Assume the magnetic field declination at your location is 22 degrees east of north. How should you orient a current carrying wire near a compass so that the compass will point directly toward geographic north, assuming the wire produces a field strong enough to deflect the compass

Most hair dryers are in the 1800 watt range and lets assume you turned it to the highest speed and heat setting.

It will heat up the air in the box, failry quickly. Rules of thermodynamics says that the heat will increase the temperature according to the rate of heat input minus the rate of heat loss from the box… due to conduction, radiation and convection (in this case none of the latter because the box is sealed).

The loss rate depends in the temperature difference and the thermal resistance. THicker and better insulation increases the resistance; larger area decreases the resistance. Eventually the heat (temperature) will rise and the difference between inside and outside will be so large that the temperature will rise no more. But in the case of a 1800 W hairdryer the temperature will exceed the melting point of plastics and wire insulation and if allowed to come to heat equilibrium will probably short out and catch fire or blow an external fuse.

Most modern hair dryers have a internal thermal fuse that cuts out at temperatures below the melting point and probably this will cut off the dryer before catastrophic meltdown. Its a one time fuse and not readily available, mostly you toss the dryer when the fuse goes

Since the cruise ship makes its way from one island to another. The ship is in motion is compared with the reference point of option C. a lighthouse on a nearby island.

Sine the cruise ship from one Island  is relocating, the lengthy duration by the passengers within the ship makes it simple to detect the ship motion.

The skipper is the only one who is aware of the ship's direction and speed. Even though the cargo loads are stable and related to the ship's course, we cannot take them into account when calculating the traveling distance of the ship.

Since it is in the sea and relatively close to it, the lighthouse on the island can be used to calculate the distance traveled by ships. Due to its stability, we can use it as a reference point in the water that can be seen from a great distance.

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See full question below

A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

A.

the passengers on the ship

B.

the captain of the ship

C.

a lighthouse on a nearby island

D.

the cargo inside the hull of the ship

Answer:

All acids contain hydrogen. On reacting with metals, all acids produce hydrogen gas. All acids produce hydrogen ions in water.

Explanation:

Answer:

6 km is the right answer

hope it helps you

Explanation:

please mark me as brainliest

Answer:

146.27 m

Explanation:

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

Velocity of tourist (vₜ) = 4.65 m/s

Distance travelled by tourist (dₜ) = d

Velocity of bear (v₆) = 5.68 m/s

Distance travelled by bear (d₆) = 32.4 + d

Value of d =?

Next, we shall determine the time taken for tourist and the bear to get to the car.

For the tourist:

Velocity of tourist (vₜ) = 4.65 m/s

Distance travelled by tourist (dₜ) = d

Time (tₜ) =?

vₜ = dₜ / tₜ

4.65 = d/tₜ

Cross multiply

4.65 × tₜ = d

Divide both side by 4.65

tₜ = d / 4.65

For the bear:

Velocity of bear (v₆) = 5.68 m/s

Distance travelled by bear (d₆) = 32.4 + d

Time (t₆) =?

v₆ = d₆ / t₆

5.68 = (32.4 + d) / t₆

Cross multiply

5.68 × t₆ = (32.4 + d)

Divide both side by 5.68

t₆ = (32.4 + d) / 5.68

NOTE: Both the tourist and the bear have the same time. Thus, to obtain the value of d, we shall equate both the time taken for the tourist and the bear together. This is illustrated below:

Time taken by tourist (tₜ) = time taken by the bear (t₆)

tₜ = d / 4.65

t₆ = (32.4 + d) / 5.68

tₜ = t₆

d / 4.65 = (32.4 + d) / 5.68

Cross multiply

d × 5.68 = 4.65 (32.4 + d)

5.68d = 150.66 + 4.65d

Collect like terms

5.68d – 4.65d = 150.66

1.03d = 150.66

Divide both side by 1.03

d = 150.66 / 1.03

d = 146.27 m

Thus, the maximum value of d is 146.27 m

The velocity, measured in the Earth reference frame, of an inertial reference frame in which the cat's kinetic energy does not change is equal to the velocity of the skier before the collision. The velocity of the skier before the collision is 15 m/s.

According to the law of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision. This can be expressed as m1*v1 + m2*v2 = (m1 + m2)*vf, where m1 and m2 are the masses of the skier and the cat respectively, v1 is the velocity of the skier, and vf is the velocity of the skier and the cat after the collision.

The kinetic energy converted to internal energy in the Earth reference frame can be determined by applying the law of conservation of momentum.

