you have a choice to drill your water well into a shallow, unconfined aquifer or a deep, confined aquifer. drilling deeper will cost more money; however, you chose to drill into the confined aquifer. why?

Answer:

Solution given:

Wight [W]=500N

gravity [g]=10m/s²

mass[m]=?

we have

W=mg

500=m*10

m=500/10

m=50kg

the mass of the truck is 50kg.

Answer:

It's "C"

Explanation:

Friction slows things down.

"Friction always slows a moving object down. The amount of friction depends on the materials from which the two surfaces are made. The rougher the surface, the more friction is produced. Friction also produces heat. "

-Google

Answer:

F = G M m / R^       force between m and earth where R >= radius of earth

a = F / m = G M / R^2        acceleration at radius R

am / ae = (Re / Rm)^2     acceleration of meteor to that of earth

am / ar = (4 / 9.8) = (Re / Rm)^2

Rm = (9.8 / 4)^1/2 Re

Rm = 1.56 Re   = 1.56 * 6.4E106 m = 10E6

10E6 - 6.4E6 = 3.6E6 m above surface of earth

(You only need mass of earth if you are calculating a at the surface of the earth)

Calculate g = G M / R^2 =  6.67E-11 * 6E24 / 6.4E6)^2  = 9.77 m/s^2

This is close to the value we used - 9.8 m/s^2

Answer:

1)  10 kg-m/s

2)  2 kg

3)  20 m/s

Explanation:

The momentum of an object can be calculated using the equation:

\(\large\boxed{p=mv}\)

where:

\(\hrulefill\)

For this question we need to find the momentum of a bowling ball whose mass is 4.0 kg is rolling at a rate of 2.5 m/s.

Given values:

Substitute the given values into the momentum formula and solve for p:

\(p=4.0\;\text{kg} \cdot 2.5\;\text{m/s}\)

\(p=10\;\text{kg m/s}\)

Therefore, the momentum of the bowling ball is 10 kg-m/s.

\(\hrulefill\)

For this question we need to find the mass of a skateboard rolling at a velocity of 3.0 m/s with a momentum of 6.0 kg-m/s.

Given values:

As we want to find mass, rearrange the momentum formula to isolate m:

\(\large\boxed{m=\dfrac{p}{v}}\)

Substitute the given values into the formula and solve for m:

\(m=\dfrac{6.0\; \text{kg m/s}}{3.0\; \text{m/s}}\)

\(m=2\;\text{kg}\)

Therefore, the mass of the skateboard is 2 kg.

\(\hrulefill\)

For this question we need to find the velocity of a baseball with a mass of 0.5 kg and a momentum of 10 kg-m/s.

Given values:

As we want to find velocity, rearrange the momentum formula to isolate v:

\(\large\boxed{v=\dfrac{p}{m}}\)

Substitute the given values into the formula and solve for v:

\(v=\dfrac{10\; \text{kg m/s}}{0.5\; \text{kg}}\)

\(v=20\;\text{m/s}\)

Therefore, the velocity of the baseball is 20 m/s.

Answer:

vjad zdmenrmdnbeanssshsnaaamrnrnealatkujejdshdendddddjnyejeynarnDhdndndddnN\nnddzNHNdnnkddkddzmdddhmeeenNDN

Answer:....

Explanation:.....

In the dry and extremely hot conditions of Southern California during summer, at approximately 1 pm, one would expect to see that the stomata of plants are closed. This is because to avoid excessive water loss through transpiration, plants tend to close their stomata, which are tiny openings on the surface of leaves responsible for gas exchange. Closing the stomata helps to conserve water by reducing the loss of water vapor from the plant's leaves.

Stomata play a crucial role in gas exchange during photosynthesis. However, when plants are exposed to hot and dry conditions, the closure of stomata becomes essential to prevent excessive water loss. By closing their stomata, plants can reduce transpiration and conserve water. Transpiration is the process through which water vapor escapes from the plant through the stomata.

