the smallest grains of dust stick together in an accretion disk by which force?

The correct answer is A, the brake pedal.

Before starting the engine, it is essential to ensure that the car is in a stationary position. Placing your foot on the brake pedal helps to keep the car stationary and prevent any accidental movement. It is a safety measure that must be followed at all times. Placing your foot on the accelerator pedal is not recommended as it could cause the car to move unexpectedly and potentially result in an accident. Placing your foot on the floor is also not recommended, as it does not serve any purpose in terms of safety or starting the engine. Therefore, the correct answer is A, the brake pedal, which is the standard practice recommended by car manufacturers and driving instructors.

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The  engine performs 9400 J of work over the 100 cycles.

The amount of work performed by a heat engine in one cycle can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

For a heat engine operating between two reservoirs at temperatures T1 and T2 (with T1 > T2), the maximum efficiency is given by:

η = 1 - (T2 / T1)

where η is the efficiency of the engine.

In this problem, the cold reservoir is a large insulated tank of ice water, which is at a temperature of 0°C (273 K). Let's assume that the hot reservoir is at a temperature of 27°C (300 K).

The heat added to the system in one cycle is given as 8000 J. The heat rejected to the cold reservoir melts 0.0180 kg of ice, which requires a heat of fusion of:

Qf = m * Lf

where Qf is the heat of fusion, m is the mass of ice melted, and Lf is the latent heat of fusion of water, which is 334 kJ/kg.

Substituting the given values, we get:

Qf = (0.0180 kg) * (334 kJ/kg) = 5.99 kJ

The total heat rejected to the cold reservoir in one cycle is the sum of the heat of fusion and the remaining heat that is transferred to the ice water:

Qc = Qf + (8000 J - Qf) = 5.99 kJ + 1940 J = 7.94 kJ

The efficiency of the engine is:

η = 1 - (273 K / 300 K) = 0.09

The work done by the engine in one cycle is:

W = Qh - Qc = (8000 J) - (7.94 kJ) = -94 J

(Note that the negative sign indicates that work is done on the engine, rather than by the engine.)

Therefore, the work performed by the engine in 100 cycles is:

100 cycles * (-94 J/cycle) = -9400 J

So the engine performs 9400 J of work over the 100 cycles.

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The intensity level of the sound from the siren decreases as the distance from the source increases. The intensity level heard by an observer 12 m away from the source is 102 dB.

The inverse square law for sound intensity states that the intensity level of sound decreases proportionally to the square of the distance from the source. This can be expressed as:

I₂ = I₁ × (r₁/r₂)²

Where I₁ is the initial intensity level, I₂ is the new intensity level, r₁ is the initial distance, and r₂ is the new distance.

We can plug in the given values:

I₁ = 120 dB

r₁ = 2.0 m

r₂ = 12 m

I₂ = 120 dB × (2.0 m/12 m)²

I₂ = 102 dB

Therefore, the intensity level heard by an observer 12 m away from the source is 102 dB.

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From the Hertzsprung-Russell (H-R) the correct option is A) It is very hot compared to other stars.

In the event that a star is found within the upper cleared-out corner of the Hertzsprung-Russell (H-R) chart, it is likely a hot star.

The upper cleared out corner of the H-R graph speaks to stars that are exceptionally brilliant and have tall temperatures, such as blue monsters, blue supergiants, and O-type stars.

These stars are much hotter than stars just like the Sun, which are found within the center of the H-R graph. In this manner, in the event that a star is found within the upper cleared out corner of the H-R graph, it is likely a hot star.

In any case, the particular properties and characteristics of the star, such as its age, estimate, and developmental organize, would depend on its correct area on the H-R chart. 

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The maximum kinetic energy of the photoelectrons is 3.10 eV.

When light of a certain frequency or wavelength is incident on a metal surface, it can cause electrons in the metal to be emitted. This process is called the photoelectric effect. The work function of a metal is the minimum amount of energy required to remove an electron from the metal surface.

