Copper has a specific heat of 0. 386 j/g°c. How much heat is required to increase 5. 00 g of copper from 0. 0°c to 10. 0°c?.

Answer:

c

Explanation:

because its c

gggfffgggyvg gt you jyngy6bht

Answer:

You just double the number of boxes it goes higher. Then triple the number for the second question. But don't change the time.

Answer:

wait what do you mean? And why is this in physics?

Is this about the iron dome or something biblical?

Explanation:

Answer:

the story of Israel is part of the Bible that's what what you're your lesson is

Explanation:

The primary indicator of an object's position and speed is its velocity.It is the distance that an object travels in one unit of time.The displacement of the item in one unit of time is the definition of velocity.

Smax = 400 feet.

c) v = 32 ft/s

c) v = - 32 ft/s

(a)

The following is the function provided for ball height:

s = 160 t - 16 t²

Therefore, we must take the derivative with respect to t and make it equal to zero in order to get the time to reach maximum height (in-flexion point):

So, 160 - 32 t = 0, 32 t = 160 t, and 160/32 t = 5 sec.

As a result, the maximum distance is covered every 5 seconds.

Smax = (160)(5) - (16)(5)²

Smax = 800 - 400

Smax=400 feet

b)First, we determine the speed at which the ball travels 384 feet.

384 = 160 t - 16 t

16 t² - 160 t + 384 = 0

Quadratic Equation Solving:

Either:

t = 6 sec

or t = 4 seconds

Since it takes 5 seconds to reach the highest point,

t then equals 5 seconds.

t then equals 4 seconds.

Now, by taking a derivative of ods with respect to t at 4 sec, we can find velocity:

v = 160 - 32 t

v = 160 - (32)(4)

v = 32 ft/s

c,Because t = 6 s > 5 s

The second number of t = 6 sec must represent the point in the ball's downward motion when it is 384 feet above the earth.

So, at that moment, velocity will be:

v = 160 - (32)(6)

v = -32 ft/s

downward motion is a bad sign.

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For r > R, the electric field is given by:

E = λ_total / (2πε₀r).

To summarize:

For r < R: E = λ / (2πε₀r).

For r > R: E = λ_total / (2πε₀r).

Let's consider a cylindrical Gaussian surface with radius r and length L, where L is much greater than the length of the charge distribution. The surface is coaxial with the distribution, meaning it shares the same axis.

For r < R:

Inside the charge distribution, we can imagine a cylindrical Gaussian surface of radius r and length L, which is entirely enclosed within the charge distribution. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀). Since the electric field is radially symmetric, it will have a constant magnitude at every point on the Gaussian surface.

The charge enclosed by the Gaussian surface is given by:

Q_enclosed = λ * L,

where λ is the charge per unit length of the distribution.

By Gauss's law, the electric flux through the Gaussian surface is:

Φ = E * A,

where E is the magnitude of the electric field and A is the area of the Gaussian surface.

Since the electric field is constant over the entire surface, we can write:

Φ = E * A = E * 2πrL,

According to Gauss's law, the electric flux is also equal to the charge enclosed divided by ε₀:

Φ = Q_enclosed / ε₀ = (λ * L) / ε₀.

Setting the two expressions for Φ equal to each other:

E * 2πrL = (λ * L) / ε₀,

Simplifying:

E = λ / (2πε₀r).

Therefore, for r < R, the electric field is given by:

E = λ / (2πε₀r).

For r > R:

Outside the charge distribution, the enclosed charge is the total charge of the distribution. However, instead of using the charge per unit length (λ), it is more convenient to use the total charge (Q) divided by the length of the distribution (L). This quantity is the linear charge density, denoted as λ_total.

The charge enclosed by the Gaussian surface is:

Q_enclosed = λ_total * L.

Applying Gauss's law again, the electric flux through the Gaussian surface is:

Φ = E * A = E * 2πrL,

Equating Φ to Q_enclosed / ε₀:

E * 2πrL = (λ_total * L) / ε₀.

Simplifying:

E = λ_total / (2πε₀r).

Therefore, for r > R, the electric field is given by:

E = λ_total / (2πε₀r).

To summarize:

For r < R: E = λ / (2πε₀r).

For r > R: E = λ_total / (2πε₀r).

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Shawna's hockey coach is committed to helping her players succeed in their upcoming game against their biggest rival.

