Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount x, a force along the x-axis with x-component Fx=kx−bx2+cx3 must be applied to the free end. Here k=100N/m, b=700N/m2, and c=12000N/m3. Note that x>0 when the spring is stretched and x<0 when it is compressed.

An external agent must do 0.04 J of work. The amount of work done depends on the spring constant and the displacement.

The formula for the work done by an external agent to stretch a spring is:

W = (1/2)kx^2. Where W is the work done, k is the spring constant, and x is the displacement. Assuming the spring is ideal and obeys Hooke's law, the spring constant remains constant as it is stretched. Let's assume the spring constant is 50 N/m, and the displacement is 4.00 cm = 0.04 m.

W = (1/2)(50 N/m)(0.04 m)^2 = 0.04 J

Therefore, an external agent must do 0.04 J of work to stretch the same spring 4.00 cm from its unstretched position.

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A. The average drift speed of the electrons in the wire is  1.19 x \(10^{-3}\)cm/s.

B. The time it would take an electron to travel 8.84 m is 7.43 x \(10^{5}\)sec

(A) To find the average drift speed of electrons in a wire, we can use the equation:

I = nAvq

Where:
I is the current in Amperes
n is the electron density in electrons per \(cm^{3}\)
A is the cross-sectional area of the wire in \(cm^{2}\)
v is the average drift velocity of electrons in cm/s
q is the charge of an electron, which is 1.6 x \(10^{-19}\) Coulombs

First, we need to find the cross-sectional area of the wire. The formula for the area of a circle is:

A = π\(r^{2}\)

Where:
A is the area of the circle
r is the radius of the circle

Given that the wire diameter is 1.63 mm, we can find the radius by dividing it by 2:

r = 1.63 mm / 2 = 0.815 mm = 0.0815 cm

Now, we can calculate the area:

A = \(π(0.0815 cm)^{2}\) = 0.0209 \(cm^{2}\)

Next, we can rearrange the equation to solve for v:

v = I / (nAq)

Given that the current is 20 A and the electron density is 8.47 x \(10^{24}\)electrons per \(cm^{3}\), we can substitute these values into the equation:

v = 20 A / (8.47 x \(10^{24}\) electrons per cm^3 * 0.0209 cm^2 * 1.6 x \(10^{-19}\) C)

Simplifying the expression:

v = 1.19 x \(10^{-3}\) cm/s

Therefore, the average drift speed of electrons in the 14 gauge wire carrying a maximum current of 20 A is approximately 1.19 x \(10^{-3}\) cm/s.

(B) To calculate the time it would take for an electron to travel a distance of 8.84 m, we can use the formula:

t = d / v

Where:
t is the time in seconds
d is the distance in meters
v is the average drift velocity of electrons in meters per second

Given that the distance is 8.84 m and the average drift velocity is 1.19 x 10^-3 cm/s, we need to convert the velocity to meters per second:

v = 1.19 x \(10^{-3}\) cm/s * 0.01 m/cm = 1.19 x \(10^{-5}\) m/s

Now, we can substitute the values into the formula:

t = 8.84 m / 1.19 x \(10^{-5}\)m/s

Simplifying the expression:

t = 7.43 x \(10^{5}\)s

Therefore, it would take approximately 7.43 x \(10^{5}\) seconds for an electron to travel a distance of 8.84 m.

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The intensity of sound is inversely proportional to the square of the distance from the source. The intensity of sound at 4 m away from the source is approximately \(0.993W/m^{2}\) (watts per square meter).

The intensity of sound is defined as the power per unit area. According to the inverse square law for sound propagation, the intensity of sound is inversely proportional to the square of the distance from the source. Mathematically, it can be expressed as

\(I=P/(4\pi r^{2})\)

where I is the intensity, P is the power, and r is the distance from the source.

Given that the power of sound at the source outlet is 100 W and the distance from the source is 4 m, we can substitute these values into the formula. Thus, we have

\(I = 100 W / (4\pi (4 m)^2)\)

Simplifying this expression, we find that the intensity of sound at 4 m away from the source is approximately \(0.993W/m^{2}\) (watts per square meter).