The maximum kinetic energy (K.E.) of the photoelectrons is given by K.E. = hν - φ, where h is Planck's constant (6.626 × 10⁻³⁴ J⋅s), ν is the frequency of the incident light, and φ is the work function of the metal. We can use the relationship between frequency and wavelength, ν = c/λ, where c is the speed of light, to write K.E. = hc/λ - φ.

Using the given range of wavelengths (400 nm to 700 nm) and converting them to meters, we have λ = 4.00 × 10⁻⁷ m to 7.00 × 10⁻⁷ m. Plugging in these values, we get the maximum kinetic energy of the photoelectrons to be:

K.E. = hc/λ - φ

      = (6.626 × 10⁻³⁴ J⋅s)(3.00 × 10⁸ m/s)/(4.00 × 10⁻⁷ m) - 1.3 eV

      = 3.10 eV.

As a result, the photoelectrons maximal kinetic energy is 3.10 eV.

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The answer is d. all of these. Tree rings provide valuable information about past climates as they grow in response to the environmental conditions of their surroundings.

The thickness of growth rings is affected by factors such as temperature, precipitation, and soil moisture, and can indicate the overall climate conditions during the growing season. The density of growth rings is also influenced by these factors, and can provide information on the severity of drought or other environmental stresses.

Additionally, the presence of frost rings, which are formed when a tree is damaged by frost, can provide evidence of past cold weather events. By analyzing all of these factors together, scientists can gain a comprehensive understanding of past climate conditions, which can inform our understanding of climate change and its potential impacts on the environment.

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The minimum allowable number of lighting branch circuits required for a residence 90 feet by 60 feet is 9.

To determine the minimum allowable number of lighting branch circuits required for a residence 90 feet by 60 feet, use the National Electrical Code (NEC) guidelines.

Using these guidelines, calculate the minimum allowable number of lighting branch circuits as follows:

Calculate the total wattage of the lighting load:

Assume 3 watts per square foot for general lighting:

90 ft x 60 ft = 5,400 sq ft

5,400 sq ft x 3 watts/sq ft = 16,200 watts

Add 50 watts for each fixed appliance:

Assume 4 appliances (refrigerator, stove, oven, dishwasher)

4 x 50 watts = 200 watts

Total wattage = 16,200 watts + 200 watts = 16,400 watts

Calculate the total amperage of the lighting load:

Total amperage = total wattage / voltage = 16,400 watts / 120 volts = 136.67 amperes

Calculate the minimum number of circuits required:

Divide the total amperage by the maximum allowable load per circuit:

136.67 amperes / 16 amperes = 8.54 (round up to 9)

Therefore, the minimum allowable number of lighting branch circuits required for a residence 90 feet by 60 feet is 9.

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The moment of inertia of the rigid body about the axis passing through the midpoints of the other two sides is approximately 3252.89 g cm^2.  

The moment of inertia of a rigid body about an axis that is parallel to one side of the triangle and passes through the respective midpoints of the other two sides, we need to first determine the distance between the axis of rotation and the centroid of the triangle.

The centroid of an equilateral triangle can be found using the formula:

C = (a/3) + (b/3) + (c/3)

here a, b, and c are the lengths of the sides of the triangle.

In this case, a = b = c = 60 cm, so the centroid is:

C = (60/3) = 20 cm

Next, we need to find the distance between the centroid and the axis of rotation. Let's call this distance d. The distance is equal to the square root of the sum of the squares of the three distances between the centroid and each of the particle positions.

The distances between the centroid and each particle can be calculated as follows:

distance to particle 1 = 60 cm - 20 cm = 40 cm

distance to particle 2 = 60 cm - 20 cm = 40 cm

distance to particle 3 = 60 cm - 20 cm = 40 cm

The sum of the squares of the three distances is:

[tex](40^2) + (40^2) + (40^2) = 4000 + 1600 + 400 = 7000[/tex]

The square root of this sum is:

(7000) ≈ 23.42 cm

Therefore, the moment of inertia of the rigid body about the axis passing through the midpoints of the other two sides is:

[tex]I = d^2 * (3m/12)[/tex]

here m is the mass of each particle.

