Question 7
Identify the correct Earth System in order from #1-4.



Reorder answers
1.Atmosphere
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2.Geosphere
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3.Hydrosphere
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4.Biosphere

Answer: The symbol representing  the result of a  gas expanding against a constant pressure is W

Explanation:

According to first law of thermodynamics, energy can neither be destroyed nor created but rather transferred to one place to another and converted to and from other energy forms.

The mathematical form of first law of thermodynamics is:

\(\Delta U=q+W\)

where \(\Delta U\) = change in internal energy

q = heat absorbed or released

W = work  \((W= P\Delta V)\)

P = pressure

\(\Delta V\) = volume change

Thus the changes in heat or work leads to change in internal energy.

Answer:

θ = 16.2 rad

Explanation:

First we find the angular acceleration by using first equation of motion in angular form:

ωf = ωi + αt

where,

ωf =final angular speed = (110 rev/min)(2π rad/1 rev)(1 min/60 s) = 11.5 rad/s

ωi =initial angular speed = (45 rev/min)(2π rad/1 rev)(1 min/60 s) = 4.7 rad/s

α = angular acceleration = ?

t = time = 2 s

Therefore,

11.5 rad/s = 4.7 rad/s + α(2 s)

α = (6.8 rad/s)/(2 s)

α = 3.4 rad/s²

Now, we use 2nd equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = rotation = ?

Therefore,

θ = (4.7 rad/s)(2 s) + (1/2)(3.4 rad/s²)(2 s)²

θ = 9.4 rad + 6.8 rad

θ = 16.2 rad

The pressure at the upper level of a water flow into the house by means of pipe is 1081 kPa.

Calculate the cross-sectional area of the lower pipe:

A₁ = πr₁²

where:

A₁ = cross-sectional area of the lower pipe (m²)

π = mathematical constant (3.14)

r₁ = radius of the lower pipe (m)

A₁ = π(0.12 m)² = 0.0452 m²

Calculate the cross-sectional area of the upper pipe:

A₂ = πr₂²

where:

A₂ = cross-sectional area of the upper pipe (m²)

π = mathematical constant (3.14)

r₂ = radius of the upper pipe (m)

A₂ = π(0.06 m)² = 0.0113 m²

Calculate the flow rate per unit area:

q = Q/A

where:

q = flow rate per unit area (m³/s)

Q = flow rate (m³/s)

A = cross-sectional area (m²)

q = 6 m³/s / 0.0452 m² = 13.28 m²/s

Calculate the velocity of the water in the lower pipe:

v₁ = q/A₁

where:

v₁ = velocity of the water in the lower pipe (m/s)

q = flow rate per unit area (m³/s)

A₁ = cross-sectional area of the lower pipe (m²)

v₁ = 13.28 m²/s / 0.0452 m² = 29.3 m/s

Calculate the velocity of the water in the upper pipe:

v₂ = q/A₂

where:

v₂ = velocity of the water in the upper pipe (m/s)

q = flow rate per unit area (m³/s)

A₂ = cross-sectional area of the upper pipe (m²)

v₂ = 13.28 m²/s / 0.0113 m² = 117.0 m/s

Calculate the head loss:

hL = (v₁² - v2₂²) / 2g

where:

hL = head loss (m)

v₁ = velocity of the water in the lower pipe (m/s)

v₂ = velocity of the water in the upper pipe (m/s)

g = acceleration due to gravity (9.8 m/s²)

hL = (29.3 m/s)² - (117.0 m/s)² / 2(9.8 m/s²) = 23.2 m

Calculate the pressure at the upper level:

p₂ = p₁ + ρghL

where:

p₂ = pressure at the upper level (Pa)

p₁ = pressure at the lower level (Pa)

ρ = density of water (1000 kg/m³)

g = acceleration due to gravity (9.8 m/s²)

hL = head loss (m)

p₂ = 400 kPa + 1000 kg/m³(9.8 m/s²)(23.2 m) = 1081 kPa

Therefore, the pressure at the upper level is 1081 kPa.

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It is not recommended to fire a gun straight up into the air.

When a bullet is fired into the air, it will eventually come down and can pose a danger to people and property below. The bullet can still be lethal when it reaches the ground, especially if it lands on a hard surface or hits someone directly.

Additionally, firing a gun in a residential area can be illegal and can result in legal consequences. In general, guns should only be fired in designated shooting ranges or in self-defense situations where there is an immediate threat to life. It is important to handle firearms responsibly and follow all safety guidelines to prevent accidents and injuries.