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Answer:

The mass of the apple is 0.49kg

Explanation:

Potential energy=mgh

P=mgh

6=m×1.22×10

6=12.2m

divide both sides by 12.2

m=6/12.2

m=0.49kg

i. The elasticity at point \( g \) is the measure of responsiveness or sensitivity of a quantity to changes in another variable at point \( g \).

ii. To determine the elasticity at point \( h \), we first need to establish what quantity we are referring to at point \( h \). Once we have identified the relevant quantity, we can then calculate its elasticity by measuring the responsiveness or sensitivity to changes in another variable at point \( h \).

i. To determine the elasticity at point \( g \), we need specific information about the variables and their relationship. Elasticity is typically calculated as the percentage change in one variable divided by the percentage change in another variable.

ii. Without additional information about the specific variables and their relationship at point \( h \), it is difficult to provide a precise answer. The concept of elasticity requires specific context and variables to be defined in order to calculate or describe it accurately.

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Field strength is 0.03234 N/kg. The formula to determine the field strength is given by:

F = mg Here, F is the field strength, m is the mass, and g is the gravitational field strength.

Substituting the values given: Weight = 0.96 N Mass = 3.3 g = 0.0033 kg = 9.8 m/s² Therefore, F = mg = 0.0033 kg × 9.8 m/s² = 0.03234 N the field strength is the gravitational force acting on a unit mass. It is measured in newtons per kilogram. The field strength is an expression of the strength of a gravitational field. In this case, the mass of the object is 3.3 g, which can be converted to kilograms by dividing by 1000.

The weight of the object is given as 0.96 N. Using the formula

F=mg, where m is the mass and g is the gravitational field strength, we can calculate the field strength as 0.03234 N/kg.

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Answer:

2 m long, how wide is it? SOLUTION. Given: F = 60.0 N. W = 2.05 × 105. J x = 9.60 × 102 m. Unknown: d = ? ... moved? 3. A hummingbird has a mass of about 1.7 g. Suppose a hummingbird does 0.15 J of work against gravity, so that it ascends straight up with a net acceleration of 1.2 m/s. 2 . How far up does it move? 4.

Explanation: hope this helps plz help me with my question

The acceleration of the wagon along the ground is 3.6 m/s².

To solve the problem above, we need to use the formula of acceleration as related to force and mass.

Acceleration: This can be defined as the rate of change of velocity.

⇒ Formula:

⇒ Where:

⇒ make a the subject of equation 1

From the question,

⇒ Given:

⇒ Substitute these values into equation 2

Hence, The acceleration of the wagon along the ground is 3.6 m/s²

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Answer: 3.60 m/s^2

Explanation:

QUESTION-

What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 N at an angle of 35° above the horizontal?

ANSWER

3.60 m/s^2 = 3.6 m/s^2

Answer:

b

Explanation:

there is no net force on the basketball

The persοn is mοving at a speed οf 55 miles per hοur in the Sοuth directiοn when the persοn is 50 miles West and 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur.

Tο determine the speed at which the persοn is mοving, we can use the cοncept οf relative velοcity.

Let's cοnsider the hοrizοntal and vertical cοmpοnents separately:

Hοrizοntal Cοmpοnent:

The persοn is walking due East, which is perpendicular tο the Nοrth directiοn οf the car. Therefοre, the hοrizοntal cοmpοnent οf the persοn's velοcity dοes nοt affect the speed at which the persοn is mοving away frοm the car.

Vertical Cοmpοnent:

The persοn is 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur. This indicates that the persοn's vertical pοsitiοn is changing with time. Since the persοn is mοving in the Sοuth directiοn and the distance is increasing, the persοn's speed can be determined by the rate οf change οf the vertical distance.

Given that the distance is increasing at a rate οf 55 miles per hοur, the persοn's speed in the Sοuth directiοn is 55 miles per hοur.

Therefοre, the persοn is mοving at a speed οf 55 miles per hοur in the Sοuth directiοn when the persοn is 50 miles West and 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur.

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Answer:

C)  Total displacement = 1275 m or 1.275 km

Explanation:

We'll break this into two parts:  

1) Displacement during constant speed, and

2) Displacement during the acceleration phase.

Constant Speed

The displacement for this segment is already stated:  1 km.  But the information includes the time, which will allow us to calculate the speed of travel.  We'll need this for the next phase of acceleration.

Distance = Speed x Time

Speed = Distance/Time

Time = 1 minute + 20 sec, or 80 sec.

Speed = 1 km/80 sec

Speed = 0.0125 km/sec

Since we'll be using meters (not km) in the next step, convert 0.0125 km/sec to m/sec with a conversion factor:  

  (0.0125 km/sec)*(1000m/1 km) = 12.5 m/sec

Acceleration Phase

The obect is moving at a speed of 12.5 m/sec (from above) and then accelerates at a rate of 0.125 m/sec^2 for 20 seconds.