Since there are three particles, each with a mass of 80 g, the total mass of the rigid body is:

m = 3 * 80 g = 240 g

[tex]I = d^2 * (3m/12)\\= (23.42 cm)^2 * (3/12) * 240 g\\= 3252.89 g cm^2[/tex]

Therefore, the moment of inertia of the rigid body about the axis passing through the midpoints of the other two sides is approximately 3252.89 g [tex]cm^2.[/tex]

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The physics concept of wave speed measures how quickly a wave moves through a medium. It is measured in meters per second (m/s) and is defined as the distance covered by a wave in one unit of time.

To find the wave speed, we can use the formula:

wave speed = wavelength x frequency

We are given that the distance between two sequential troughs (which is the wavelength) is 2.20 meters. We are also given that six troughs pass a fixed point along the direction of travel every 12.5 s, which means the frequency is:

frequency = number of troughs/time
frequency = 6 / 12.5 s
frequency = 0.48 Hz

Now we can plug in the values into the formula:

wave speed = 2.20 m x 0.48 Hz
wave speed = 1.056 m/s

Therefore, the wave speed for this particular transverse wave is 1.056 m/s.

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The minimum coating thickness of the material to minimize reflection at 700 nm will be approximately 0.05 mm.

The minimum coating thickness of a material to minimize reflection depends on various factors such as the refractive index of the material, the wavelength of the incident light, and the angle of incidence of the light.

In general, a thicker coating can provide better reflection control, but it may also increase the cost and weight of the coating. Therefore, the optimal coating thickness will depend on the specific application and trade-offs between reflection control and other factors.

Assuming that the material has a refractive index of 1.5 and the wavelength of interest is 700 nm, the minimum coating thickness can be calculated using the formula:

d = 1 / (n * λ)

where d is the coating thickness, n is the refractive index of the material, and λ is the wavelength of the incident light.

For a wavelength of 700 nm, the minimum coating thickness can be calculated as:

d = 1 / (1.5 * 700) ≈ 0.05 mm

Therefore, the minimum coating thickness of the material to minimize reflection at 700 nm will be approximately 0.05 mm. However, it is important to note that this is only an approximate value and the actual minimum coating thickness may be different depending on the specific material and other factors.  

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When comparing two waves, if one wave is going up when the other is going down, the two waves are said to be completely out of phase.

Out of phase refers to the relationship between the oscillations of two waves. In the case of waves being completely out of phase, the peaks of one wave align with the troughs of the other wave, and vice versa. This results in a complete cancellation or destructive interference between the two waves. As a result, the net amplitude of the combined wave is zero at every point, leading to a flat line or no discernible wave pattern.

Being completely out of phase indicates a phase difference of 180 degrees or an odd multiple of 180 degrees between the waves. It means that the waves are in exact opposition to each other in terms of their oscillations. This phenomenon can occur with various types of waves, including sound waves, light waves, and electromagnetic waves. When two waves are completely out of phase, they do not exhibit any constructive interference, and their combined effect is the absence of a discernible wave pattern.

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A series RC circuit contains a 0.01 microfarad capacitor and a 2,000-ohm resistor and has a frequency of 500 Hz. The impedance of the series RC circuit is approximately 1416 ohms.

The impedance Z of a series RC circuit can be calculated using the formula:

Z = sqrt(R^2 + (1/ωC)^2)

where R is the resistance, C is the capacitance, and ω is the angular frequency.

Given that the capacitance of the circuit is 0.01 microfarads and the resistance is 2,000 ohms, we can calculate the angular frequency as:

ω = 2πf = 2π(500 Hz) = 1000π rad/s

Substituting the values into the formula, we get:

Z = sqrt((2000 Ω)^2 + (1/(1000π rad/s * 0.01 μF))^2)

= sqrt(4,000,000 + 10^12/(π^2))

≈ 1416 Ω

Therefore, the impedance of the series RC circuit is approximately 1416 ohms.

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The smaller recoil speed of the cannon when a cannonball is fired is due to differences in masses. According to Newton's third law of motion, for every action, there is an equal and opposite reaction.

