What process provides the sun with its energy

Explanation:

Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector

Check attached photo

Check attached photo

Solution :

Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.

And thus the bulb flashes for the moment as opposed to continuous glow.

Answer:

D) 1003 N​

Explanation:

Given the following data;

Mass of man = 85 kg

Acceleration of elevator = 2 m/s²

Acceleration due to gravity, g = 9.8 m/s²

To find the force exerted by the man on the floor;

Force = mg + ma

Answer:

Explanation:

Given force between 2 currents carrying

wires = F₀

Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L

where I₁=Current in wire 1

           I₂= Current in wire 2

           L= Length of the wire

when one current is doubled and the other is halved

I₁= 2 I₁

I₂=    I₂/2

F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians

Element is

[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]

[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]

And

[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

[tex] A = l*w [/tex]

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]                            

And the largest area is:

[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]  

Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.            

I hope it helps you!                        

Answer:

the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m

the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

[tex]A_l[/tex]= (L+0.5)(W+0.5)

(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m

To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

[tex]A_s[/tex]= (L-0.5)(W-0.5)

(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

The given parameters include:

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]

To learn more about projectile calculations please visit: https://brainly.com/question/14083704

Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

Answer:

5 m/s I hope it will help you

Explanation:

mark me as a brainlist answer

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

Answer:

1.84 hours

I hope it's helps you

Answer:

Types of tidal turbines

Axial turbines.

Crossflow turbines.

Flow augmented turbines.

Oscillating devices.

Venturi effect.

Tidal kite turbines.

Turbine power.

Resource assessment.

Answer:

Axial turbines

Crossflow turbines

flow augmented turbines

Answer:

Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.

Answer:

W = 641.52 J

Explanation:

The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:

[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]

where,

W = Work Done = ?

m = mass = 6 kg

v = speed = 4.2 m/s

g = acceleration dueto gravity = 9.81 m/s²

h = height = 10 m

Therefore,

[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]

W = 52.92 J + 588.6 J

W = 641.52 J

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length [tex]l[/tex] = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance [tex]R_{sol[/tex] = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

[tex]R_o[/tex] = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) =  [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]

so

ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )

ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]

ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

Answer:

1.28 *10^-5 A

Explanation:

Same work as above answer. Needs to be more precise

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface  called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

Answer:

the change in the kinetic energy of the system is -42.47 J

Explanation:

Given;

mass A, Ma = 2 kg

initial velocity of mass A, Ua = 15 m/s

Mass M, Mm = 4 kg

initial velocity of mass M, Um = 7 m/s

Let the common velocity of the two masses after collision = V

Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;

[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]

The initial kinetic of the two masses;

[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]

The final kinetic energy of the two masses;

[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]

The change in kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]

Therefore, the change in the kinetic energy of the system is -42.47 J

Answer:

960 nm

Explanation:

Given that:

wavelength = 640 nm

For the second (2nd) dark spot;  the order of interference m = 1

Thus, the path length difference is expressed by the formula:

[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]

[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]

[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]

dsinθ = 960 nm

Answer:

The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.

A

Explanation:

1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.

Answer:

c) kinetic energy

Explanation:

Answer: C)  kinetic energy

Explanation:

Answer:

[tex]h=10m[/tex]

Explanation:

From the question we are told that:

Area [tex]a=70 x 10^{-6}[/tex]

Force [tex]F=7N[/tex]

Generally the equation for Pressure is mathematically given by

Pressure = Force/Area

[tex]P=\frac{F}{A}[/tex]

[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]

[tex]P= 1*10^{5} Pa[/tex]

Generally the equation for Pressure is also mathematically given by

[tex]P=hpg[/tex]

Therefore

[tex]h=\frac{P}{hg}[/tex]

[tex]h=\frac{10000}{1000*9.8}[/tex]

[tex]h=10m[/tex]

Answer:

So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.

Answer:

9 Brainly hahaha ............huh

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

Answer:

X = 10.8387 cm²/s

Explanation:

In this exercise, you're required to convert a value from one unit to another.

Converting 42 ft²/hr to cm²/s;

Conversion:

1 ft² = 929.03 cm²

42 ft² = X cm²

Cross-multiplying, we have;

X = 42 * 929.03

X = 39019.26 cm²

Next, we would divide by time in seconds.

1 hour = 3600 seconds

X = 39019.26/3600

X = 10.8387 cm²/s