The amount of kinetic energy converted to internal energy can be calculated as follows:

m1*v1 = (m1 + m2)*vf

vf = (m1*v1)/(m1 + m2)

KE = (1/2)*m2*v2²

KE converted = KE initial - KE final

KE converted = (1/2)*m2*v2² - (1/2)*m2*((m1*v1)/(m1 + m2))²

KE converted = (1/2)*m2*v2² - (1/2)*m2*((60*15)/(60 + 5))²

KE converted = (1/2)*5*3.8² - (1/2)*5*(15²/65)

KE converted = 28.8 - 22.15

KE converted = 6.65 J

The velocity, measured in the Earth reference frame, of an inertial reference frame in which the cat's kinetic energy does not change is equal to the velocity of the skier before the collision.

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Answer: The answer is C i believe

9.8 N is the magnitude of the force that maintains circular motion acting on the puck.

32.53 m/sec is the linear speed of the puck.

The term "force" has a specific meaning in science. At this level, it is quite acceptable to refer to a force as a push or a pull. An item does not have a force inside of it or within it. Another item applies a force to another. The concept of a force is not restricted to living or non-living entities. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

As we know, from free body diagram, when the system is in equillibrium, the force acting on the puck on the table is as follows:

When the air puck of mass is revolving round the circle, then the tension (T) in the top string and the bottom of the string is same.

F = T

F = 9.8 N

Now, calculate the velocity of the puck:

From the given data, the air puck moves in a circular path. Hence, the tension in the top string is equal to the centripetal force acting on the puck.

Thus, the expression for the tension in the top string is as follows:

T = mv² / R

The velocity of the puck is as follows:

Mg = mv² / R

v² = MgR / m

v² = (2.7 × 9.8 × 1.2 ) / 0.030

v² = 1058.4

v = \(\sqrt{1058.4}\)

v = 32.53 m/sec

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Answer: C. Determine how long the reaction will take.

Answer:

a) 103.176 m / min

b) 1751.28 meters

Explanation:

Given:-

- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:

                         \(E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t\)

- Where, E ( t ) and L ( t ) are given in m / min

- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:

                         ( 0 ≤ t ≤ 15 ) mins                  

- It is known that Emma is 10 meters ahead of Lily at time t = 0.

Find:-

a) Find the value of \(\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\) using correct units, interpret the meaning of

b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?  

Solution:-

- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:

                          \(f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx\)

- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:

                         \(E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\\)

- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:

                         \(E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\\)

- Complete the square in the denominator:

                          \(E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\\)

- Use the following substitution:

                          \(u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du\)

- Substitute the relations for (u) and (dt) in the above E_avg expression.

                          \(E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du\)

- Use the following standard integral:

                          \(arctan(u) = \int {\frac{1}{u^2 + 1} } \, du\)

- Evaluate:

                         \(E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |\)

- Apply back substitution for ( u ):

                        \(E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\\)

- Plug in the limits and find Emma's average velocity:

                        \(E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}\)

Answer: Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.

- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of  0≤t≤15 is given by the relation:

                    \(S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\\)

- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of  0 ≤ t ≤ 15 is given by the relation:

                    \(S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\\)

 Apply integration by parts:

  \(S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\\)

 Re-apply integration by parts 2 more times:

 \(S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\\)             \(S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\\)    

\(S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\\)

- The distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting S ( L )  from S ( E ):

                    \(S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\\)

- The maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.

- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:

              \(\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671\)

- We will plug in each value of t and evaluate the displacement function S(t) for each critical value:

Explanation:

A rainbow is a meteorological phenomenon that is caused by reflection, refraction and dispersion of light in water droplets resulting in a spectrum of light appearing in the sky. It takes the form of a multicoloured circular arc. Rainbows caused by sunlight always appear in the section of sky directly opposite the sun.

Hope it helps You °_°

Answer:

The current will be 18 A

Explanation:

Given;

potential difference, V = 10 V

current between the resistor, I = 2 A

Apply ohm's law;

V = IR

R = V / I

R = 10 / 2

R = 5Ω

Resistance is given as;

\(R = \frac{\rho l}{A}\)

where;

ρ is resistivity

l is length

A is area

\(R = \frac{\rho l}{A} \\\\R = \frac{\rho l}{\pi r^2} = \frac{\rho l}{\pi (\frac{d}{2}) ^2} = \frac{\rho l}{\pi (\frac{d^2}{4}) }\\\\R = \frac{4*\rho l}{\pi d^2} \\\\R = (\frac{4*\rho l}{\pi } )\frac{1}{d^2} \\\\R = (k)\frac{1}{d^2} \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = \frac{R_1d_1^2}{d_2^2}\)

When the diameter of the resistor is tripled

d₂ = 3d₁

\(R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms\)

The current is now calculated as;

Apply ohms law;

V = IR

I = V / R

I = 10 / 0.556

I = 17.99 A

I = 18 A

Therefore, the current will be 18 A

Answer:

making sense of a 3-d world from 2-d data

Explanation:

The total resistance of the circuit is 49 Ω. The current strength in the circuit, when connected to a voltage source of 56 V, is approximately 1.14 A.