Closing the stomata helps in minimizing water loss by reducing the evaporation of water from the plant's leaves into the surrounding dry air. This is particularly important in drought conditions when water availability is limited. By conserving water, plants can better withstand the dry and hot environment.

While the closure of stomata helps in water conservation, it also has an impact on photosynthesis. When the stomata are closed, the exchange of gases, including carbon dioxide (CO2) for photosynthesis, is restricted. As a result, the rate of photosynthesis decreases. However, in extremely hot and dry conditions, the priority for the plant is to conserve water rather than carry out photosynthesis. Therefore, it is expected that the stomata will be closed to limit water loss, even though this reduces the availability of CO2 for photosynthesis.

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The pattern of bright and dark fringes that emerges on a viewing screen when light passes through a single slit is known as a diffraction pattern.

When light encounters a single slit, it diffracts or spreads out due to the wave nature of light. This diffraction leads to the formation of a pattern of alternating bright and dark regions on a screen placed after the slit.

The central bright region is called the central maximum, and it is surrounded by a series of alternating bright and dark fringes, known as interference fringes.

The diffraction pattern arises due to the interference of light waves that have been diffracted by different parts of the slit. The waves emerging from different portions of the slit interfere with each other constructively or destructively, resulting in the pattern of bright and dark fringes.

The width of the slit plays a crucial role in determining the characteristics of the diffraction pattern. If the slit width is smaller compared to the wavelength of light, the diffraction pattern will exhibit a broader central maximum and narrower fringes.

Conversely, if the slit width is larger, the central maximum will be narrower, and the fringes will be wider.

The diffraction pattern produced by a single slit is an important phenomenon in physics and has applications in various fields such as optics, spectroscopy, and wave analysis.

By studying the characteristics of the diffraction pattern, scientists and researchers can gain valuable insights into the properties of light and the behavior of waves.

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Answer:

car 1

Explanation:

The final volume of the air in the compressor is approximately 7.981 cubic inches.

To determine the final volume of the air in the compressor, we can use the ideal gas law, which states that the pressure times the volume divided by the temperature is equal to a constant.

Given:

Initial pressure (P1) = 14.7 psia

Initial volume (V1) = 5 cubic inches

Initial temperature (T1) = 530 degrees Rankine

Final pressure (P2) = 100 psia

Final temperature (T2) = 640 degrees Rankine

Using the ideal gas law equation: P1 * V1 / T1 = P2 * V2 / T2

We can rearrange the equation to solve for the final volume (V2):

V2 = (P1 * V1 * T2) / (P2 * T1)

Substituting the given values into the equation:

V2 = (14.7 psia * 5 cubic inches * 640 degrees Rankine) / (100 psia * 530 degrees Rankine)

Calculating the value:

V2 ≈ 7.981 cubic inches

Therefore, the final volume of the air in the compressor is approximately 7.981 cubic inches.

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The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction. Heavier atoms are products of a fusion reaction.

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.

For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.

Hence heavier atoms are e products of a fusion reaction.

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Answer:

B C D

Explanation:

The maximum height an object will reach if it is thrown vertically upwards at 24.0 m/s will be equal to 29.35 meters.

Kinetic energy is the term used in physics to describe the force that a moving item has.

It is described as the amount of effort necessary to accelerate anybody with a particular mass from rest to a given velocity. Except for variations in speed, the body retains the kinetic energy it gains during acceleration.

As per the given information in the question,

The formula of kinetic energy is,

K.E = 1/2 mv²   (i)

The formula of gravitational potential energy is,

u = mg     (ii)

Equate equations (i) and (ii)

1/2 mv² = mgh

h = 1/2 v²/g

h = 1/2 (24)²/9.81

= 576/19.62

h = 29.35 meters.

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Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.

Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.

The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.

The phenomenon of Doppler effects is generally applicable to both sound and light.

An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.

Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

This should be the correct answer.

Answer: 25.2m/s

Explanation:

as a = vf - vi /t

vf-vi=t*a

vf-vi= 1.2*21

vf-vi= 25.2m/s

Answer:

The change in velocity is 25.2m/s.