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Answer:

D

Explanation:

The maximum height reached by a stone thrown straight up with an initial speed of 35 m/s can be found using the kinematic equation:

v^2f = v^2i - 2gh

where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (35 m/s, the magnitude of the velocity with which the stone is thrown upwards), g is the acceleration due to gravity (-9.8 m/s^2), and h is the maximum height reached by the stone.

Rearranging the equation, we get:

h = (vi^2)/(2g)

Substituting the given values, we have:

h = (35 m/s)^2 / (2 * 9.8 m/s^2)

= 62.6 m

Therefore, the maximum height reached by the stone is approximately 63 m.

The answer is (d).

Answer:

When the cold weather fills hard surfaces with water, it causes it to freeze and expand, causing cracks in the surface.

Answer:

answer is 50m/s because velocity requires a quantity and a direction (vector quantity)

There are meters created to precisely measure the fundamental values of voltage, current, and resistance for analyzing and testing circuits.

Any apparatus designed to precisely detect and present an electrical quantity in a way that can be understood by a human is referred to as a meter. The majority of the current flows via one branch, while the remainder flows through the other branch. At the point of meter insertion, an ammeter is used to measure the amount of current (in amps or milliamps) flowing in a circuit. The circuit element you are measuring is always linked in series with an ammeter. The most frequent measurements in an electrical circuit are typically made with a meter. Current is measured by an ammeter.

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Given:

The current drawn by the hair dryer is: I = 9 A

The voltage applied to the hair dryer is: V = 120 v

The time for which the charge passes is: t = 15 min = 15 × 60 s = 900 s

To find:

a) The resistance

b) Charge passing through the resistor in 15 mins

Explanation:

a)

The expression for potential difference (V) in terms of current (I) and resistance (R) is given as:

Rearranging the above equation, we get:

Substituting the values in the above equation, we get:

b)

The electric current (I) through the conductor is defined as the amount of charge (Q) passing through the conductor per unit time (t). It is given as:

Rearranging the above equation and substituting the values in it, we get:

Final answer:

a) The resistance is 13.33 Ω

b) The charge passing for 15 mins is 8100 C.

Please find attached photograph for your answer. Please do comment whether it is useful or not.

Answer:

get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it

Explanation:

When the monitoring instrument develops an inability to average, steps need to be taken including all except C: 'Consult with the anesthesiologist anesthesia changes'.

Anesthesiologists are medical doctors like family doctors and surgeons. They specialize in pain management, anesthesia care, and critical care medicine as well as have the knowledge vital to understanding and treating the human body as a whole.

So when the monitoring instrument develops an inability to average, there is no need to consult with the anesthesiologist. The required steps that need to be taken would  include:

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The mass of the lighter star is 2.15 x 10³⁰ kg.

The mass of the lighter star is obtained by applying the formula force gravitational potential energy between the two stars.

According to Newton's law of universal gravitation, every two objects in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects.

U = F x r

U = Gm₁m₂/r²   x   r

U = Gm₁m₂/r

where;

r = 3 x 10⁸ m/s  x 3.154 x 10⁷ s/yr

r = 9.46 x 10¹⁵ m

let the lighter star = m₁

then, the heavier star = 2m₁

U = (G x m₁ x 2m₁)/r

U = 2Gm₁²/r

Ur = 2Gm₁²

m₁² = Ur/2G

m₁ = √(Ur/2G)

m₁ = √(6.5 x 10³⁴ x 9.46 x 10¹⁵ / 2 x 6.626 x 10⁻¹¹)

m₁ =  2.15 x 10³⁰ kg

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Answer:

Emotional and Social Development

During the preschool years, your little one is learning to read emotional cues and is getting better at relating to other people, according to the New York University Child Study Center 1⭐

⭐This is a verified and trusted source

New York University Child Study Center: The Preschool Years: (Ages Four and Five) Expectations and Challenges

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. He can empathize with others and make friends with other children his own age. By his fifth birthday, your preschooler often seeks friends of the same gender and shows a preference for playing with other children, rather than adults. Your preschooler also engages in pretend play and can probably tell the difference between reality and fantasy, according to the American Academy of Pediatrics.