When the cannonball is fired, it exerts a force on the cannon in one direction. As a result, the cannon experiences a reactive force in the opposite direction, causing it to recoil.The recoil speed of the cannon is determined by the conservation of momentum. The momentum of an object is defined as the product of its mass and velocity. Since the mass of the cannonball is typically much smaller than the mass of the cannon, the change in velocity (recoil speed) of the cannon will be larger due to the inverse relationship between mass and velocity.Therefore, the smaller recoil speed of the cannon when a cannonball is fired is primarily due to the difference in masses between the cannonball and the cannon itself.

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The mass per unit length of the string, which is a measure of its thickness:
μ = (ρ * A) / L   'where μ (mu) is the mass per unit length, ρ (rho) is the density of the string material (which we'll assume is constant), A is the cross-sectional area of the string (which we can calculate from its diameter), and L is the length of the string.

When you pluck a string on a musical instrument, it vibrates back and forth, creating sound waves that travel through the air and reach our ears. The frequency of these sound waves determines the pitch we hear - higher frequencies produce higher pitches, while lower frequencies produce lower pitches.

The fundamental frequency of a string is the lowest frequency at which it will vibrate. This frequency is determined by several factors, including the length, thickness, and tension of the string.
the wave equation to calculate the speed of the sound wave traveling through the string:
v = f * λ

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The total impedance has a real component and no imaginary component, it is purely resistive. Therefore, the statement is true: if e = 50 V ∠ 20° and i = 25 A ∠ 20°, then the total impedance zt is purely resistive.

Assuming that the circuit is in steady state, we can use Ohm's Law and Kirchhoff's laws to find the total impedance. In a series-parallel circuit, we first calculate the total resistance of the series part and the total impedance of the parallel part, and then add them together.

Let's assume that the circuit has two parallel branches, each containing a resistor and an unknown impedance element. We'll call the impedance of the first branch Z₂and the impedance of the second branch Z₂. We'll also call the resistance of the two resistors R₁ and R₂, respectively.

Using Ohm's Law, we can calculate the resistances:

R₁= [tex]V_{1} /I_{1} = |e|/|i| = 50 V/25 A = 2[/tex] Ω

R₂= [tex]V_{2} /I_{2} = |e|/|i| = 50 V/25 A = 2[/tex] Ω

Using Kirchhoff's laws, we can calculate the total impedance of the parallel part:

[tex]1/Zp = 1/Z_{1} + 1/Z_{2}[/tex]

Since the total impedance is purely resistive, the imaginary component must be zero. Therefore, we can set the imaginary parts of Z₁ and Z₂ to zero:

Im{Z₁} = Im{Z₂} = 0

We can also express Z₁ and Z₂ in polar form:

Z₁= R₁ + jX₁

Z₂= R₂ + jX₂

where X₁ and X₂are the reactive components of the impedances.

Substituting these expressions into the equation for the total impedance of the parallel part and setting the imaginary part to zero, we get:

1/Zp = (1/R₁+ j/X₁) + (1/R₂ + j/X₂)

0 = j/X₁ + j/X₂

Since the imaginary parts must be equal and opposite, we have:

X₁ = -X₂

Therefore, the total impedance of the parallel part is purely resistive:

Zp = R₁R₂/(R₁ + R₂) = 2 Ω

Now, let's calculate the total impedance of the series-parallel circuit:

Zt = Zs + Zp

where Zs is the total impedance of the series part of the circuit.

Since the current is the same in both the series and parallel parts, we can use Ohm's Law to express the total impedance of the series part:

Zs = [tex]V/I = |e|/|i| = 50 V/25 A = 2[/tex]Ω

Therefore, the total impedance of the circuit is:

Zt = Zs + Zp = 2 Ω + 2 Ω = 4 Ω

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For a 22Na source which is labeled 2. 50 mCi, but its present activity is found to be 2. 31 ✕ 107 Bq (a) the present activity in mCi is 0. 623 mCi and (b) the time it actually had a 2. 50-mCi activity is 19 years ago.