To calculate the total resistance of the circuit, we need to determine the equivalent resistance of the resistors connected in a series.

Given:

R1 = 4 Ω

R2 = 30 Ω

R3 = 10 Ω

R4 = 5 Ω

Calculate the equivalent resistance (RT) of R1 and R2, as they are connected in series:

RT1-2 = R1 + R2

RT1-2 = 4 Ω + 30 Ω

RT1-2 = 34 Ω

Calculate the equivalent resistance (RTotal) of RT1-2 and R3, as they are connected in parallel:

1/RTotal = 1/RT1-2 + 1/R3

1/RTotal = 1/34 Ω + 1/10 Ω

1/RTotal = (10 + 34) / (34 * 10) Ω

1/RTotal = 44 / 340 Ω

1/RTotal ≈ 0.1294 Ω

RTotal ≈ 1 / 0.1294 Ω

RTotal ≈ 7.74 Ω

Calculate the equivalent resistance (RTotalCircuit) of RTotal and R4, as they are connected in series:

RTotalCircuit = RTotal + R4

RTotalCircuit = 7.74 Ω + 5 Ω

RTotalCircuit ≈ 12.74 Ω

Therefore, the total resistance of the circuit is approximately 12.74 Ω.

To determine the current strength (I) when connected to a voltage source of 56 V, we can use Ohm's Law:

I = V / RTotalCircuit

I = 56 V / 12.74 Ω

I ≈ 4.39 A

Therefore, the current strength in the circuit, when connected to a voltage source of 56 V, is approximately 4.39 A (or 1.14 A, considering significant figures).

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The efficiency of the leaf blow is 48%.

         Here, E = Energy transferred

                   m = Mass

                   Ф = Temperature change

            Here, E = Energy transferred

                      P = Power

                      t = time

        Here, v = speed

                   f = frequency

                  ∧ = wavelength

       Efficiency =  (Useful power out / Total power in ) = 360/750 = 0.48

So, the efficiency of the leaf blower is 48%.

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The weight of the object at distance 3R from the Earth is 10 N if the weight at center of the Earth is 90N.

According to Newton's law of gravitation,

Weight W = m × g; where m is the mass of the object and

g is the acceleration due to gravity.

Also,

\(g = \dfrac{GM}{R^{2}}\)

where G=gravitational constant, M=mass of earth, R=radius of earth.

Hence,

\(W = m \times \dfrac{GM}{R^{2}}\)

Weight is inversely proportional to the square of radius R.

Weight of objects at the surface of the earth is 90 N.

Let W be the weight at a distance of 3 R from the center.

\(\dfrac{W}{90} = \dfrac{R^{2}}{(3R)^{2}}\\\dfrac{W}{90} = \dfrac{1}{9}\\W = 10\)

Weight at a distance 3 R is 10 N.

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Answer:

61.0168 g/mol Explanation:

The equivalent resistance between points A and B is 0.837Ω.

Resistors in series are connected end-to-end so that the current flows through them in sequence. The equivalent resistance of resistors in series is the sum of their individual resistances.

The formula for equivalent resistance of resistors in series: R_eq = R_1 + R_2 + ... + R_n

Resistors in parallel are connected across each other so that the voltage is the same across each resistor. The equivalent resistance of resistors in parallel is the reciprocal of the sum of the reciprocals of their individual resistances.

The formula for equivalent resistance of resistors in parallel: 1/R_eq = 1/R_1 + 1/R_2 + ... + 1/R_n

Here in the Fig.

we can simplify the second set of resistors in parallel (4.8 Ω, 3.3 Ω, and 8.1 Ω) using the same formula:

1/Req1 = 1/4.8 + 1/3.3 + 1/8.1

Req1=1.575Ω

This Req1 connected series with 6.3Ω, then Req of this two resistance given by:

Req2= 1.575Ω+ 6.3Ω

Req2=7.875Ω

Once again this req2 makes the parallel with the other two resistance i. e 1.5Ω and 2.5Ω

Their equivalent resistance  is given by,

1/Req3=1/1.5 + 1/2.5 + 1/7.875

Req3=0.837Ω

Hence, The equivalent resistance between points A and B is 0.837Ω

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According to the give value of the mass of lemonade and change in the temperature, the specific heat capacity of the lemonade is 3.81 J/kg/k.

The formula for the specific heat capacity (C) = Q/ m × ΔT

Q is energy added and the value, which is 334J

m is the mass of the lemonade, which is 300g = 0.3 kg

Δ T is the change in temperature, which is 19°C =(273+19)= 292 K

So, C = 334/ 0.3 × 292

         = 3.81 J/kg/k

So the specific heat capacity of the lemonade is 3.81 J/kg/k.