Physical Development

There's a big difference between a 2-year-old and a 5-year-old. During the preschool years, your child develops in many physical ways. She goes from toddling around unsteadily to being able to jump, stand on one foot, walk up and down stairs and walk backward, according to the New York University Child Study Center 1⭐

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New York University Child Study Center: The Preschool Years: (Ages Four and Five) Expectations and Challenges

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. By the time she's 5, your little one should also be able to ride a tricycle, build a block tower, paint on paper, skip, throw a ball, dress and undress and begin drawing recognizable objects, adds the University of Illinois Extension.

A representative element is an element whose properties can be used to discuss the chemistry of the group into which the element falls.

A representative element is an element whose properties can be used to discuss the chemistry of the group into which the element falls. Now we will try to answer each of the questions.

1) The atoms of arsenic and nitrogen have five valence electrons

2) The statement that correctly describes the two representative elements are in the same period of the periodic table is that they belong to the same period and have the same number of shells.

3) Helium is different from the other members of group 18 because it has 2 and not eight valence electrons

4) The element that has  atoms with valence electrons in a higher energy level than those of calcium (Ca) is barium

5) The statements that best describes the arrangement of electrons in fluorine is that the the electronic configuration is [He] 2s2 2p5.

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The new angular velocity becomes  0.286 rev/s.

Given parameters:

Mass of the marry go round: M = 120 kg.

Radius of the marry go round: r = 1.80 m.

Mass of the boy: m = 27 kg.

Initial angular velocity of the marry go round: ω₁ = 0.350 rev/s.

Final angular velocity of the marry go round: ω₂ = ?

From the principle of conservation of angular momentum;

Initial angular momentum = final angular momentum

⇒ 1/2 Mr²ω₁ = 1/2( M+m)r²ω₂

⇒ ω₂ = Mω₁/(m+M)

= 120×0.35/(120+27) rev/s

= 0.286 rev/s.

So, final angular velocity of the marry go round: ω₂ = 0.286 rev/s.

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5.8 x 10-8 N of force is needed to maintain the electron in its elliptical orbit.

In his description of the hydrogen atom, Bohr used the analogy of an electron orbiting a nucleus in a circle. According to his theory, the electron was confined to discrete orbits with different energies. Photons are absorbed or emitted during transitions between these legal orbits.

\(Fc = (9 x 10^-31 kg)(3.2 x 10^6 m/s)^2/(5 x 10^-11 m)\)

Simplifying this expression, we get:

\(Fc = 5.8 x 10^-8 N\)

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ThereExplanation:

We can represent the situation with the following figure

Therefore, we need to calculate the electric field generated by each charge. The electric field can be calculated as

Where k = 9 x 10^9 N m²/C², q is the charge, and r is the distance.

For the first charge, we need to replace q = 10 nC = 10 x 10^(-9) C and r = 2.205 cm = 0.02205 m, then

In the same way, we can calculate the electric field for the second charge, replacing q = 3.62nC = 3.62 x 10^(-9) C and r = 0.02205 m

Then, the electric field strength at the midpoint of the two charges is

Therefore, the answer is 2.47 x 10^(5) N/C

Answer:

the suppose the widtge frist is 6.5 inches and the lenth of your arm is 320 and 500

Answer:

a

Explanation:

its right

To increase the electromagnet's strength, the student must enlarge the nail, wrap more wire around the nail, and increase the voltage of the battery. Option c is the correct answer.

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The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J

First, we shall obtain the initial kinetic energy. Details below:

KE₁ = ½mu²

= ½ × 1500 × 55²

= 41250 J

Next, we shall final kinetic energy. Details below:

KE₂ = ½mv²

= ½ × 1500 × 65²

= 3168750 J

Finally, we shall determine the net work done. Details below:

W = KE₂ - KE₁

= 3168750 - 41250

= 3127500 J

Thus, the net work done is 3127500 J

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When a ball is dropped from the same height and at the same time as the ball that was shot horizontally, both balls will hit the ground at the same time. This is because both balls experience the same acceleration due to gravity, which is a constant 9.81 m/s² near the surface of the Earth.

The horizontal velocity of the second ball has no effect on its vertical motion. This is because the two motions (horizontal and vertical) are independent of each other, and do not interfere with each other.
The vertical motion of the second ball is determined solely by the force of gravity acting on it, which causes it to accelerate downwards at a constant rate. The time it takes for the ball to hit the ground is determined by its initial height and the acceleration due to gravity.

The first ball that is dropped vertically, travels only vertically. It doesn't have any horizontal velocity. Hence it comes down straight, whereas the other ball comes down in a parabolic path. However, both the balls will hit the ground at the same time as the horizontal motion doesn't affect the vertical motion.