Mass is a fundamental physical property of every object that exists in the universe. It is the measure of an object's inertia and resistance to acceleration. Mass is measured in kilograms or pounds, and is commonly used to measure the weight of an object. Mass is an intrinsic property of an object that does not depend on its location or the gravitational field in which it is situated. Mass is related to the amount of matter, or material, an object contains, and it is not affected by gravity. Mass is an invariant quantity, meaning it is unaffected by any external forces or conditions.

To this end, she emphasizes the importance of fundamental hockey skills and proper passing techniques. She encourages her players to make quick, accurate passes to teammates in order to create more scoring opportunities; the goal is to always have an open man. To achieve this, the coach instructs the players to pass the puck to a spot slightly ahead of their teammate, the teammate covered by two defenders, their teammate's back skate, or their teammate's front skate. By focusing on the basics and utilizing these strategies, the coach hopes to give her team the best chance of coming out on top.

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The net electric field at:

(a) x=-4.0cm is 2.8 x 10^5 N/C

(b) x=+4.0cm is 0 N/C

We can use Coulomb's law to find the electric field due to each charge, and then add them vectorially to get the net electric field at a given point.

(a) At x = -4.0 cm, the distance from the origin to the point is r1 = 4.0 cm, and the distance from the point charge at x = 10.0 cm is r2 = 14.0 cm. The electric field due to each charge is:

E1 = kq1/r1^[2], where q1 = 6.2 μC and k = 9.0 x 10^[9] N*m^[2]/C^[2]

E2 = kq2/r2^[2], where q2 = -9.5 μC

Substituting in the given values, we have:

E1 = (9.0 x 10^[9] Nm^[2]/C^[2]) * (6.2 x 10^[-6] C) / (0.04 m)^[2] = 4.8 x 10^[5] N/C

E2 = (9.0 x 10^[9] Nm^[2]/C^[2]) * (-9.5 x 10^[-6] C) / (0.14 m)^[2] = -2.0 x 10^[5] N/C

The net electric field at x = -4.0 cm is the vector sum of E1 and E2:

E = E1 + E2 = (4.8 x 10^[5] N/C) + (-2.0 x 10^[5] N/C) = 2.8 x 10^[5] N/C, directed towards the positive x-axis.

(b) At x = +4.0 cm, the distance from the origin to the point is r1 = 4.0 cm, and the distance from the point charge at x = 10.0 cm is r2 = 6.0 cm. Using the same formulae as before, we get:

E1 = (9.0 x 10^[9] Nm^[2]/C^[2]) * (6.2 x 10^[-6] C) / (0.04 m)^[2] = 4.8 x 10^[5] N/C

E2 = (9.0 x 10^[9] Nm^[2]/C^[2]) * (-9.5 x 10^[-6] C) / (0.06 m)^[2] = -4.8 x 10^[5] N/C

The net electric field at x = +4.0 cm is:

E = E1 + E2 = (4.8 x 10^[5] N/C) + (-4.8 x 10^[5] N/C) = 0 N/C, since the electric fields due to the two charges are equal in magnitude but opposite in direction, and they cancel out at this point.

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The term that describes the impact of thick walls and ceilings on radio wave signals is attenuation. Attenuation refers to the loss of signal strength or energy as it travels through a medium, such as a wall or ceiling.

The thicker the material, the higher the level of attenuation, resulting in weaker radio wave signals on the other side of the barrier.

Attenuation refers to the reduction in the strength or intensity of a signal as it travels through a medium or encounters obstacles. In the context of radio wave signals, thick walls and ceilings can act as barriers or obstacles that obstruct the transmission of radio waves. As radio waves encounter these physical barriers, they can be absorbed, reflected, or refracted, leading to a decrease in signal strength.

Thick walls and ceilings tend to be denser and more opaque to radio waves compared to open spaces or thinner materials. This increased density and opacity result in greater attenuation of the radio waves passing through them. The higher the attenuation, the weaker the signal becomes after passing through the barriers.

The extent of attenuation depends on factors such as the thickness and composition of the walls and ceilings, the frequency of the radio waves, and the distance between the transmitter and receiver. In some cases, the attenuation caused by thick walls and ceilings may require the use of additional signal amplification or the installation of repeaters or signal boosters to maintain an adequate signal strength.

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Answer:

transmission: the passing of a wave through an object

Explanation:

. For a diode with Jsc = 42 mA/cm² at 120 mW solar power, the estimated conversion efficiency at 300K is 19.5%.

In order to calculate the solar cell conversion efficiency, we need to know the value of Voc (open circuit voltage) and the value of Isc (short circuit current) of the diode as a solar cell.

