Carlos is playing darts. He throws a dart at the bullseye, but it hits the outer ring of the dartboard instead. He aims for the bullseye again and hits it with a second dart. Carlos threw three more darts which landed close to his first shot. What does the second throw indicate about his accuracy and precision? The second throw shows poor accuracy but high precision. The second throw shows accuracy but poor precision. The second throw shows both accuracy and precision. The second throw shows poor accuracy and poor precision.

The computer model would rely on the labeled pH values of various hand soap brands however the labled pH value might not be entirely accurate, which could comprise the accuracy of the computer model.

Inaccuracies in the simulation's underlying chemistry model could be a potential source of mistake in an experiment involving the pH values of several liquid hand soap solutions and water.

The outcomes could differ from actual behavior if the model does not adequately represent the intricate chemical interactions that control pH fluctuations. Furthermore, because they might not exactly match the varied properties of real hand soap solutions, the simulation's assumptions and parameters could introduce bias. Disparities between simulation results and empirical findings may also result from inadequate consideration of temperature impacts and ambient factors. To reduce such possible mistakes, careful simulation validation against empirical data and chemistry model modification are required.

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Answer:

To convert 400 mm to m you can apply the formula [m] = [mm] / 1000; use 400 for mm. Thus, the conversion 400 mm m is the result of dividing 400 by 1000. 0.4

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Answer:

light reflects off the objects and enter our eyes

Answer:

A = 150 cm at 120 degc

Where A + B = C       find B

B = C - A          add -A to both sides

Ax = 150 cos 60 = -75

Ay = 150 sin 60 = 129.9

Likewise

Cx = 114.7

Cy = 80.3

Bx = Cx - Ax = 114.7 + 75 = 189.7

By = Cy - Ay = 80.3 -129.9 = -49.6

B = (189.7^2 + 49.6^2)^1/2 = 196.7       length of B vector

tan B = By / Bx = -49.6 / 189.7       B = -14.65 deg

Also

sin B = By / B = -49.6 / 196.1 = -14.65 deg

So B is 196.1 cm at -14.65 deg

Answer:

The maximum current which can flow through a fuse without melting it, is called its rating. ... If current higher than 8 A flows through the fuse, it would melt and circuit gets broken. ... Hence, fuse acts as a safety device

Answer:

The guy above me is right

Explanation:

Answer:

Velocity is the speed of something given in a direction. the SI unit of velocity is metre per second or m/s.

Answer:

The Planck constant, or Planck's constant, is a fundamental physical constant denoted h, and is of fundamental importance in quantum mechanics. A photon's energy is equal to its frequency multiplied by the Planck constant. Due to mass–energy equivalence, the Planck constant also relates mass to frequency.

Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.

E = hf

[tex] \sf \: h = 6.626 \times {10}^{ - 34} \: kg \: {m}^{2} {s}^{ - 1} [/tex]

The current flowing through a resistor at a constant temperature is directly proportional to the potential difference across it. This is called Ohm's law

Answer:

La fuerza resultante sobre q₃ es  -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N

Explanation:

q₁ = -45 μC = -45 × 10⁻⁶ C

r₁₂ = 30 mm = 30 × 10⁻³ m

q₂ = 25 μC = 25 × 10⁻⁶ C

r₂₃ = 50 mm = 50 × 10⁻³ m

q₃ = 20 μC = 20 × 10⁻⁶ C

k = 9×10⁻⁹ N·m²/C²

Por lo tanto;

r₁₃ = √(50² + 30²) = 10·√(34)

F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴

F₁₂ = -1.125 × 10⁻¹⁴ N

F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j

F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵

Los componentes de F₁₃ son;

-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j

-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i

La fuerza resultante sobre la carga q₃, [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = [tex]\underset{F_{13}}{\rightarrow}[/tex] + [tex]\underset{F_{23}}{\rightarrow}[/tex]

∴ [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j

La fuerza resultante sobre q₃ es [tex]\left | \underset {F_3} \rightarrow \right |[/tex]  = -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃,

[tex]\left | F_3 \right |[/tex] = √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵

La magnitud de la fuerza resultante sobre q₃, [tex]\left | F_3 \right |[/tex] ≈ 1.25 × 10⁻¹⁵ N.

Answer:

nods 40th anniversary rid off e 49en9 snns

Answer:

Jet stream would be displaced southwards causing heavy rain and flooding.

Explanation:

The other options of the question were A) Jet stream would be displaced northwards causing drought. B) Jet stream would be displaced southwards causing drought. D) Jet stream would be displaced northwards causing heavy rain and flooding,

The statement that is a likely impact of stronger than normal trade winds in the Pacific Northwest to the United States is "Jet stream would be displaced southwards causing heavy rain and flooding."