(a) Given, 1 mCi = 3. 7 ✕ 10^10 Bq

Therefore, 2. 50 mCi = 2. 50 × 3. 7 ✕ 10^10 = 9. 25 ✕ 10^10 Bq

So, Present activity in mCi = 2. 31 ✕ 10^7 / 3. 7 ✕ 10^10 = 0. 623 mCi

(b) Let's assume, after time t, the activity of the source is 2. 50 mCi. Then, at present, the activity of the source = 2. 31 ✕ 10^7 Bq

Let, λ be the decay constant and A₀ be the initial activity of the source at time t = 0.

The activity of a radioactive substance at any time t can be represented by the formula,

A = A₀ e^(-λt)

Given, A₀ = 2. 50 mCi = 2. 50 × 3. 7 ✕ 10^10 = 9. 25 ✕ 10^10 Bq

A = 2. 31 ✕ 10^7 Bqe^(-λt)

∴ λ = ln(A₀/A) / t = ln(9. 25 ✕ 10^10 / (2. 31 ✕ 10^7)) / t ≈ 2. 27 ✕ 10^-9 s^-1

Therefore, the time taken for the activity to reduce from 2. 50 mCi to 0. 623 mCi is given by

2. 50 e^(-2. 27 ✕ 10^-9 t) = 0. 623

e^(-2. 27 ✕ 10^-9 t) = 0. 623 / 2. 50 = 0. 2492

Taking natural logarithm on both sides, we get

-2. 27 ✕ 10^-9 t = ln(0. 2492)

t = - ln(0. 2492) / 2. 27 ✕ 10^-9≈ 0. 60 × 10^9 s ≈ 19 years

Therefore, the source actually had a 2.50-mCi activity about 19 years ago.

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The development of a hurricane typically begins with an Arctic disturbance, which is essentially a low-pressure zone in the atmosphere. As this disturbance draws in a cluster of thunderstorms with weak surface winds, the storm begins to gather strength and develop a more defined structure. However, it's worth noting that not all Arctic disturbances will develop into hurricanes - it often depends on a variety of factors such as wind shear, temperature, and moisture levels.

When it comes to where hurricanes typically form, you're right that the equator is a common location. This is because the seawater in this region tends to be the warmest, which provides the necessary energy for a hurricane to form and grow. Additionally, the Coriolis effect - which is the force that causes objects to move to the right in the Northern Hemisphere and to the left in the Southern Hemisphere - decreases as you move away from the equator. This effect is important in creating the swirling motion that we associate with hurricanes, so a weaker Coriolis effect can make it easier for a storm to develop into a hurricane.
I hope that helps answer your question! Let me know if you have any further questions.
I understand that you'd like an explanation of hurricane development, incorporating the terms Arctic disturbance, low-pressure zone, thunderstorms, surface winds, equator, seawater, and Coriolis effect. Here's my answer:
The development of a hurricane begins with a low-pressure zone called a tropical disturbance, rather than an Arctic disturbance. This tropical disturbance draws in a poorly organized cluster of thunderstorms with weak surface winds. Most hurricanes form near the equator, typically between 5 and 20 degrees latitude, because the seawater is warmest in these areas, providing the necessary heat and moisture for hurricane development. Additionally, the Coriolis effect, which is essential for cyclone formation, increases with distance from the equator, ensuring that hurricanes do not form directly on the equator.

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The pattern on the wall has a broad central spot, with a series of dimmer spots on either side. This pattern is known as the diffraction pattern, and it occurs due to the wave-like nature of light.

When light passes through a narrow slit, it diffracts, spreading out in all directions. The central spot corresponds to the direct beam passing through the center of the slit, while the dimmer spots on either side correspond to the first-order diffraction maxima. These are locations where the light waves constructively interfere with each other.

The pattern is also affected by the width of the slit and the wavelength of the laser light. The broader the slit, the broader the central spot, and the narrower the dimmer spots. The wavelength of the laser determines the spacing between the spots. This pattern is an example of a fundamental concept in physics called diffraction.

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An observer on earth measures speed c for the light. This is because the speed of light is constant for all observers regardless of the motion of the source or the observer. Option D is Correct.