Explanation:

Given acceleration is \(1.2m/s^{2}\),

given time is \(21 sec\)

The displacement travelled in one second is called velocity, denoted by (v).

The rate of change of velocity is called an acceleration, denoted by (a).

Formula used:  a=Δv/Δt,

putting values: 1.2=Δv/21,

so change in velocity Δv=25.2m/s.

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Answer:

The function that describes the acceleration of the particle at time t is:

a(t) = -6 m/s^2.

Explanation:

To find the function that describes the acceleration of the particle at time t, we need to take the second derivative of the position function s(t).

Given:

s(t) = -3t^2 - 4t + 1

Taking the derivative of s(t) with respect to t will give us the velocity function v(t):

v(t) = d/dt [-3t^2 - 4t + 1]

= -6t - 4

Now, we can take the derivative of v(t) with respect to t to find the acceleration function a(t):

a(t) = d/dt [-6t - 4]

= -6

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Answer:

B

Explanation:

The weight of object with a mass of 8 kg is 78.4 N.

The weight of object with a mass of 0.04 kg is 0.392 N.

The weight of object with a mass of 760 mg is 7.45 N.

The weight of an object is the force acting on the object due to gravity.

Weight, W = mg

where;

W = mg

W = 8 kg x 9.8 m/s²

W = 78.4 N

W = mg

W = 0.04 kg x 9.8 m/s²

W = 0.392 N

W = mg

W = 0.76 kg x 9.8 m/s²

W = 7.45 N

Thus, the weight of object with a mass of 8 kg is 78.4 N.

The weight of object with a mass of 0.04 kg is 0.392 N.

The weight of object with a mass of 760 mg is 7.45 N.

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Answer:

1a: Lead atoms are heavier than aluminum atoms, which means that a given mass of lead contains fewer atoms than the same mass of aluminum. Since the heat required to raise the temperature depends on the number of atoms, it takes less heat to raise the temperature of the lead than the temperature of the aluminum.

Ari's final vertical jump height is 5.2 m.

Ari's final vertical jump height is calculated from the law of conservation of energy as shown below.

P.E ( of Ari ) = P.E ( of Bari )

m1gh1 = m2gh2

where;

h1 = ( m2h2 ) / ( m1 )

h1 = ( 70 x 4 ) / ( 50 )

h1 = 5.6 m

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Answer:

3.141592653 im  king of NYC

Explanation:

(a)The pressure at B is 0.1248 atm.

(b)The temperature at C is 727.1 K.

(c)The total work done by the gas in one cycle is -1979J

General calculation:

We can use the First Law of Thermodynamics to analyze the heat engine cycle:

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 complete cycle, ΔU = 0, so:

Q = W

We can also use the ideal gas law to relate the pressure, volume, and temperature of the gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature.

(a)How to find the pressure at B segment?

To find the pressure at B, we can use the fact that the segment AB is an isothermal expansion. This means that the temperature remains constant, so:

PV = nRT

PB = (nRT)/(2V) = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/(2V) = (0.0821 L·atm/mol)(600 K)/V

Since the pressure at A is 5.00 atm, we can use the fact that the temperature is constant to find the volume at A:

PV = nRT

VA = (nRT)/P = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/5.00 atm = 197.76 L

Since the volume at B is twice the volume at A, we have:

VB = 2VA = 395.52 L

Substituting into the expression for PB, we get:

PB = (0.0821 L·atm/mol)(600 K)/395.52 L = 0.1248 atm

Therefore, the pressure at B is 0.1248 atm.

To find the temperature at C, we can use the fact that the segment BC is an adiabatic expansion. This means that no heat is added or removed from the system, so:

\(PV^\gamma\)= constant

where γ is the ratio of specific heats (for a monoatomic gas, γ = 5/3). We can use the fact that the volume at C is equal to the volume at A to find the pressure at C:

\(PAV^\gamma = PCV^\gamma\)

PC =  \(PA(V/A)^\gamma\) = 5.00 atm\((1/2)^(^5^/^3^)\) = 1.556 atm

Since the segment BC is adiabatic, the temperature changes but no heat is added or removed from the system. Using the ideal gas law, we can relate the pressure, volume, and temperature:

PV = nRT

TC = (PCVC)/(nR) = (1.556 atm)(197.76 L)/(2.00 mol)(0.0821 L·atm/mol·K) = 727.1 K

Therefore, the temperature at C is 727.1 K.