Therefore, it can be concluded that if a ball is dropped from the same height and at the same time as the ball that was shot horizontally, both balls will hit the ground at the same time. This is because the time taken by both balls is the same as their motion is independent of each other.

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Based on the information, it should be noted that the resistance R is 0.5 Ω.

When S1 and S2 are open, i leads e by 30°. In this case, the circuit consists of only the inductor (L) and the capacitor (C) in series. Therefore, the impedance of the circuit can be written as:

Z = jωL - 1/(jωC)

Since i leads e by 30°, we can express the phasor relationship as:

I = k * e^(j(ωt + θ))

Z = jωL - 1/(jωC) = j(120π)L - 1/(j(120π)C)

Re(Z) = 0

By equating the real parts, we get:

0 = 0 - 1/(120πC)

Let's assume that there is a resistance (R) in series with the inductor and capacitor. The impedance equation becomes:

Z = R + jωL - 1/(jωC)

Z = R + jωL

Im(Z) = ωL > 0

Substituting the angular frequency and rearranging the inequality, we have:

120πL > 0

L > 0

This condition implies that the inductance L must be greater than zero.

When S1 and S2 are closed, i has an amplitude of 0.5 A. In this case, the impedance is:

Z = R + jωL - 1/(jωC)

Since the amplitude of i is given as 0.5 A, we can express the phasor relationship as:

I = 0.5 * e^(j(ωt + θ))

By substituting this phasor relationship into the impedance equation, we can determine the value of R. The real part of the impedance must be equal to R:

Re(Z) = R

Since the amplitude of i is 0.5 A, the real part of the impedance must be equal to 0.5 A: 0.5 = R

Therefore, the resistance R is 0.5 Ω.

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To calculate the current in a circuit, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). So, in this case, the current would be:

I = V / R

I = 9.0V / 5.00 ohms

I = 1.8 A

So, 1.8 A of current would flow in a closed circuit that has a 9.0V battery and 5.00 total resistance.

(A) The water will shrink when is heated above 4°C. (B).water at an initial temperature of 33.3°C will be expand by twice as much when it is raised by 1°C compared to water at 30°C.

The anomalous expansion of water refers to the fact that its volume increases upon cooling from 4°C to 0°C, and then contracts upon further cooling to 0°C, and continues to contract upon further cooling. Similarly, when water is heated, its volume first contracts until it reaches 4°C, and then expands upon further heating.

To determine at what temperature water shrinks when heated, we need to find the point at which the coefficient of volume expansion, β, becomes negative. The coefficient of volume expansion is defined as the fractional change in volume per degree Celsius change in temperature, i.e.,

β = (1/V)  (dV/dT)

where V is the volume of the water and dV/dT is the rate of change of volume with respect to temperature.

At temperatures below 4°C, the coefficient of volume expansion is positive, indicating that water expands upon heating. However, at temperatures above 4°C, the coefficient of volume expansion becomes negative, indicating that water contracts upon heating.

Therefore, water will shrink when heated above 4°C.

To determine the initial temperature at which water will expand by twice as much when raised by 1°C, we can use the formula for the coefficient of volume expansion:

β = (1/V)  (dV/dT)

We want to find the initial temperature T such that

(dV/dT)T = 2 (dV/dT)30

where (dV/dT)T is the rate of change of volume with respect to temperature at temperature T, and (dV/dT)30 is the rate of change of volume with respect to temperature at 30°C.

Using the coefficient of volume expansion for water, we have

β = 3α

where α is the coefficient of linear expansion, which is approximately constant for small temperature changes. Therefore, we can write

(dV/dT) = V × 3α

Substituting this into the equation above and simplifying, we get

T = 30 + 10/3 = 33.3°C

Therefore, water at an initial temperature of 33.3°C will expand by twice as much when raised by 1°C compared to water at 30°C.

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The kinetic energy affects when there is an increase in the mass and speed of the object.

When the object is in motion, Kinetic energy is observed. Kinetic energy is the product of the mass and square of the speed of an object. It depends on both the mass and speed of the object.

K.E = 1/2 (mv²), where m is the mass of the object and v is the speed of the object. Thus, the kinetic energy is directly proportional to the mass and speed of the object.

When speed increases two times, the kinetic energy increases by four times. When speed increases 3 times, the kinetic energy increases by 9. If the object has more mass, the kinetic energy also increases, and vice-versa.

Hence, the kinetic energy increases with the increase in mass and speed of the object.

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