In this scenario, a rocket is flying towards Earth at 0.5c (half the speed of light) and the captain shines a laser light beam in the forward direction. Here are the statements and their correctness:
1. An observer on Earth measures the speed 1.5c for the light. (Incorrect)
2. The captain measures speed 0.5c for the light. (Incorrect)
3. The captain measures speed c for the light. (Correct)
4. An observer on Earth measures speed c for the light. (Correct)
According to the theory of relativity, the speed of light is always measured to be c (approximately 299,792 kilometers per second), regardless of the observer's frame of reference. So both the captain and an observer on Earth would measure the speed of the light beam as c.

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The Complete question is

A rocket flies toward the earth at 0.5c and the captain shines a laser light beam in the forward direction. Which of the following statements about the speed of this light are correct? (There may be more than one correct answer.)

A. An observer on earth measures speed 1.5c for the light. B. The captain measures speed 0.5c for the light. C. The captain measures speed c for the light. D. An observer on earth measures speed c for the light.

A rocket flies toward the Earth at 0.5c (half the speed of light) and the captain shines a laser light beam in the forward direction, the correct statements about the speed of this light are:
           •  The captain measures speed c for the light.
           •  An observer on Earth measures speed c for the light.

The speed of light (c) is constant for all observers, regardless of their relative motion, according to the theory of special relativity.

Therefore, the captain on the rocket and the observer on Earth would both measure the speed of the laser light as c (approximately 299,792 km/s), not 1.5c or 0.5c.

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D. the region is full of gases that become ionized after they are released from volcanoes on io. The reason for the intense radiation in the region that traces Io's orbit around Jupiter (the Io torus) is due to Io's gravity.

Lo's gravity causes it to interact strongly with Jupiter's powerful magnetic field, and as a result, the region becomes a trap for charged particles from the solar wind. These charged particles are accelerated to high energies and become trapped in the magnetic field around Io, creating intense radiation. In addition to Io's gravity, the strong interaction between Io, Europa, and Ganymede also plays a role in creating the intense radiation in the Io torus.

This interaction creates a resonance effect, where the three moons' gravitational fields work together to create a synchronized orbital motion, which intensifies the trapping of charged particles in the region. The combination of these factors creates a harsh radiation environment that would be extremely hazardous to any spacecraft or living organisms.

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Complete Question:

why is the radiation so intense in the region that traces io's orbit around Jupiter (the io torus)?

group of answer choices

A. an orbital resonance between io, Europa, and ganymede makes the radiation intense.

B. io's gravity allows this region to capture huge numbers of charged particles from the solar wind.

C. Jupiter's strong magnetic field makes the radiation intense everywhere, and the region around io is no different than any other region.

D. the region is full of gases that become ionized after they are released from volcanoes on io.

We first need to understand what happens during contact between the two spheres. Charge on Sphere B after contact = Total charge / 2 = -1.8 nC / 2 = -0.9 nC

When two objects with different electric charges come into contact, electrons can transfer from one object to the other until both objects have the same charge. In this case, sphere A has a positive charge of 1.9 nc (nanocoulombs) and sphere B has a negative charge of -3.7 nc. When they come into contact, electrons will transfer from sphere A to sphere B until they both have the same charge. To determine the final charge on sphere B, we need to calculate the total charge of both spheres before contact. The total charge is the sum of the charges on each sphere:
Total charge before contact = 1.9 nc - 3.7 nc = -1.8 ncelectrons
Since the total charge is negative, we know that there are more  on sphere B than on sphere A before contact. During contact, electrons will transfer from sphere A to sphere B until they both have the same charge.

To calculate the final charge on sphere B, we need to divide the total charge before contact by 2 (since the charges will be split equally between the two spheres after contact):
Final charge on sphere B = -1.8 nc / 2 = -0.9 nc
Therefore, the final charge on sphere B after contact is -0.9 nc.
When two spheres come into contact, they redistribute their charges evenly between them due to the principle of charge conservation. To find the charge on sphere B after contact, we can calculate the total charge and then divide it by 2, as both spheres will have the same charge after contact.
Total charge = Charge on Sphere A + Charge on Sphere B = 1.9 nC + (-3.7 nC) = -1.8 nC.

