3 times a day. What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg)

Answers

Answer:

Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].

Explanation:

Assume that the radius of this orbit is [tex]r[/tex].

Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].

The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].

The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].

In other words:

[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].

The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:

[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].

Substitute this value into the expression for [tex]r[/tex] and evaluate:

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].

(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)

How is faraday's law applicable in Electronic Drum

Answers

Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum

What is Faraday's law?

Faraday's law states that a changing magnetic field through an area or equivalently, a changing area with constant field will cause a voltage.

So therefore, Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum

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