The total work done by the gas in one cycle is the sum of the work done in each segment of the cycle:

W = WAB + WBC + WCD + WDA

For segment AB, the work done is:

WAB = -QAB = -∫PdV = -nRT∫(1/V)dV = -nRT ln(VB/VA) = -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2) = -602 J

For segment BC, the work done is:

WBC = -QBC = -∫PdV = -nγRT∫(1/V)dV = -nγRT

We know that VB = 2VA and VC = 2VD, so we can express the ratio VB/VC in terms of VA/VD:

VB/VC = (2VA)/(2VD) = VA/VD

Substituting into the expression for WBC, we get:

WBC = -nγRT ln(VA/VD)

For segment CD, the work done is:

WCD = -QCD + PCDΔV = -nCpΔT + PCDΔV

where Cp is the specific heat at constant pressure, ΔT is the change in temperature, and ΔV is the change in volume. We know that the segment CD is isobaric, so ΔV = VB - VA = (2VA) - VA = VA. We can also use the ideal gas law to relate the pressure, volume, and temperature:

Substituting into the expression for WCD, we get:

WCD = -nCpΔT + (nRT/VD)VA = -nCp(TC - TD) + (nRT/VD)VA

For segment DA, the work done is:

WDA = -QDA + ΔU = -nCvΔT

where Cv is the specific heat at constant volume. We know that the segment DA is isovolumetric, so ΔV = 0. Using the First Law of Thermodynamics, we know that ΔU = 0 for a complete cycle, so:

QDA = -WDA = nCvΔT

Substituting into the expression for WDA, we get:

WDA = -nCvΔT

Adding up the work done in each segment, we get:

W = WAB + WBC + WCD + WDA

= -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(Cp)(TC - TD) + (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(Cv)(TC - TA)

We know that Cp and Cv for a monoatomic gas are related by Cp = Cv + R, so we can express Cp in terms of Cv:

Cp = Cv + R = (3/2)R + R = (5/2)R

Substituting and simplifying, we get:

W = (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(727.1 K)+ (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(600 K)

W = -966.2 J - 4957 J - 7476 J + 5154 J

    = -1979 J

Therefore, the total work done by the gas in one cycle is -1979 J

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6.102×10¹⁶ Joules of energy of other forms are produced when a mass of 0.678 kg is converted completely into the energy of other forms.

Mass-energy equivalence or E = mc² equation is given by Albert Einstein's theory of special relativity which expresses that mass and energy are the same physical entity and they can be changed into each other. Mass-energy equivalence implies that the total mass of a system may change but the total energy and momentum remain constant.

According to Einstein’s equation,

E=mc²

Where :

E = energy

m = mass =  0.678 kg

c = the speed of light in a vacuum  = 3 × 10⁸ m/s

Substuting the m and c values in the formula we get:

E = mc²

E = 0.678 kg × ( 3 × 10⁸ m/s)²

= 6.102×10¹⁶ Joules of energy.

Therefore, the energy produced is 6.102×10¹⁶ Joules of energy.

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Answer:

mparing this value with the weight of the vehicle we see    W₂> 1.6 10⁴ N

therefore the woman can lift the car

Explanation:

For this exercise on Pascal's principle, we use that the pressure in every incompressible liquid is the same

         P₁ = P₂

   the press is defined by

         P = F / A

for the woman's side

  the area of ​​a circle is

         A = π D₁² / 4

         P₁ = W (4 /π d₁²)

for the car side

         A₂ =π d₂²

        P₂ = W₂ (4 /π d₂²)

we substitute in the first equation

         W (4 /π d₁²) = W₂ 4 /π d₂²)