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Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

Answer:

\(\boxed {\tt s=133 \frac{1}{3} \ or \ 133.333333 \ km/hr}\)

Explanation:

We want to find speed, so we can use the following formula.

\(s=\frac{d}{t}\)

where \(d\) is the distance traveled and \(t\) is the time.

We traveled 400 kilometers in 3 hours. Therefore,

\(d= 400 \ km \\t= 3 \ hrs\)

Substitute the values into the formula.

\(s=\frac{400 \ km }{3 \ hrs}\)

Divide

\(s= 133.33333 \ km/hr\)

\(s= 133 \frac{1}{3} \ km/hr\)

The speed is 133 1/3 or 133.33333 kilometers per hour.

Answer:

Power = 2.5 [W]

Explanation:

To be able to solve this problem we must remember that power is defined as the relationship of the work done in a given time interval.

\(P=W/t\)

where:

P = power [W] (units of watts)

W = work = 50 [J]

t = time = 20 [s]

Now replacing:

\(P=50/20\\P=2.5[W]\)

Answer:

2.5

Explanation:

A dust particle with a charge of 250 pg has a charge of -250 e. At point 1, where the electric potential is v1=2000 v, its speed is 2.0 m/s. It will have a speed of 1.238 m/s at point 2, where the potential is v2=-5000 v.

Let the speed be v₂at point 2.

Given a mass of m=250 pg.

At point 1, the speed isv₁=2.0 m/s.

Q=250e for the charge.

v₁=1200V is the electric potential at point 1.

v₂=-6500 V potential at point 2.

We can conclude from energy conservation that

                                    \(\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}=\left ( V_{2}- V_{1} \right )Q or\: \: \: \: \: \: \: \:v_{2}^{2}=\frac{\left ( V_{2}- V_{1} \right )Q\times 2}{m}+v_{1}^{2}\)

                       \(or\: \: \: \: \: \: \: \:v_{2}^{2}=\frac{\left ( \left ( -6500 \right )- 1200 \right )\times \left ( 1.602\times 10^{-19}\times 250 \right )\times 2}{250\times 10^{-15}}+2^{2} or\: \: \: \: \: \: \: \:v_{2}^{2}=\frac{-15400\times \left ( 1.602\times 10^{-19}\times 250 \right )}{250\times 10^{-15}}+2^{2}\)

                         \(or v2 = -2.46708 + 4 = or\: \: \: \: \: \: \: \:v_{2}^{2}=1.53292 \therefore \: \: \: \: \: \: \: \:v_{2}=1.238111465\approx 1.238\: \: \: \: m/s\)

Electric potential, also known as voltage, is a measure of the electric potential energy per unit charge of an electric field at a point in space. It is a scalar quantity that describes the work required to move a unit positive charge from a reference point to a specific point in the field, without any acceleration. Electric potential is measured in volts (V), and its direction is determined by the direction of the electric field. The electric potential can be either positive or negative depending on the sign of the charge and the direction of the electric field. It plays an important role in many electrical applications, including the design and operation of electronic devices and power systems.

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umm actually is 10 points

A VfL (Vertical Field LED) backlighting system is commonly used in LCD (Liquid Crystal Display) televisions.

LCD TVs rely on a backlight to illuminate the liquid crystal layer, which controls the passage of light to create the visual image. The VfL technology is a specific type of LED backlighting arrangement used in certain LCD TV models. In a VfL backlighting system, the LEDs (Light-Emitting Diodes) are positioned vertically along the edges of the LCD panel.

The light emitted by these LEDs is directed across the panel using light guides or optical films, illuminating the liquid crystal layer uniformly. One advantage of VfL backlighting is its ability to provide consistent illumination across the LCD panel, reducing any potential inconsistencies in brightness or color uniformity. The vertical orientation of the LEDs allows for more precise control over light distribution, improving overall image quality.

Additionally, VfL backlighting offers potential advantages in terms of power efficiency. By selectively dimming or turning off specific zones of LEDs, local dimming techniques can be employed to enhance contrast and black levels, resulting in improved picture quality while conserving energy. It's important to note that VfL backlighting is just one of several backlighting technologies available for LCD TVs.

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Answer:analize a afirmacao a seguir e tudo que envolve o gerenciamento da marca e que ultrapassa as acoes com objetivos economicos e refere se a cultura principios e valores

Explanation:

Answer:

The expression for the time taken by an object to move with a speed at some distance is,

t= d/v

Here, t is the time taken by the opponent to react, d is the length of the court, andis the speed of the ball.

Explanation:

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A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).

9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.

The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.

The following formula may be used to determine how long 1.5 million radiation cycles take to complete:

Time is equal to the frequency of cycles.

Plugging in the numbers, we get:

time = 1.5 million / 9.192631770 × 10^9 Hz

time = 1.632995101 × 10^-7 seconds

So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.

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