Assuming that the diode has an ideal diode equation, we can use the following equations to calculate Voc and Isc:

Voc = (kT/q)ln(Iph/I0 + 1)
Isc = Iph

where k is the Boltzmann constant, T is the temperature in Kelvin, q is the elementary charge, Iph is the photocurrent generated by the solar radiation, and I0 is the diode saturation current.

Since we know that Jsc = 42 mA/cm² at 120 mW solar power, we can calculate the photocurrent as:

Iph = Jsc * (area of diode) = 42 * 10⁻³ * (1 cm²) = 0.042 A

Assuming an ideality factor of 1, we can estimate I0 as:

I0 = Isc / exp(qVoc / kT) = 0.042 / exp(qVoc / kT)

Solving for Voc, we get:

Voc = (kT/q)ln(Isc / I0 + 1) = (kT/q)ln(1 + Isc / 0.042)

Substituting the values, we get:

Voc = (1.38 * 10⁻²³ * 300 / 1.6 * 10⁻¹)ln(1 + 42 / 0.042) = 0.56 V

Therefore, the solar cell conversion efficiency can be calculated as:

Efficiency = (Pout / Pin) = (Isc * Voc / Pin) = (0.042 * 0.56 / 0.12) = 19.5%



The solar cell conversion efficiency of a diode used as a solar cell can be calculated using the values of open circuit voltage (Voc) and short circuit current (Isc). Assuming an ideal diode equation, we can estimate the values of Voc and Isc using the values of photocurrent generated by the solar radiation and the diode saturation current. Once we have these values, we can calculate the conversion efficiency as the ratio of output power to input power. For a diode with Jsc = 42 mA/cm² at 120 mW solar power, the conversion efficiency at 300K is estimated to be 19.5%.


In conclusion, a diode can be used as a solar cell and the solar cell conversion efficiency can be calculated using the values of Voc and Isc. Assuming an ideal diode equation and estimating the values of photocurrent and diode saturation current, we can calculate the values of Voc and Isc and then estimate the conversion efficiency as the ratio of output power to input power. For a diode with Jsc = 42 mA/cm² at 120 mW solar power, the estimated conversion efficiency at 300K is 19.5%.

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The centripetal force that must be exerted against the wings to keep the plane moving in a circle is 3.71 × 10⁴ N.

Given that,

Time = 30 min = 30 × 60 sec = 1.8 × 10³ s

Radius = 50 km = 50× 10³ m = 5× 10⁴ m

We know the expression for velocity as,

v = 2πr/(T) = (2π×5× 10⁴) /(1.8 × 10³) = (π×10⁵)/(1.8 × 10³) = (100× 3.14)/1.8 = 174.44 m/s ≈ 175 m/s

Mathematically, centripetal force can be written as,

F = mv²/r = (60,500× 175²)/(5× 10⁴) = 3.71 × 10⁴ N

Thus, required centripetal force is 3.71 × 10⁴ N.

The question is incomplete. The complete question is ' Captain Chip, the pilot of a 60,500-kg jet plane, is told that he must remain in a holding pattern over the airport until it is his turn to land. If Captain Chip flies his plane in a circle whose radius is 50.0 km once every 30.0 min, what centripetal force must the air exert against the wings to keep the plane moving in a circle?'

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a) The radial wave functions for the 3s, 3p, and 3d orbitals in a hydrogenic atom depend on the specific mathematical expressions, which are complex functions involving spherical harmonics and radial components.

These functions describe the probability density of finding an electron at different distances from the nucleus. b) The number of radial nodes in each wave function can be determined by the quantum numbers. For example: The 3s orbital has 2 radial nodes. The 3p orbital has 1 radial node. The 3d orbital has 0 radial nodes. The locations of the radial nodes in terms of the radial distance, r, can be determined by solving the respective radial wave functions. However, the exact values would depend on the specific mathematical form of the wave functions. c) The angular nodes refer to the regions where the wave function changes sign. For hydrogenic orbitals, the number of angular nodes can be determined by the azimuthal quantum number, l. For example: The 3s orbital has no angular nodes (l = 0). The 3p orbital has 1 angular node (l = 1). The 3d orbital has 2 angular nodes (l = 2). d) The orbital angular momentum of an electron in each orbital can be determined by the product of the Planck's constant (h-bar) and the square root of the azimuthal quantum number, l. For example: The 3s orbital has an orbital angular momentum of √0 = 0. The 3p orbital has an orbital angular momentum of √1 = 1. The 3d orbital has an orbital angular momentum of √2 ≈ 1.414.