The distance (s) an object travels during this acceleration (a) is gven by:

  s = vi*t + (1/2)a*t^2,

where vi is the initial velocity (12.5 m/sec in this case), a is the acceleration (0.125 m/sec^2), and t is the time (20 seconds).

s = (12.5 m/sec)(20 sec) + (1/2)(0.125 m/sec^2)*(20 sec)^2

s = 275 meters

Now add the two distances:

1) Constant Speed Phase = 1000 m

2)  Acceleration Phase    =  275 m

Total displacement = 1275 m or 1.275 km

Answer:

B

Explanation:

Answer:

B. scientific laws have been proven without exception under certain stated conditions.

Explanation:

No, magnetic fields do not exist on every planet. The magnetic fields of the four gas giants are extremely strong, while those of the Earth and Mercury are fairly strong and Venus and Mars have essentially no detectable magnetic fields.

Mars and Venus don't have much of a magnetic field, according to probes. The magnetic fields of Jupiter, Saturn, Uranus, and Neptune are all far stronger than those of the Earth. Jupiter is the winner since it has the biggest magnetic field. The magnetosphere grows in size as the magnetic field gets stronger. Jupiter's magnetic field, which is around 20,000 times stronger than Earth's magnetic field, generates a magnetosphere that extends over 3 million kilometers from Jupiter before it starts to deflect solar wind.

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The air you feel next to Earth's surface in the shade is heated by two main factors: indirect solar radiation and redirected solar radiation.

Indirect solar radiation refers to the process by which sunlight is absorbed by the Earth's surface and then re-emitted as heat. When the sun's rays reach the Earth, some of the energy is absorbed by buildings, pavement, and other objects. As these objects heat up, they release the absorbed energy as heat, which warms the surrounding air. This is why the air feels warm when you walk around the ASU Campus in the shade.

Redirected solar radiation also plays a role in heating the air near the Earth's surface. This occurs when sunlight is scattered or reflected by the atmosphere, clouds, or nearby objects, and then reaches the shaded areas. The redirected solar radiation contributes to the overall heating of the air, making it feel warm.

In conclusion, the air you feel next to Earth's surface in the shade is heated by indirect solar radiation, as the absorbed energy from the sun is released as heat, and redirected solar radiation, as sunlight is scattered or reflected and contributes to the warming of the air.

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Answer: (a) The speed is zero.

(b) The speed of the electrons ejected by light of wavelength 350 nm is 6.0 × 10⁵ m/s.

a) Let's find out the energy of the incident light by using E = hc/λ

where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the light. Here we have λ = 740 nm = 7.4 × 10⁻⁷m

So, E = hc/λ

= (6.63 × 10⁻³⁴ J.s × 3 × 10⁸ m/s)/(7.4 × 10⁻⁷m)

= 2.69 × 10⁻¹⁹ J.

Since the work function is given in electron volts, we need to convert the energy into eV.1 eV = 1.6 × 10⁻¹⁹ J2.69 × 10⁻¹⁹ J = (2.69 × 10⁻¹⁹ J / 1.6 × 10⁻¹⁹ J/eV) = 1.68 eVThe kinetic energy of the ejected electron can be calculated as the difference between the energy of the incident light and the work function.KE = E - Φ = 1.68 eV - 2.14 eV = -0.46 eV. Since the electron has a negative kinetic energy, it is not ejected. Therefore, the speed is zero.

b) To find the speed, we can use the formula: KE = 1/2 mv²

v = √(2KE/m)

The mass of the electron is 9.11 × 10⁻³¹ kg. So,

v = √(2 × 1.41 eV × 1.6 × 10⁻¹⁹ J/eV / 9.11 × 10⁻³¹ kg)= 6.0 × 10⁵ m/s.

Therefore, the speed of the electrons ejected by light of wavelength 350 nm is 6.0 × 10⁵ m/s.

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The volume flow rate of the petroleum along the pipe is 0.267 m³/s.

The pipe's diameter at the refinery is 0.96 m.

The volume flow rate of the petroleum along the pipe is calculated by applying the following flow rate equation.

Q = AV

where;

A = πd²/4

where;

A = π(0.583)²/4

A = 0.267 m²

Q = 0.267 m³/s

The diameter of the pipe is calculated as follows;

Q = AV

Where;

A = Q/V

A = (2.91) / (5.05)

A = 0.576 m²

A = πd²/4

d² = 4A/π

d = √(4A/π)

d = √(4 x 0.576 / π)

d = 0.96 m

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The that is happening inside a star while it expands into a subgiant  is that it is fusing hydrogen into helium in a shell outside the core.

The expansion of the outer layers of the star into the subgiant size  can be seen as one that is balance and bring aboput  the increase of energy generated by the hydrogen shell fusion wehich do make the star to nearly maintain its surface temperature.

It shopuld be noted that it causes the spectral class of the star to change very little  and this can be felt from the lower end of this range of star mass.

In conclusion, the Fusion as described can be seen as one that do occurs when two atoms slam together so the formation of the heavier atom,  can be acheied as seen in the way the hydrogen fuss into helium in a shell outside

Therefore, the second option is correct.

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Answer:  10 minutes

Explanation:

30 gal / 2 min = 15 gal/min  (this is the fill-rate)

x = # minutes to fill a 150 gal pool at a rate of 15 ga/min

(15 gal/min)(x min) = 150 gal

x min = (150 gal) / (15 gal/min) = 10 min

Answer:

1 N/C

Explanation:

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Using the concepts of torque, we got that angular speed increases when we apply an upward torque to the disk that is spinning counterclockwise.

We know very well that Torque=Power × Angular velocity

means that torque is directly proportional to angular velocity. Here Power is defined as the amount needed to rotate the disk.

So, when we apply external torque on the disk, it will rotate more fastly, due to that angular velocity of the disk also increases,

Hence, if I am looking down on a disk that is spinning counterclockwise. and if I apply an upward torque to the disk, the angular speed will increase.

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The term Young Modulus is used to describe the stress to strain of a material.  and is a measure of the strength of a material.

The term Young Modulus is used to describe the strength of a material. It is a ratio of the stress to strain of a material. The unit of Young Modulus is N/m2.

The question is incomplete hence the average Young's modulus, in units N/m2 can  not be computed. This value helps an engineer to make decision regarding the use of materials for construction.

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Hi! I believe your answer is 6.3 seconds because it takes 0.45 seconds for an object to fall 1 meter, so just multiply 0.45 by 14 and you'll get 6.3 seconds. I hope this helps you! Good luck and have a great day/night. ❤️✨

Answer:

0,146 L

Explanation:

Según la ley de Charles; el volumen de una determinada masa de gas es directamente proporcional a su temperatura a presión constante.

A partir de los datos proporcionados;

V1 = 0,2 L

T1 = 100 ° C + 273 = 373 K

V2 =

T2 = 0 ° C + 273 = 273 K

V1 / T1 = V2 / T2

V1T2 = V2T1

V2 = V1T2 / T1

V2 = 0,2 × 273/373

V2 = 0,146 L

Answer:

40

Explanation:

vi=0

a=2

vf=?

d=400

vf^2=vi^2+2ad

vf^2=0+2×2×400=1600

vf=√1600=40m/s

Answer:

\(559\).

Explanation:

Integrate \(a(t)\) with respect to time \(t\) to find an expression for velocity:

\(\begin{aligned} v(t) &= \int a(t)\, d t \\ &= \int (6\, t + 8)\, d t && (\text{power rule}) \\ &= 3\, t^{2} + 8\, t + C_{v} \end{aligned}\).

Note that since this integral is indefinite, the expression for \(v(t)\) includes a constant \(C_{v}\).

Find the value of \(C_{v}\) using the fact that \(v(0) = 2\). Specifically, substitute \(t = 0\) into the expression \(v(t) = 3\, t^{2} + 8\, t + C_{v}\) and solve for \(C_{v}\!\):

\(v(0) = 3\, (0)^{2} + 8\, (0) + C_{v} = C_{v}\).

\(v(0) = 2\).

\(C_{v} = 2\).

In other words, \(v(t) = 3\, t^{2} + 8\, t + 2\).

Similarly, integrate \(v(t)\) with respect to \(t\) to find an expression for position:

\(\begin{aligned} s(t) &= \int v(t)\, d t \\ &= \int (3\, t^{2} + 8\, t + 2)\, d t\\ &= t^{3} + 4\, t^{2} + 2\, t + C_{s} \end{aligned}\).

Similarly, find the value of constant \(C_{s}\) using the fact that \(s(0) = 6\):

\(s(0) = (0)^{3} + 4\, (0)^{2} + 2\, (0) + C_{s} = C_{s}\).

\(s(0) = 6\).

\(C_{s} = 6\).

In other words, \(s(t) = t^{3} + 4\, t^{2} + 2\, t + 6\). Substitute in \(t = 7\) and evaluate to find the position of the particle at that moment:

\(s(7) = 7^{3} + 4\, (7)^{2} + 2\, (7) + 6 = 559\).

The pοsitiοn of the particle at time t = 7 is 559 units.

Tο find the pοsitiοn at time t = 7, we need tο integrate the given acceleratiοn functiοn tο οbtain the velοcity functiοn and then integrate the velοcity functiοn tο οbtain the pοsitiοn functiοn.

Given:

Acceleratiοn functiοn: a(t) = 6t + 8

Initial pοsitiοn: s(0) = 6

Initial velοcity: v(0) = 2

First, let's integrate the acceleratiοn functiοn tο οbtain the velοcity functiοn:

v(t) = ∫(a(t)) dt

= ∫(6t + 8) dt

= 3t^2 + 8t + C

Tο find the cοnstant οf integratiοn (C), we can use the initial velοcity v(0) = 2:

2 = 3(0)² + 8(0) + C

C = 2

Sο, the velοcity functiοn becοmes:

v(t) = 3t² + 8t + 2

Next, let's integrate the velοcity functiοn tο οbtain the pοsitiοn functiοn:

s(t) = ∫(v(t)) dt

= ∫(3t² + 8t + 2) dt

= t³ + 4t² + 2t + C'

Tο find the cοnstant οf integratiοn (C'), we can use the initial pοsitiοn s(0) = 6:

6 = (0)³ + 4(0)² + 2(0) + C'

C' = 6

Sο, the pοsitiοn functiοn becοmes:

s(t) = t³ + 4t² + 2t + 6

Finally, we can find the pοsitiοn at time t = 7:

s(7) = (7)³+ 4(7)² + 2(7) + 6

= 343 + 196 + 14 + 6

= 559

Therefοre, the pοsitiοn at time t = 7 is 559 units.

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The voltage across the capacitor is VC = 0 - j0.003654V.

The circuit given on the right is a series circuit comprising a resistor R, an inductor L, and a capacitor C. The applied voltage is given as δV = δVmax sin(ωt) where δVmax = 51V and ω = 15rad/s. The values of L, R and C are 0.6H, 25Ω and 0.011F respectively.

The complex impedance of the circuit is given as follows;

Z = R + jωL - j/ωC = 25 + j(15)(0.6) - j(1/15)(0.011)

= 25 + j9 - j0.000733=

25 + j8.99927Ω

The amplitude of the current in the circuit is given as;

I = V/Z where V is the amplitude of the applied voltage.

Substituting the values of V and Z;

I = δVmax/Z= 51/(25 + j8.99927)

The current in the circuit is thus;

I = 1.661 - j1.8355A.

The voltage drop across each component can be obtained by multiplying the amplitude of the current by the impedance of each component. The voltage across the resistor is given as;

VR = IR = (1.661)(25) = 41.525V

The voltage across the inductor is given as;

VL = jωLI = j(15)(0.6)(1.661) = j14.98V

The voltage across the capacitor is given as;

VC = -j(1/ωC)I = -j(1/15)(0.011)(1.661)

= -j0.003654V

The voltage across the capacitor is thus;

VC = 0 - j0.003654V.

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As the pressure on the gas increases, the volume will decrease and the temperature will increase. Thus option c is correct.

A gas that obeys the gas laws regardless of the pressure and temperature is known as an ideal gas. The particles of an ideal gas have no intermolecular forces and have negligible volume.

A model of an ideal gas is shown here. A student designs an experiment to test the impact of increasing the pressure on the system. Which prediction is most likely to be supported by the data?

A) As pressure on the gas increases, the volume and temperature will both decrease.

B) As the pressure on the gas increases, the volume and temperature will also increase.

C) As pressure on the gas increases, the volume will decrease and the temperature will increase.

D) As pressure on the gas increases, the volume will increase and the temperature will decrease.

Here,

The variation of volume and temperature according to the variation in pressure can be obtained from the ideal gas equation,

            PV = nRT

From the equation we can write that,

P ∝ 1/V

P ∝ T

Therefore, the increase in pressure can cause the decrease in volume and increase in temperature.

Hence the correct option is

(c) As the pressure on the gas increases, the volume will decrease and the temperature will increase.

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