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A burette filled with water to a volume of 28 cm cubic is represented as a long cylindrical tube with a narrow neck, a stopcock, and a water level at 28 cm.

A burette is a laboratory equipment that is used to measure the volume of a liquid with a high degree of accuracy. It is usually made of glass and has a long,  cylindrical shape with a narrow neck and a stopcock at the bottom.

The burette is commonly used in chemistry experiments, particularly in titrations, to measure the volume of a liquid that is being added to a solution.

To draw a burette filled with water to a volume of 28 cm cubic, first, you will need to set up the burette on a stand. The burette should be clamped securely to the stand, and the stopcock should be closed to prevent any liquid from flowing out.

Next, you will need to fill the burette with water. To do this, you can use a funnel and slowly pour the water into the burette through the funnel. Make sure that there are no air bubbles in the burette and that the water level is below the zero mark on the burette.

To fill the burette to the desired volume of 28 cm cubic, you will need to slowly open the stopcock and let the water flow out until the water level reaches 28 cm cubic on the burette scale. It is important to take your time when filling the burette to ensure that you do not overshoot the desired volume.

Once you have filled the burette to the desired volume, you can close the stopcock to stop the flow of water. You should now have a burette filled with water to a volume of 28 cm cubic. Remember to record the volume measurement accurately to ensure the accuracy of your experiment.

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While global warming refers to the overall increase in Earth's average temperature, it doesn't necessarily mean that all areas will get warmer.

In fact, certain regions may actually experience colder temperatures as a result of changing weather patterns. One reason for this is the melting of Arctic sea ice, which can lead to altered ocean currents and ultimately cause colder temperatures in some areas of the northern hemisphere.

Additionally, as the Earth's climate continues to change, extreme weather events such as polar vortexes and heavy snowfall can occur in areas that may not have experienced them before.

It's important to note that while certain areas may experience colder temperatures, the overall trend is still towards global warming and its negative impacts on the planet.

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Answer: The correct answer is False.

Explanation: Gray matter derives its color from a high concentration of neurons cell bodies.

The myelinated axons gives white matter its color.

The block's speed is 1.0 m/s at positions 10 cm and 30 cm. and The spring constant is approximately 61.7 N/m.

a. First, we need to determine the maximum speed (Vmax) of the block. To do this, we can use the formula Vmax = 2πA/T, where A is the amplitude (20 cm or 0.2 m) and T is the period (0.80 s).
Vmax = \frac{2π(0.2 m) }{ 0.80 s} ≈ 1.57 m/s
Now that we have the maximum speed, we can find the positions where the block's speed is 1.0 m/s. Since the speed of the block is related to its position, we can use the formula v = Vmax * cos(2πx / λ), where v is the speed (1.0 m/s), x is the position, and λ is the wavelength (equal to 2A, or 0.4 m).
1.0 m/s = 1.57 m/s * cos(2πx / 0.4 m)
Solving for x, we get:
x ≈ 0.1 m and 0.3 m
So the block's speed is 1.0 m/s at positions 10 cm and 30 cm.
b. To determine the spring constant (k), we can use the formula T = 2π√(m/k), where m is the mass (0.5 kg) and T is the period (0.80 s).
0.80 s = 2π√(0.5 kg / k)
Solving for k, we get:
k ≈ 61.7 N/m
So, the spring constant is approximately 61.7 N/m.

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Removing  100 kg of excess weight   from the car saves approximately 1,470,000 joules of energy each time the car is driven up a 3 km high mountain, assuming the above assumptions and calculations.

The amount of energy saved by removing 100 kg of excess weight from a car depends on a few factors, including the efficiency of the engine and the specific driving conditions. However, we can make some reasonable assumptions to estimate the energy savings.

Let's assume that the car in question has an average fuel efficiency of 10 km per liter of gasoline, and that driving up a 3 km high mountain requires 1 liter of gasoline. This means that the car consumes 0.1 liters of gasoline for every kilometer driven on flat ground.