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The exit temperatures of the fluids are 35.9°C and 19.3°C for the solution and water, respectively. The heat transfer rate between the fluids is 46.15 kW.

To determine the exit temperatures of the fluids, we can use the energy balance equation:

\($m_1 C_{p1} (T_{1,\text{inlet}} - T_{1,\text{outlet}}) = UA(T_{2,\text{outlet}} - T_{1,\text{outlet}})$\)

\($m_2 C_{p2} (T_{2,\text{inlet}} - T_{2,\text{outlet}}) = UA(T_{2,\text{outlet}} - T_{1,\text{outlet}})$\)

Where m1, \($C_{p1}$\), and T₁,inlet are the mass flow rate, specific heat, and inlet temperature of the solution, respectively, and m₂, \(, $C_{p2}$\), and T₂,inlet are the corresponding properties of water. T₂,outlet and T₂,outlet are the exit temperatures of the solution and water, respectively.Solving the above equations, we get T₁,outlet = 35.9°C and T₂,outlet = 19.3°C.

To evaluate the heat transfer rate between the fluids, we can use the formula:

\($Q = m_1 C_{p1} (T_{1,\text{inlet}} - T_{1,\text{outlet}}) = m_2 C_{p2} (T_{2,\text{outlet}} - T_{2,\text{inlet}})$\)= 46.15 kW.

Therefore, the heat transfer rate between the solution and water is 46.15 kW. In summary, the solution is cooled in a parallel flow heat exchanger with water, and the exit temperatures and heat transfer rate between the fluids are calculated using the given values and equations.

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The light from the bottom slit travels one wavelength (500nm) farther compared to the light from the top slit to reach the second bright fringe.

In a two-slit interference experiment, the bright fringes are formed due to constructive interference between the light waves from the two slits. The path difference between the waves from the two slits determines the position of the fringes. For the second bright fringe below the center line, the path difference is equal to one wavelength. Since the light has a wavelength of 500nm, the light from the bottom slit travels an additional 500nm compared to the light from the top slit to reach this fringe. The light from the bottom slit travels one wavelength (500nm) farther compared to the light from the top slit to reach the second bright fringe below the center line in a two-slit interference experiment using 500nm light.

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Answer:

876m/s

Explanation:

Using the formula

1/2mv² = 3/2nRT

mv² = 3nRT

m is the mass = 0.02kg

v is the average velocity of atom

n is the mole = 2.00atm

T is the temperature = 308K

R is the Boltzmann constant = 8.31 J/(mol-K).

Get the velocity

v² = 3nRT/m

Substitute the given values into the formula

v² = 3×2(8.31)(308)/0.02

v² = 3×5,118.96/0.02

v² = 15356.88/0.02

V = √767844

v = 876.27m/s

Hence the average velocity of the atom us 876m/s

Organ function is related to cell size which determines the function of tissues explains the relationship between cells, tissues and organs in the human body

Define a cell.

The smallest structural and operational unit of an organism is a cell. Although it is often tiny, macroscopic cells can also exist. Your body's basic building blocks are these cells, which are present in many different forms.

A tissue is made up of a collection of comparable cells and the connective tissue that binds them together to serve a common purpose. An organ is a collection of tissues found in an organism that cooperate to carry out a specific task. The lungs, kidneys, and stomach are a few examples. The bronchi, bronchioles, and alveoli all work together to assist the lungs exchange oxygen and carbon dioxide from the blood.

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Answer:

Car B

Explanation:

speed = distance/time

speed for car A = 20 m/s

speed for car B = 50 m/s

hence car B is faster than car A

Equilibrium temperature is the temperature of a planet in perfect thermal balance with its surroundings. The order of the planets by their equilibrium temperature from hottest to coldest is Venus, Mercury, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. Venus has the highest equilibrium temperature of 735 K, followed by Mercury with 440 K, Earth with 255 K, and Mars with 210 K. Jupiter, Saturn, Uranus, and Neptune have significantly lower equilibrium temperatures due to their distance from the sun.

the planets by their equilibrium temperature. Equilibrium temperature refers to the temperature at which a planet absorbs and emits the same amount of energy from its star, maintaining a stable temperature. Here is the order of the planets from hottest to coldest based on their equilibrium temperatures:
1. Mercury
2. Venus
3. Earth
4. Mars
5. Jupiter
6. Saturn
7. Uranus
8. Neptune
This order is determined by the planets' distance from the Sun, with the closest planet (Mercury) being the hottest and the furthest planet (Neptune) being the coldest. The closer a planet is to the Sun, the more energy it absorbs, resulting in a higher equilibrium temperature.

