Our galaxy, the ______________, contains more than 200 billion stars.

Answer:

Refer to the attachment!~

Answer:

Explanation:

The sensor contains an LDR which has a resistance of 10kohlms in daylight and 100kohlms in the dark.

If the resistor in the circuit is 1 megaohlm, the total resistance in daylight and darkness will be 1.01 megaohms and 1.1 megaohlms.

The percentage difference = (1.1-1.01)/1.1*100% = 8.18%

If the resistor in the circuit is 25 kohlm, the total resistance in daylight and darkness will be 35 kohms and 125 kohlms.

The percentage difference = (125-35)/125*100% = 72%

With the input p.d to the sensing circuit fixed at 12 v, the sensing current will change according to the total resistance. A 72% difference is much more detectable. So the 25 kohm resistor is the better choice.

Answer:

Explanation:

V=IR

I=12/(R of resistor + R of LDR)

R of LDR = 10kohm in light and = 100kohm in dark

R1 =  25kohm

R2 = 1Mohm

solve 4 current

                       light                                       dark

R1        12/(25+10)=0.343mA          12/(25+100)=0.096mA

R2       12/(1000+10)=0.012mA       12/(1000+100)=0.011mA

so R1 is better as its easier 2 tell its light or dark

Answer:

ya bestie Anda benar terima kasih nanti sekarang selamat tinggal

Answer:

0.6 resistance

Whichever object has more mass will need more energy to slow it down.

We don't know the mass of the car or the mass of the large.

None of this has anything to do with "difficult". We don't know what's difficult for you, and we don't know the SI unit of "difficult".

Answer:

Time, t = 28.87 seconds

Explanation:

Given the following data;

Acceleration = 1.20 m/s²

Distance = 500 meters

Initial velocity = 0 m/s

To find the time to cover the distance, we would use the second equation of motion;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

[tex] 500 = 0*t + \frac {1}{2}*(1.2)*t^{2}[/tex]

[tex] 500 = 0 + 0.6*t^{2} [/tex]

[tex] 500 = 0.6t^{2} [/tex]

[tex] t^{2} = \frac {500}{0.6} [/tex]

[tex] t^{2} = 833.33 [/tex]

Taking the square root of both sides, we have;

[tex] t = \sqrt{833.33} [/tex]

Time, t = 28.87 seconds

Answer:

1. A

2. B

Explanation:

Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.

Sound travels faster through solids than it does through either liquids or gases.

Wavelength can be defined as the distance from one peak to the next peak of a waveform. This distance is also equal to the distance from one trough of a wave to another.

Diffraction is a wavelength property that makes it possible for sound to be heard around a bend. Thus, it's a property that causes a change in the direction of a wave as it travels through or around a barrier (obstacle) in its path.

Simply stated, diffraction involves the bending of waves around small obstacles with their dimensions smaller than or approximately equal to the wavelength of the sound wave.

On the other hand, reflection can be defined as a phenomenon which typically involves a change in the direction of a wave (bounces back) as it falls on a surface.

Hence, reflection is the property of sounds which animals such as bats use to detect their paths as they navigate from one point to another. This phenomenon is generally referred to as echolocation and it typically involves the use of reflection (echo) of sound to gain information about the nature and location of surrounding objects.

Answer:

condensation?

Explanation:

Explanation:

hope it helps

pls mark me as brainliest thanks❤

Answer:

D

Explanation:

Let's analyze each option.

A "Multiple copies of the stored data take up very little space."

This is actually true, and it is a positive aspect, as stored digital data does not need any "physical space", it only needs memory and not a lot of it.

So storing data digitally is way more efficient than storing analog data.

B: "They lose quality when they are copied several times."

The data shouldn't change when it is copied, so you should not see a lose in quality.

C: "Stored data are made up of only two different values."

True, but as we know, we can define a lot of things with only two values (zeros and ones), so this is not really a disadvantage.

D: "They are vulnerable to hackers and viruses."

This is true, when you store your data digitally you become vulnerable to hackers stealing your data, so you need to get informatic security in order to protect your data. The same thing with viruses, if you have all your data stored in a given device, and the device becomes infected, there is a chance that you just lost all your data, so you need to have multiple backups of your important information, and again, some protection against viruses.

The correct option is D.

Answer:

D

trust me i just did it

Explanation:

Explanation:

Newton's second law of motion gives the relation between mass, force and acceleration.

We know that,

Force, F = mass (m) × acceleration (a)

or

[tex]m=\dfrac{F}{a}[/tex]

or

[tex]a=\dfrac{F}{m}[/tex]

Hence, this is the required solution.

Answer:

I. Aceleración, a = -2 m/s²

II. Distancia, S = 100 metros

Explanation:

Dados los siguientes datos;

Velocidad inicial = 72 km / h

Tiempo = 10 segundos

Velocidad final = 0 m/s

Conversión:

72 km/h a metros por segundo = 72 * 1000/3600 = 72000/3600 = 20 m/s

I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;

[tex] V = U + at[/tex]

Dónde;

Sustituyendo en la fórmula, tenemos;

[tex] 0 = 20 + a*10 [/tex]

[tex] -20 = 10a [/tex]

[tex] Aceleracion = \frac{-20}{10}[/tex]

Aceleración, a = -2 m/s²

Nota: el signo negativo indica desaceleración o retraso.