Removing 100 kg of excess weight from the car reduces its mass by approximately 5%. This means that the car requires 5% less energy to climb the mountain than it did before the weight reduction. Therefore, the energy savings can be estimated as:

Energy savings = 5% x Energy required to climb the mountain

The energy required to climb the mountain can be calculated as the potential energy difference between the base of the mountain and the top, which is given by:

Potential energy = mass x gravity x height

Assuming a mass of 1000 kg for the car (before weight reduction), a gravitational acceleration of 9.8 m/s^2, and a height of 3000 m, we can calculate the potential energy required to climb the mountain as:

Potential energy = 1000 kg x 9.8 m/s^2 x 3000 m = 29,400,000 joules

Using the formula for energy savings, we can now calculate the amount of energy saved by removing 100 kg of excess weight:

Energy savings = 5% x 29,400,000 joules = 1,470,000 joules

Therefore, removing 100 kg of excess weight   from the car saves approximately 1,470,000 joules of energy each time the car is driven up a 3 km high mountain, assuming the above assumptions and calculations.

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The height h of the fluid in the capacitor can be determined using the energy conservation principle. The potential energy of the fluid is equal to the work done by the electric field to lift the fluid against gravity.

The height h of the fluid in the capacitor can be determined using the energy conservation principle. The potential energy of the fluid is equal to the work done by the electric field to lift the fluid against gravity. Since the capacitor is connected to a battery with voltage V, the electric field strength inside the capacitor is E = V/d, where d is the distance between the plates. The potential energy per unit volume of the fluid is U = 1/2 * ε * E^2. The mass of the fluid lifted is m = ρ * A * h, where A is the area of the capacitor plates. Therefore, equating the potential energy and the work done, we get:

1/2 * ε * E^2 * A * h = m * g * h

Substituting the values, we get:

1/2 * ε * (V/d)^2 * A * h = ρ * A * h * g

Simplifying, we get:

h = (2 * V^2)/(d^2 * g * ε * ρ)

Therefore, the height of the fluid in the capacitor is directly proportional to the square of the voltage, and inversely proportional to the square of the distance between the plates, the gravitational acceleration, the dielectric constant, and the density of the fluid.

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The height of the lunar cliff is 256 ft when an object is dropped from a lunar cliff and its speed in ft/sec. after t seconds is given by v(t)=5.3t.

We can use the formula for distance traveled by an object under constant acceleration to find the height of the lunar cliff. The acceleration of the object is due to gravity and is constant. We can use the formula: d = [tex](1/2)at^2[/tex], where d is the distance traveled, a is the acceleration and t is the time taken.
In this case, we know the time taken is 4 seconds and the acceleration is due to gravity, which is approximately 32 ft/sec^2 on the moon. Therefore, we have:
d = [tex](1/2) * 32 * (4)^2[/tex]
d = 256 ft
It is interesting to note that the speed of the object is directly proportional to the time taken, as given by v(t) = 5.3t. This means that after 4 seconds, the object would be traveling at a speed of 5.3 x 4 = 21.2 ft/sec. This is relatively high considering the weaker gravitational pull on the moon compared to the Earth.

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The phenomenon that occurs at the surface of the paper is called diffuse reflection or scattering.

When a student looks into a plane mirror, she sees a virtual image of herself because the mirror reflects the light rays coming from her body and forms an image behind the mirror. Since the image is formed by the reflected light, it is a virtual image, which means that the light rays do not actually converge at the position of the image.

On the other hand, when the student looks into a sheet of paper, no such image forms because the paper does not reflect enough light to form a clear image. The surface of the paper appears diffused and irregular due to the scattering of light by the surface irregularities and fibers of the paper.

The phenomenon that occurs at the surface of the paper is called diffuse reflection or scattering. When light falls on a rough or irregular surface, it is scattered in different directions instead of reflecting in a single direction, as in the case of a mirror. The scattered light produces a diffused image of the object, which does not have a well-defined shape or position. This is why the student cannot see a clear image of herself when looking at the paper.

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