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Answer:

Friction is another force that energy gets "wasted" on in the dragging. Therefore, the machine gives you less mechanical advantage when too much friction is involved because it wastes energy. It also makes it less efficient!

Predation involves one individual gaining and one individual losing

In a predation, one organism, the predator, consumes all or a portion of the body of another, the prey. An example of herbivory is when a plant serves as the prey organism.

Predation lowers the number of individuals in the prey population at the level of the community. Carnivorous interactions, in which one animal consumes another, are the most well-known types of predation. Consider wolves hunting moose, owls catching mice, or shrews catching worms and other insects.

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A ball is thrown vertically upward with a velocity of 20m/s from the top of a multistory building. The height of the point where the ball is thrown is 25 m from the ground. The height will the ball rise is 20m. 5s long will it be before the ball hip the ground.

The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.

A. for solving how high will ball go , velocity at the top point will be 0 this gives ,

s = v² - u² / 2a

= 0 - 20² / -2 × 10

= 20m

B. Time to reach maximum height can be obtained from v = u + at

0 = 20 + ( − 10 ) t

t = 2s

s = ut + 0.5at²

= 20 ( 2 ) + 0.5 ( −10 ) ( 2 )²

= 20m

Therefore, the total distance for maximum height is 45 m

s = ut + 0.5at²

45 = 0 + 0.5 ( 10 ) ( t )²

t = 3s

Then,

Total time = 3+2

= 5s

Thus, A ball is thrown vertically upward with a velocity of 20m/s from the top of a multistory building.The height will the ball rise is 20m. 5s long will it be before the ball hip the ground.

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The wavelength of the 21-cm-line if it was coming from a distance of 8,000,000,000 (8*10⁻⁹) light-years would be 21 cm. The correct answer is C.

Wavelength only varies when relative velocity does. The wavelength won't alter as a result. The distance between two identical locations (adjacent crests) in successive cycles is known as the wavelength, and it is used to describe waveform signals that are transmitted over wires or into space.

Typically, in wireless systems, this length is specified in meters (m), centimeters (cm), or millimeters (mm) (mm). Inverse proportionality governs the relationship between frequency and wavelength. The wave with the highest frequency has the shortest wavelength. Twice the frequency is corresponding to a wavelength half. The longest wavelength is found in the red spectrum.

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Answer:

The silk has become negatively charged

In ancient times, Astronomy was predicting the motion of the stars, while Astrology was predicting how those stars affected us.

 If a star is moving away, its light waves get stretched out to longer, redder, wavelengths, producing a redshift. The faster the star, the greater this shift, so observers can measure the line-of-sight speed from the Doppler shift.the astronomer William Herschel (1738-1822) systematically measured the proper motion of stars in the sky. He found that stars were moving apart in one region and coming closer together in another, and concluded that the Sun was moving toward the region where stars are moving apart.

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Answer:

Polonium(Po)

Explanation:

Answer:

B. polonium (Po)

Explanation:

i took the test

Answer:

it can be stored using

I hope it helps

Answer: 22 batches.

Explanation:

Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.

How many batches of cookies can you make with one bag of flour

Let's first convert 11 kg into grams (g) by multiplying it by 1000

11 × 1000 = 11000 g

Divide 11000 by 500

11000/500 = 22

Therefore, 22 batches of cookies can be made with one bag of flour.

ANSWER

EXPLANATION

Parameters given:

Refractive index, n = 1.15

Wavelength of the light, λ = 410.74 nm

To find the minimum thickness of the non-reflective coating, apply the formula:

Therefore, the minimum thickness of the non-reflective coating is:

That is the answer.

The example of electric force in the given question is a Plastic wrap sticks to glass bowl containing leftovers. Option A is the right answer.

ELECTRIC FORCE

According to Coulomb's law, the electric force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

The electric force can be:

The electric force can be exhibited in both static and current electricity. Example of static electricity are;

The example of electric force in the given question is a Plastic wrap sticks to glass bowl containing leftovers.

The plastic wrap is able to stick to glass bowl container because of the force of attraction between the plastic wrap and the glass bowl container.

Therefore, the correct answer is the option A

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