We are talking about climate or weather terminology. In this case, we are referring to the "El Niño" (the Children) effect. Its presence affects the weather in North America. This phenomenon combines with the "La Niña) effect and it presents itself every two to seven years, ad they last from 8 to 12 months, affecting the weather conditions of the region.

Answer:

jet stream should be displaced southwards which causes heavy rainfall and flood.

Answer:

I. Vs = 8.0 Volts.

II. Is = 4.5 Amperes.

Explanation:

Given the following data;

Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex]  ..... equation 1

Ip = 3.0 A

Vp = 12 V

To find the voltage and current on the secondary coil;

I. For the voltage in the secondary coil (Vs), we would use the following formula;

[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex]  ...... equation 2.

Substituting eqn 1 into eqn 2, we have;

[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]

[tex] \frac {12}{V_{S}} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] V_{S} * 1.5 = 12 [/tex]

[tex] V_{S} = \frac {12}{1.5} [/tex]

Vs = 8.0 V

II. For the current in the secondary coil (Is), we would use the following formula;

[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex]  .... equation 3

Substituting eqn 1 into eqn 3, we have;

[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]

[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] I_{S} = 1.5 * 3.0 [/tex]

Is = 4.5 A

Answer:

Explanation:

[tex]a=\frac{v_f-v_0}{t}[/tex] We don't yet need the mass of the car, as we can see by this equation. What we do need is the velocity of the car, the initial velocity, in meters per second, and right now it's in km/hr. Not good. We need to convert. The conversion is as follows:

[tex]54\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex] Ok, that's good. Now we have everything we need but the time element. If the car traveled a distance of 40 meters at 15 m/s, then we can use the d = rt equation to solve for t, and when we find t we plug it into the acceleration equation:

40 = 15t and

t = 2.7 seconds. The car traveled for 2.7 seconds to go that 40 meters. That's the only reason we were given the displacement. We need it for nothing else but that.

Filling in the acceleration equation now:

[tex]a=\frac{0-15}{2.7}=-5.6\frac{m}{s^2}[/tex] and the negative indicates we are in fact slowing down. That's the answer for the acceleration portion of the problem; now we need the force, F, applied to the brakes.

F = ma where m is mass (we get to use that value now!) and a is -5.6 m/s/s.

F = 500(-5.6) and

F = -2800 N and the negative here means that the force of the brakes is acting against the motion of the car: the brakes are pulling the car "backwards" to stop while the car's motion is forward. The negative indicates the direction the force is being applied.

Tenemos un problema de termodinamica sobre gases ideales, los cuales pueden ser descritos por la ecuación de estado del gas ideal

[tex]P*V = n*R*T[/tex]

donde:

P = presión

V = volumen

n = número de moles

R = constante de los gases

T = temperatura.

1) Para el primer problema podemos pensar que el gas dentro del globo es un gas ideal:

Originalmente sabemos que n = 4 y V = 5.0 L

Entonces tendremos:

[tex]P*(5.0 L) = (4 moles)*R*T[/tex]

Ahora decidimos agregar 8 moles más, dejando constante la presión y la temperatura, entonces podemos reescribir la ecuación de arriba como:

(5.0L)/(4 moles) = R*T/P

Y como R, T y P son constantes, entonces:

R*T/P es una constante.

Esto quiere decir que cuando agreguemos 8 moles, para tener un total de 12 moles, tendremos que:

V'/(12 moles) = R*T/P = (5.0L)/(4 moles)

Donde V' es el nuevo volumen final del globo, y es lo que queremos obtener.

V'/(12 moles) =  (5.0L)/(4 moles)

V' =   (5.0L)*(12 moles)/(4 moles) = (5.0 L)*3 = 15.0 L

2) Queremos obtener la densidad en gramos sobre litro de un gas con una masa molar de 60g/mol, a P = 0.75 atm y T = 25°C.

Primero pasemos la temperatura a grados kelvin:

T = 25°C = (25 + 273.15)°K = 298.15°K

Planteamos la ecuación del gas ideal:

[tex]P*V = n*R*T[/tex]

reemplazando R = 0.082 (atm*L/mol*k), además de la presión y la temperatura para un solo mol de gas:

[tex](0.75 atm)*V = (1 mol)*(0.082 (atm*L/mol*k))*(298.15°K)[/tex]

Resolviendo para V obtenemos que:

[tex]V = (1 mol)*(0.082 (atm*L/mol*k))*(298.15°K)/(0.75 atm) = 32.597 L[/tex]

Así, vemos que un mol de este gas ocupa un volumen de  32.597 L

Y también sabemos que un mol de este gas pesa 60 gramos.