          W / d₁² = W₂ / D₂²

           W₂ = (d₂ / d₁)² W

the weight of the woman is

           W = mg

we calculate

          W₂ = (24/3)₂ 50 9.8

          W₂ = 3,136 10⁴ N

when comparing this value with the weight of the vehicle we see

          W₂> 1.6 10⁴ N

therefore the woman can lift the car

The duck can always reach the shore without being eaten by the fox.
Assume that the duck is swimming in a circular path around the centre of the lake and that the fox is waiting at a fixed point on the shore. Since the fox is four times faster than the duck, it can run along the shore at a constant speed that is also four times the speed of the duck's swimming.

Now, imagine that the duck swims around the lake one full time, starting at the point farthest from the shore where the fox is waiting. The duck will take some amount of time to complete this circuit, during which the fox will also have travelled some distance around the lake.

However, since the duck is swimming in a circular path, it will eventually cross the point on the opposite side of the lake from where it started. At this point, the duck is closer to the shore than the fox is. Furthermore, since the duck can now fly, it can reach the shore before the fox catches up to it.

So, it will eventually reach a point where it is closer to the shore than the fox and can fly the rest of the way to safety. Therefore, the duck can always reach the shore without being eaten by the fox.

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Answer:

ok

Explanation:

Have a nice day!

The mass of the parachutist doesn't decrease to 0 kg. It is still 70 kilos and always will be. The parachutist's mass is constant at 70 kg regardless of where they are on Earth, in free fall, or in outer space. Mass remains constant.

You might be considering how the parachutist loses all sense of weight while falling. That's accurate. When they fall freely to the Earth, their weight will be nil (or any other body).

The problem is that mass and weight are two different concepts. Your scales here on Earth confuse you by displaying kilograms (kg). They are actually gauging your weight, which is expressed in Newtons and is, in essence, the downward force your mass experiences as a result of the Earth's gravity.

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i. When the regular 3 kg lump of material is fully submerged in a fluid, and the fluid that previously occupied the lump's space has a mass of 3 kg, the lump will remain in place when released.

ii. If the lump is subsequently fully submerged in a more dense fluid and released, it will move upward.

i. When the lump is submerged in the fluid, the buoyant force acting on the lump is equal to the weight of the fluid displaced by the lump. In this case, since the mass of the fluid displaced (3 kg) is equal to the mass of the lump itself (3 kg), the buoyant force will be equal to the weight of the lump. As a result, when released, the lump will experience a net force of zero and will remain in place.

ii. If the lump is then fully submerged in a more dense fluid, the fluid that displaces the lump will have a greater mass than the lump. As a result, the buoyant force will exceed the weight of the lump. When released, the lump will experience a net upward force (the difference between the buoyant force and the weight of the lump) and will move upward.

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Answer:

To find the work done in moving the particle from the origin to x = 2, we need to integrate the force over the given interval.

The work done (W) is calculated by integrating the force function with respect to displacement (dx) from the initial position (0) to the final position (2):

W = ∫(0 to 2) (10x³ - 5) dx

Integrating the force function, we get:

W = ∫(0 to 2) (10x³ - 5) dx = [2.5x⁴ - 5x] evaluated from 0 to 2

Now, substituting the upper limit (2) and lower limit (0) into the equation:

W = [2.5(2)⁴ - 5(2)] - [2.5(0)⁴ - 5(0)]

 = [2.5(16) - 10] - [0 - 0]

 = 40 - 10

 = 30

Therefore, the work done in moving the particle from the origin to x = 2 is 30 units of work.

Explanation:

The cost of boiling 500cm3 of water using the 3kW kettle is 1.35 P.

The cost is calculated as follows;

1 unit = 9p /kWh

Total energy consumed by 3 kW kettle, E = P x t

where;

E = 3 kW x (3 mins/60 mins/hr)

E = 0.15 kWh

Energy cost = 9 p/kWh x 0.15 kWh = 1.35 P

Thus, the cost of boiling 500cm3 of water using the 3kW kettle is 1.35 P.

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