II. To find the acceleration, we would use the third equation of motion;

[tex] V^{2} = U^{2} + 2aS [/tex]

Dónde;

Sustituyendo en la fórmula, tenemos;

[tex] 0^{2} = 20^{2} + 2*(-2)*S [/tex]

[tex] 0 = 400 - 4S [/tex]

[tex] 4S = 400 [/tex]

[tex] S = \frac {400}{4}[/tex]

Distancia, S = 100 metros

Answer:

1- 31.25

2-?

3-?

Explanation:

Answer:

Speed = 1.6 m/s

Explanation:

Formula,

Speed = Distance ÷ Time

Answer:

a. Energy = 28800000 Joules

b. Energy = 8 Kwh

c. Cost = $4

Explanation:

Given the following data;

Power = 1000 Watts

Time = 8 hours

Cost = $0.5 per KWh

Conversion:

1 hour = 60 * 60 = 3600 seconds

8 hour = 8 * 3600 = 28800 seconds

a. To find much electrical energy, in joules;

Energy = power * time

Energy = 1000 * 28800

Energy = 28800000 Joules

b. To find how much energy is that in kilowatt hours;

Energy = power * time

Energy = 1000 * 8

Energy = 8000 Watt-hour = 8000/1000 = 8 Kwh (1 Kilowatts is equal to 1000 watts).

c. To calculate the cost of using this heater;

Cost = 8 * 0.5

Cost = $4

Answer:

Y

Explanation:

Acceleration = (v-u)/t

W = (29-11)/6 = 3

X = (34-10)/12 = 2

Y = (40-12)/7 = 4

Z = (28-20)/8 = 1

So the answer is Y = 4.

Hope this will help. Please mark me brainliest.

Answer:

Option (c)

Explanation:

According to question,

R= 60 ohm

V=120 V

Current can be calculated as :

[tex]I=\frac{V}{R}[/tex]

Where,

I  is current

V is voltage

R is resistance

[tex]I=\frac{120}{60} \\I=2 A[/tex]

Current is flowing across 60 ohm will be 2.0 A

Therefore, option c is correct

Answer:

Student C

Explanation:

order from heaviest to lightest is...

9 kg (A) , 8000g (8 kg) (D) , 2800g (2.8kg) (B), 2 kg (C)

Answer:

total distance dibide by total time

Answer:

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

Explanation:

The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x - axis is

E = Qx/4πε₀[√(x² + R²)]³

Since the rings are separated by a distance L, the electric field at point x due to the second ring is E' = -Q(L - x)/4πε₀[√((L - x)² + R²)]³. It is negative since it points in the negative x - direction.

So, the resultant electric field at x is E" = E + E' = Qx/4πε₀[√(x² + R²)]³ + {-Q(L - x)/4πε₀[√((L - x)² + R²)]³}

E" =  Qx/4πε₀√[(x² + R²)]³ - Q(L - x)/4πε₀√[((L - x)² + R²)]³

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[((L - x)² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L² - 2Lx + x² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L - 2x)L + (x² + R²)]³})

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

So, the electric field at points along the x axis is

E" =  Q/4πε₀√[(x² + R²)]³(x - {(L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

Answer:

False you dont repaint your hamster.

Explanation:

LOL

Answer:

C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.

Explanation:

Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.

If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.

Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

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

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.  

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

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

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

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

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

Answer:

[tex]T=62.9*10^{-9}[/tex]

Explanation:

From the question we are told that:

Index of refraction of Rinestones [tex]\gamma_1 =1.5[/tex]

Index of refraction of silicon [tex]\gamma_2 =2.0[/tex]

Wavelength  [tex]\lambda=576nm=576*10^{-9}[/tex]

Let each layer have thickness T

Therefore

Total Thickness =2T

Generally the equation for  Constructive interference is mathematically given by

[tex]2T=(m+0.5)\frac{l\lambda}{\gamma_2}[/tex]

Where

 [tex]M=0[/tex]

 [tex]2T=(0+0.5)\frac{576*10^{-9}}{2*2.0}[/tex]

 [tex]T=62.9*10^{-9}[/tex]

Explanation:

A hydrometer sinks more in a lighter liquid

Answer:

The height jumped by the person on the moon is 6 times the height jumped by the person on earth.  

Explanation:

As we know that the acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth.

So, it is false.

Let the mass of man is m and the gravity on moon is g' = g/6.

Let the height jumped on earth is h and the height jumped on moon is h'.

So,

m x g' x h' = m x g x h

g/6 x h' = g x h

h' = 6 h  

So, the height jumped by the person is 6 times the height jumped by the person on earth.

Answer:

E = 62.72 J

Explanation:

Given that,

The mass of a ball, m = 4 kg

The teacher raised the lawn bowling ball to a height of 1.6 meters before they released it.

We need to find the gravitational potential energy of the ball.

We know that,

E = mgh

Where

g is acceleration due to gravity

So,

E = 4 kg × 9.8 m/s² × 1.6 m

E = 62.72 J

So, the gravitational potential energy of the ball is equal to 62.72 J.

Answer:it’s 400 meters

Explanation: