The diagram shows the electric field due to point charge Q. The negative charge, A, is within the field. Charge Q has vectors radially inward starting perpendicular from the surface. The farther you get from the charge, the shorter the vectors. All vectors point towards the charge. A point labeled A is just to the right of the charged object. Which statements are correct? Check all that apply. Charge Q is positive. Charge Q is negative. The electric field is uniform. The electric field is nonuniform. If charge A is negative, it moves away from charge Q. If charge A is positive, it moves away from charge Q.

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

Answer:

The answer is Spring Potential

Explanation:

Because all the others are a mechanical form of energy

The correct answer is 5 km/h

Explanation:

The speed at which the duck travels can be found by using the equation that is given (speed= distance/ time). The first step to do this is to replace distance and time using the values given. Here is the process:

speed = distance / time

speed = 10 km / 2 hours

Now, solve this equation

speed = 10 km/ 2 hours

10 / 2 or 10 divided 2 = 5

Finally, use the units, in this case, the correct units are km/h

Torque is a measure of the turning force applied to an object about a rotational axis. It is calculated as the product of the force applied to the object and the distance from the axis of rotation at which the force is applied.

Use the equation slope = mod to calculate the unknown mass (mo) for parcels A and C, setting d = 1.0 m, and parcels G and H, setting d = 1.5 m.

Record all data, tables, and four graphs for analysis.

The experiment demonstrates the application of torque in determining unknown masses and provides valuable insights into the concept of torque in physics.

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Answer:

Teaching is the process of attending to people's needs, experiences and feelings, and making specific interventions to help them learn particular things.

Explanation:

Answer:

In education, teaching is the concerted sharing of knowledge and experience, which is usually organized within a discipline and, more generally, the provision of stimulus to the psychological and intellectual growth of a person by another person or artifact.

2,302.74 J is the heat generated.

Current electricity is the electricity produced by the movement of electrons. Static electricity is the electricity that accumulates on a substance's surface. Power plants and batteries are used to produce the current electricity.

Electric current in a wire, where electrons serve as the charge carriers, is a measurement of the amount of charge that moves through any point of the wire in a given amount of time.

The mass of the aluminum is m = 0.25 kg.

The difference in temperature is T=10 C.

The following is a list of aluminum's specific heat capacities:

c=921.096 J/kg∘C

Therefore, we apply the following formula in this instance:

Q=mcΔT

We replace with:

Q=(0.25kg)(921.096J/kg∘C)(10∘C)

Thus, we will receive the heat produced:

Q=2,302.74 J

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Answer:

Muhammad Ali Jinnah (Urdu: محمد علی جناح‎; born Mahomedali Jinnahbhai; 25 December 1876 – 11 September 1948) was a barrister, politician and the founder of Pakistan.

Changing the green sum of forces arrow involves manipulating applied forces, frictional forces, and normal forces, allowing for adjustments in magnitude and direction through various means.

To change the green sum of forces arrow, you can employ the following three strategies:

Adjusting Applied Forces: By modifying the magnitude or direction of the applied forces, you can alter the green sum of forces arrow. If the applied forces are increased or directed in a different way, the green sum of forces arrow will change accordingly. For instance, increasing the magnitude of a pushing force will result in a larger green sum of forces arrow in that direction.

Modifying Frictional Forces: Frictional forces play a crucial role in determining the green sum of forces. By changing the coefficient of friction or applying lubricants, you can affect the magnitude of frictional forces acting on an object. Reducing friction will decrease the green sum of forces arrow, while increasing friction will have the opposite effect.

Varying Normal Forces: The green sum of forces arrow can be influenced by adjusting the normal forces acting on an object. Normal forces are perpendicular to the surface and counteract the weight of an object. By changing the angle or surface on which an object rests, you can modify the normal forces. This alteration will subsequently impact the green sum of forces arrow.

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Answer: 520 nm; 5.77 × 1014 Hz; 3.82 × 10−19 J

Time period of a wave is the inverse of its frequency. The period of the wave with a frequency of 2.09 Hz is 0.47 seconds.

Frequency of a wave is the number of wave cycles per unit time. Frequency is the inverse of the time period of the wave. Hence, it has the unit of s⁻¹ which is equivalent to Hz.

The higher frequency of a wave indicates more number of wave cycles in a short time. Frequency is directly proportional to the energy and inversely proportional to the  wavelength.

Given the time period of the wave = 2.09 Hz

then frequency = 1/2.09 Hz = 0.47 s.

Therefore, the time period  of the wave is 0.47 seconds.

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The kinetic energy of the particle when it reaches position A will be K0 * (2d/4d)^2 = K0/4 , when the particle is released at position C.

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As the mass of the central star increases, the distance to the habitable zone increases and the size (width) of the habitable zone also increases.

The habitable zone is the distance from a star at which the liquid water could exist on the orbiting planet's surfaces. Habitable zone is also known as Goldilocks' zones, where the environmental conditions might be just right neither too hot nor too cold for life.

A star is a large luminous object which is found in space that produces heat and light energy as well as electromagnetic radiations. The habitable zone of a star is the zone where liquid water can be found. The distance to the habitable zone of a star increases with the increase in mass of the star.

Therefore, as the mass of the central star increases, the distance to the habitable zone also increases.

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The scientific study of physics focuses on the underlying concepts that underpin the laws of nature. It investigates how matter, energy, space, and time behave and interact. Physics' fundamental goal is to comprehend the underlying laws and forces that create our universe.

Physics has developed theories and rules to explain a wide variety of events, from the motion of celestial bodies to the behaviour of subatomic particles, via meticulous observation, investigation, and mathematical analysis.

These theories offer a framework for comprehending and forecasting the behaviour of physical systems, such as Newton's laws of motion and Einstein's theory of relativity.

From the tiniest particles to the biggest cosmic structures, physics has enhanced our understanding of the world and sparked a host of technological advances.

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The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

This observation that electromagnetic radiation behaves like particles is known as the photoelectric effect.

It provides evidence that electromagnetic radiation exhibits particle-like properties. When EMR with sufficient energy (above a certain threshold) interacts with a metal surface, it can cause the ejection of electrons from the metal.

This behavior indicates that EMR behaves as discrete packets of energy called photons, which transfer their energy to the electrons and cause their release. The photoelectric effect supports the particle nature of EMR and is a fundamental concept in the field of quantum mechanics.

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Answer:

V₂₋₁  = 50 [miles/h]

Explanation:

In order to solve this problem we must use the concept of relative velocities.

Where:

V1 = 50 [miles/h]

V2 = 75 [miles/h]

V₂₋₁ = Relative velocity among the two cars.

V₂₋₁  = 75 - 25 = 50 [miles/h]

That is, the car that goes at 50 miles per hour, sees how the second car approaches at 50 miles per hour.

Answer:

the water will float over mercury and the steel ball will be in it's respected place

Answer:

1.2 meter because the equa

Answer:

1.) 2 seconds

2.) 4.5 hertz

3.) it will become one third its original value

4.) 5.9 seconds

5.) 0.87 meters

Explanation:

just took the connexus quick check

Teniendo en cuenta la definición de media, mediana y moda:

La media (también llamada promedio o media aritmética) de un conjunto de datos es una medida de posición central cuyo valor se obtiene al dividir la suma de todos los números entre la cantidad de ellos. Entonces, en este caso, el promedio de las lecturas obtenidas se calcula como:

\(Promedio=\frac{5+1+3+3+6+2+6+4+5+2+1+2+5+3+2+6+1+4+4}{19}\)

Resolviendo:

Promedio= 3,42

El promedio de las lecturas obtenidas es 3,42.

Por otro lado, la moda de un conjunto de números es el número que aparece más a menudo dentro del grupo de datos.

En este caso, podes observar que la cantidad de veces que aparece cada número es:

Debido a que el 2 es el número con mayor cantidad de repeticiones, la tirada que más se repite es 2.

Por último, la mediana es valor que se encuentra en la mitad justa entre los valores máximo y mínimo de los datos con los que se está trabajando; es decir, que al ordenar los número de menor a mayor, éste se encuentra justamente en medio de los valores.

En este caso, ordenando de menor a mayor las lecturas del dado:

1,1,1,2,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6

se observa que el valor el mayor valor intermedio de todas las lecturas es 3.

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An aerial photographic image's corners, edge-centers, or both may contain a series of marks known as fiducial marks. These traces are recorded on the original film by the camera.

Fiducial markers are tiny metal objects, usually made of gold, that are positioned inside or close to a tumor to help direct the placement of radiation beams during therapy. They are about the size of a grain of rice.

Tiny metal things are called fiducial markers (about the size of a grain of rice). They assist your healthcare professionals in aligning the radiation beams and guarantee that your radiation therapy is administered consistently.

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Fiducial marks are a set of marks that can be found in the corners, edge-centers, or both of an aerial photographic image. The camera captures these traces on the original film.

Given  the Ryerson Campus is at a scale of  1:1000

Then the photographic scale (s) = 1:3000

The focal length of camera (f) = 152mm = 0152m

The format is (k) =  230 x 230mm = 0.23m

the photographic scale (s) = F/H where H is the image height

1/3000 = 0.152 /H Then H = 456m

Assume that there is longitudinal overlap as P1 = 60% and side to side overlap as 30%. Therefore margin of ground photograph = (1-P1) x k/s

(1-0.6) x 0.23/1/3000 = 276mm

Now width of ground photograph = (1-0.3) x 0.23/1/3000 = 483m

Hence the central points are plotted within the area range in 276m x 483m magnitude.

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Answer:

Below

Explanation:

Acceleration is defined as      change in velocity / change in time

Accel = 5 m/s  /  15 s =  .33 m/s^2

There are 3.37 × \(10^{22}\) gold atoms per cubic centimeter in the silver-gold alloy.

The density of a binary alloy can be calculated using the following equation:

ρ = w1ρ1 + w2ρ2

where,

ρ = density of the alloy

w1 and w2 = weight fractions of the two components (in this case, gold and silver)

ρ1 and ρ2 = densities of the pure components.

We are given that the alloy contains 10% gold and 90% silver by weight, so we can calculate the weight fractions as:

\(w_{Au}\) = 0.10

\(w_{Ag}\) = 0.90

We are also given the densities of pure gold and silver as:

ρ_Au = 19.32 g/\(cm^{3}\)

ρ_Ag = 10.49 g/\(cm^{3}\)

Now we can substitute these values into the density equation to find the density of the alloy:

ρ = \(w_{Au}\)ρ_Au +\(w_{Ag}\)ρ_Ag

ρ = (0.10)(19.32 g/\(cm^{3}\)) + (0.90)(10.49 g/\(cm^{3}\))

ρ = 11.08 g/\(cm^{3}\)

Next, we need to calculate the number of gold atoms per cubic centimeter in the alloy.

To do this, we can use Avogadro's number and the atomic weights of gold and silver:

\(N_A\) = 6.022 × \(10^{23}\) atoms/mol

Aum = 196.97 g/mol

Agm = 107.87 g/mol

The number of gold atoms:

\(n_{Au}\) = (\(w_{Au}\)ρ/ Aum) × \(N_{A}\)

Substituting the values, we get:

\(n_{Au}\) = (0.10 × 11.08 g/\(cm^{3}\)/ 196.97 g/mol) × 6.022 × \(10^{23}\) atoms/mol

\(n_{Au}\) ≈ 3.37 × \(10^{22}\) atoms/\(cm^{3}\)

Therefore, there are approximately 3.37 × \(10^{22}\) gold atoms per cubic centimeter in the silver-gold alloy.

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The initial volume of the liquid is 31.4 cm³. The volume coefficient of thermal expansion of the liquid is 0.002 per degree Celsius. A temperature increase of 109.5°C is needed for the liquid to rise to a height of 49cm.

The initial volume of the liquid can be found using the formula for the volume of a cylinder:

V = πr²h

where r is the radius (half the diameter), h is the height, and π is approximately 3.14. Plugging in the given values, we get:

V = π(0.5 cm)²(40 cm)

V = 31.4 cm³

The volume coefficient of thermal expansion (β) is defined as the fractional change in volume per degree Celsius change in temperature. It can be calculated using the formula:

β = ΔV/(VΔT)

where ΔV is the change in volume, V is the initial volume, and ΔT is the change in temperature. We can rearrange this formula to solve for ΔV:

ΔV = βVΔT

We know that ΔT = -10°C (a decrease of 10°C) and that the height decreased from 40cm to 37cm, or by 3cm. The change in volume can be found using the formula for the volume of a cylinder again, with the new height of 37cm:

ΔV = π(0.5 cm)²(40 cm - 37 cm)

ΔV = 0.59 cm³

Plugging in all the values, we get:

0.59 cm³ = β(31.4 cm³)(-10°C)

β = 0.002

To find the change in temperature needed for the liquid to rise to a height of 49cm, we can use the same formula as before, but solve for ΔT:

ΔT = ΔV/(βV)

We know that ΔV is the difference between the initial volume and the volume at the new height, which is:

ΔV = π(0.5 cm)²(49 cm - 40 cm)

ΔV = 6.86 cm³

Plugging in all the values, we get:

ΔT = 6.86 cm³/(0.002)(31.4 cm³)

ΔT = 109.5°C

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Light behaves Sometimes as a wave sometime as a particle.

It is called dual nature of light.

Answer:

the amount of carbon dioxide used is equal to the amount of oxygen formed.

Explanation:

I hope that helped you didn't put any statements so I couldn't really tell witch one was right-  

"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.

The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)

Using Newton's second law, we set up the following equations.

• p = 100 N

∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0

∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0

• P = 350 N

∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0

∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0

(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)

Solve for n and N :

n = p sin(θ) + mg cos(θ)

N = P sin(θ) - mg cos(θ)

Substitute these into the corresponding equations containing µ, and solve for µ :

µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))

µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

Next, you would set these equal and solve for m :

(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

...

Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with

m ≈ 36.5 kg

µ ≈ 0.256

The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.

From the given,

The force that opposes the crate by sliding is P = 100N

In X-axis, the sum of forces is zero.

ΣF = 0

Pcosθ - mgsinθ-Ff = 0

Ff = Pcosθ - mgsinθ

In Y-axis

Psinθ - mgcosθ - N = 0

N = Psinθ-mgcosθ

Frictional force, Ff = μN, μ is the coefficient of friction

Ff = μN

Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0

μ = mgsin30-Pcos30/Psin30+mgcos30 ------1

The block is sliding with the horizontal force, F = 350N

X-axis

P₂cosθ - mgsinθ-Ff = 0

Y-axis

P₂sinθ - mgcosθ - N = 0

N = P₂sinθ-mgcosθ

μ = P₂cos30-mgsin30/P₂sin30-mgcos30   -----2

Equate equations 1 and 2

mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30

4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49

41.7m² + 123m - 1.516×10⁴ = 0

-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)

83.4m² - 2207m -3.03×10⁴ = 0

m= 36.4 kg

Hence, the mass of the crate is 36.4 Kg.

Substitute the value of m in equation 1,

μ = 4.905(36.4) - 86.6 / 50 + 8.49

μ  = 0.256

Thus, the coefficient of static friction is 0.256.

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Answer:

True

Explanation:

I’m assuming you meant latitude:

The angular distance north or south of the earth's equator, measured in degrees along a meridian, as on a map or globe.

a region based on its distance to the equator uses latitude measurements

basically: up and down , north to south from a certain point