Recordando que la densidad es el cociente entre la masa y el volumen, podemos ver que la densidad del gas será:

[tex]d = 60g/(32.597 L) = 1.84 g/L[/tex]

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Answer

Resistivity R = K * L / A    where resistivity is constant for material, L the length of the material and A the area of the material

K = R * A / L     where R is the resistance of the material

Answer:

I don't know ???

sorry

Answer:

250N

Explanation:

According to newton second law,

\sumF = ma

Fm - Ff = ma

Since the velocity is constant, a = 0m/s

Frictional force Ff = 250N

Substitute

Fm - 250 = m(0)

Fm - 250 = 0

Fm = 250N

Hence the force to keep the box sliding at constant speed is 250N

Answer: There are multiple options:

By varying the supply voltage.

By varying the flux, and by varying the current through the field winding.

By varying the armature voltage, and by varying the armature resistance.

Helpful Link:

https://www.elprocus.com/what-are-the-best-ways-to-control-the-speed-of-dc-motor/

Answer:

sorry I don't understand this language

Explanation:

The Zeroth Law of Thermodynamics states that if two bodies are each in ... to heat, there will be no transfer of heat from one to the other.

Complete Question:

A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/hr and the rest at 12.5km/hr. What is the average speed?​

Answer:

Average speed = 9.7333 km/h

Explanation:

Given the following data;

To find the average speed;

First of all, we would determine the time taken to cover distance A in speed A by using the formula;

[tex] Time \ A = \frac {Distance \; A}{Speed \; A} [/tex]

Substituting the values into the formula, we have;

[tex] Time \ A = \frac {4.5}{9.5} [/tex]

Time A = 0.4737 hours

Total distance = distance A + distance B

5 = 4.5 + distance B

Distance B = 5 - 4.5

Distance B = 0.5 Km

Next, we would determine the time to cover distance B in speed B;

[tex] Time \ B = \frac {0.5}{12.5} [/tex]

Time A = 0.04 hours

Total time = time A + time B

Total time = 0.4737 + 0.04

Total time = 0.5137 hours

Now, we would solve for the average speed;

Mathematically, the average speed of an object is given by the formula;

[tex] Average \; speed = \frac {total \; distance}{total \; time} [/tex]

[tex] Average \; speed = \frac {5}{0.5137} [/tex]

Average speed = 9.7333 km/h

Answer:

Explanation:

Yes, According to law of conservation of energy, it has been said that energy cannot be created or destroyed.

Answer:

Đáp án là vị trí để đặt q3 tại vị trí cân bằng là cách A một đoạn 0,0411 m

và nằm trong AB

Explanation:

Do q1 và q2 trái dấu nên q3 nằm trong

gọi r1 là vị trí từ A đến q3 suy ra r2=0.08-r1

Vị trí cân bằng nên [tex]\frac{q1}{r1^{2} }[/tex] = [tex]\frac{q2}{(0.08-r1)^{2} }[/tex]

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

Answer:

ok

Explanation:

Acceleration is defined as the rate of change of velocity.

Acceleration = (Change in velocity) / time taken

Acceleration = (Final velocity - initial velocity) / time

As the object velocity changes by the same amount in each second, it means the acceleration is constant.

Hence the expression of ω in terms of m and k is

[tex]\omega = \sqrt{\frac{k}{m}[/tex]

Given the expressions;

[tex]T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}[/tex]

Equating both expressions we will have;

[tex]2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}[/tex]

Divide both equations by 2π

[tex]\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\[/tex]

Square both sides

[tex](\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}[/tex]

Take the square root of both sides

[tex]\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}[/tex]

Hence the expression of ω in terms of m and k is

[tex]\omega = \sqrt{\frac{k}{m}[/tex]

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Answer:

Mark brainliest please

answer is 5.5 degrees

Explanation:

Given,

Speed of car (v) = 50km/hr = 50×1000/60×60 m/s

Radius of circle (r) = 200m

Banking angle (θ) =?

we know ,

tanθ = v^2/rg

or , tanθ = (50×1000/60×60)^2/200×10 = 0.0965

∴ θ = tan-1(0.0965) = 5.5°

Explanation:

In scalar addition we have to do only the sum of magnitude as we know that the scalar quantities don't have direction.

since it is the scalar quantity the direction will be same for all the magnitudes.

Answer:

In scalar addition we have to do only the sum of magnitude as we know that the scalar quantities don't have direction...

since it is the scalar quantity the direction will be same for all the magnitudes.

on the other hand if it was a vector addition then it should be with two different directions which are considerable show the resultant vector needs to be calculated by vector addition method.

The temperature of a gas molecule is measured by the average translational kinetic energy

Answer:

The temperature of a gas is a measure of the average translational kinetic energy of the molecules. In a hot gas, the molecules move faster than in a cold gas; the mass remains the same, but the kinetic energy, and hence the temperature, is greater because of the increased velocity of the molecules

Explanation:

This is also from Go0gle because my explanation would've been an essay long .

but in shorter version if the gas molecules move fast it's hot an if it moves slow its cold hope this helps .

Answer:

a plumber

Explanation: