the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of sparkling cider. how much did this wedding ring cost

Answer:

h= 32059.37 m

Explanation:

Assuming the missing in formation as

A certain lead-acid storage battery has a mass of 33 kg . Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.

Now, Applying energy conservation ( Electrical to potential)

Electrical Energy E= I×V×t

I = correct , V= voltage , t= time of flow of current

E = 6×24×20×60×60.

E= 10368 KJ

Now this energy is used to lift the battery with 100% efficiency

Hence,

electrical energy E= potential energy P

P= mgh

m=mass of the battery , g= the acceleration due to gravity is 9.8 m/s^2

h= height

mgh = 10368 kJ

33×9.8×h= 10368×1000

h = 10368×1000/(33×9.8)

h= 32059.37 m

Answer:

The average diameter of a single alveolus is 0.0222 cm.

Explanation:

Volume of the lung ,V= 1.9 L

[tex]1 L = 1000 cm^3[/tex]

[tex]1.9 L=1.9\times 1000 cm^3=1900 cm^3[/tex]

Number of alveoli in a human lung = [tex]330\times 10^6[/tex]

Volume of single alveoli =v

[tex]v\times 330\times 10^6=V[/tex]

[tex]v=\frac{1900 cm^3}{330\times 10^6}[/tex]

[tex]v=5.7575\times 10^{-6} cm^3[/tex]

The alveoli are spherical.

Radius of an alveolus = r

Volume of the sphere = [tex]\frac{4}{3}\pi r^3[/tex]

[tex]v=\frac{4}{3}\pi r^3[/tex]

[tex]5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3[/tex]

[tex]r=0.0111 cm[/tex]

Diameter of the alveolus =d

d = 2r = 2 × 0.0111 cm = 0.0222 cm

The average diameter of a single alveolus is 0.0222 cm.

The question seeks to find the average diameter of a single alveolus in the human lung, using information about the total volume of the lung and the number of alveoli. By using the formula for the volume of a sphere, one can calculate the volume of a single alveolus and derive its radius and thereby its diameter.

The problem is essentially asking to find the average diameter of a single alveolus, given we know the total volume of the lung and the number of alveoli. Using the formula for the volume of a sphere, V=4/3πr³, where V is the volume and r is the radius, we can find the volume of a single alveolus by dividing the total volume of the lungs (1.9 liters, which equals to 1.9 x 10^9 cubic millimeters) by the number of alveoli (approximately 330 million). We can then calculate the radius by rearranging the sphere volume formula to r = ((3*V)/4π)^1/3. The diameter would be double the radius.

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Final answer:

Statement (c) is incorrect because if one object is negatively charged and the other is positively charged, they would attract each other, resulting in a negative potential energy as they come closer, not a positive one.

Explanation:

The question is addressing the concept of electric potential energy between two charged objects. When the electric potential energy of the system is positive as the two objects come close, we can infer that the objects have like charges, either both positive or both negative. This is due to the fact that work needs to be done against the electrical repulsion to bring like charges together, increasing the system's potential energy.

This makes statements (a) Both objects are positively charged and (b) Both objects are negatively charged possibly correct scenarios, as they would lead to a positive potential energy when the objects are brought together. Statement (c) One object is negatively charged and the other one is positively charged would be incorrect in this context, because a positive and a negative charge would attract each other, and the system would do work on the surroundings as they come closer to each other making the potential energy negative. Therefore, the incorrect statement, given that the electric potential energy is positive when they are near each other, is (c).

The potential difference between the terminals of an ordinary AA or AAA battery is usually 1.5 volts. Voltage is the work required to move a charge, while current is the charge flow rate. Batteries can provide considerable energy for their size, analogous to lifting a significant mass against gravity.

The potential difference between the terminals of an ordinary AA or AAA battery is the electromotive force (emf) when the battery is not part of a complete circuit, and is typically about 1.5 volts (V). If the battery is in a complete circuit, the potential difference is known as the terminal potential difference, which is also measured in volts but may differ slightly from the emf due to internal resistance and load on the battery. Voltage is the measure of work required to move a charge between two points, while current is the rate of charge flow, measured in amperes.

When a battery is used in a circuit, the conventional current flows from the positive terminal to the negative terminal, and this flow of charge creates the potential difference that does work in the circuit. An ammeter, which must be connected in series, is used to measure current. The common AA battery not only has a voltage of 1.5 V but also a significant amount of stored energy, which could be likened to a substantial mass being lifted against gravity, highlighting the compact energy storage capability of such batteries.

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The total amount of energy per hour is [tex]2.039\cdot 10^{16} kJ[/tex]

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

[tex]E_1 = 2073 kJ[/tex]

The total surface area of the United States is

[tex]A=9.834\cdot 10^6 km^2[/tex]

And converting into squared metres,

[tex]A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2[/tex]

Therefore, the total energy reaching the entire United States per hour is given by:

[tex]E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ[/tex]

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The definition of volumetric charge density is given as the ratio between the load per unit volume, while the linear load is the same ratio of the load but per unit length. Applying these concepts then we have that the volumetric density of charge is,

Here,

q = Charge

V = Volume

Replacing we have,

[tex]\gamma = \frac{0.51}{\pi r^2 l}[/tex]

[tex]\gamma = \frac{0.51}{\pi (4*10^{-3})^2(0.33)}[/tex]

[tex]\gamma =30475.84C/m^3[/tex]

And the linear charge density is

[tex]\rho = \frac{q}{l}[/tex]

[tex]\rho = \frac{0.51}{0.33}[/tex]

[tex]\rho = 1.54C/m[/tex]

To determine the volume and linear charge densities, divide the total charge by the rod's volume and length, respectively, after calculating the volume using the formula for a cylinder.

The volume charge density (ρ) and the linear charge density (λ) are physical quantities that represent the distribution of electric charge in a material. To find the volume charge density, we divide the total charge (Q) by the volume (V) of the rod. The formula is ρ = Q/V. The linear charge density is found by dividing the charge by the length (L) of the rod, using the formula λ = Q/L.

Given that the charge Q is 0.51 C, distributed uniformly along a cylindrical rod with a length of 33 cm (or 0.33 m) and a radius of 4 mm (or 0.004 m), we first need to calculate the volume of the cylinder using the formula V = πr²h, where r is the radius and h is the height (length) of the cylinder. The volume V is then π ×(0.004 m)² ×0.33 m.

After computing the volume, we calculate ρ as ρ = 0.51 C / V. To find λ, we simply divide the charge by the length, λ = 0.51 C / 0.33 m. These computations will yield the volume and linear charge densities for the rod.

Answer:

The answer is zero. The total change in momentum is equal to the sum of the areas within the time interview 0-4s. The area under the graph is Force x time (F being the breadth and time the length)

The sun of the areas is zero.

Explanation:

See the attachment below for the full solution.

Thank you for reading this post. I hope it is helpful to you.

Final answer:

The work done when a 185g tomato is lifted 15.0m is 27.261 J. If 82.1% of this work is converted to kinetic energy, the velocity of the tomato when it hits the ground is approximately 15.54 m/s.

Explanation:

To determine the work done when lifting a 185g tomato 15.0m, we use the formula for gravitational potential energy (GPE), which is equivalent to the work done against gravity.

Work (W) = mass (m) × gravitational acceleration (g) × height (h)
m = 185 g = 0.185 kg (since 1g = 0.001 kg)
g = 9.8 m/s²
h = 15.0 m

W = 0.185 kg × 9.8 m/s² × 15.0 m
W = 27.261 J

To determine the velocity (v) when the tomato hits the ground, assuming 82.1% of the work done is converted to kinetic energy (KE), we calculate:

KE = 0.821 × W
KE = 0.821 × 27.261 J
KE = 22.379 J

The formula for KE is:

KE = 1/2 m v²
Solving for v gives:

v = √(2 × KE / m)
v = √(2 × 22.379 J / 0.185 kg)
v ≈ 15.54 m/s

Thus, the velocity of the tomato when it hits the ground, assuming 82.1% conversion of energy, is approximately 15.54 m/s.

Answer:

The energy is [tex]9.4\times10^{-21}\ J[/tex]

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

[tex]Q=\Delta U+W[/tex]

[tex]Q=\Delta U+nRT[/tex]

Put the value into the formula

[tex] 4480=\Delta U+\dfrac{20}{28}\times8.314\times77[/tex]

[tex]\Delta U=4480-\dfrac{20}{28}\times8.314\times77[/tex]

[tex]\Delta U=4022.73\ J[/tex]

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

[tex]N=n\times \text{Avogadro number}[/tex]

Put the value into the formula

[tex]N=\dfrac{20}{28}\times6.02\times10^{23}[/tex]

[tex]N=4.3\times10^{23}[/tex]

We need to calculate the energy

Using formula of energy

[tex]E=\dfrac{\Delta U}{N}[/tex]

Put the value into the formula

[tex]E=\dfrac{4022.73}{4.3\times10^{23}}[/tex]

[tex]E=9.4\times10^{-21}\ J[/tex]

Hence, The energy is [tex]9.4\times10^{-21}\ J[/tex]

The binding energy of a nitrogen molecule in the liquid is calculated using the formula for latent heat and Avogadro's number. Once the latent heat and the number of molecules are determined, we can find the energy per molecule and hence, the binding energy. After performing the calculations, we find the binding energy of a nitrogen molecule to be approximately 5.2 x 10^-21 J.

To calculate the binding energy of a nitrogen molecule in the liquid, we can use the formula for latent heat. This is given by:

Q = mL, where,

Q = Heat energy applied (color change or state change)

m = mass

L = Latent heat

After calculating the latent heat, we can determine the number of moles of nitrogen gas formed and subsequently the number of molecules using Avogadro's number. Then, the energy per molecule is calculated by dividing the total heat absorbed by the number of molecules. Therefore, the binding energy will be the difference in energy per molecule (latent heat per molecule) between the liquid and the gaseous states.

The amount of heat given (Q) is 4480 J and the mass (m) of liquid nitrogen boiled is 20 g. The molar mass (M) of nitrogen (N2) is approximately 28 g/mol. Using this, we can rearrange the formula to find L which is the energy/mass.

Substituting the values, we get L = Q/m = 4480J / 20g = 224 J/g

The number of moles of N2(n) = m/M = 20g / 28 g/mol ~ 0.714 mol

The number of N2 molecules(N) = n x Avogadro's number = 0.714 mol x 6.022 x 10^23 mol^-1 ~ 4.3 x 10^23

Therefore, the binding energy of a nitrogen molecule (E) = L/N = 224J/g / 4.3 x 10^23 = ~5.2 x 10^-21 J.

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Answer:

  t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

[tex]y = u t + \dfrac{1}{2}at^2[/tex]

[tex]-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2[/tex]

[tex]4.9 t^2 - 2.51 t - 3.16 = 0[/tex]

using quadratic formula to solve the equation

[tex]t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}[/tex]

  t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

The car's speed at the top of the loop is approximately 0.41 m/s.

At the top of the loop-the-loop, the normal force equals the magnitude of the gravitational force, which means the net force acting on the roller coaster car is zero.

This condition occurs when the car is just about to lose contact with the track due to insufficient normal force.

Using the centripetal force formula, [tex]\( F_{\text{net}} = \frac{mv^2}{r} \)[/tex], where \( m \) is the mass of the car, v is its speed, and \( r \) is the radius of the loop.

At the top of the loop, the net force is zero, so we equate the gravitational force and the centripetal force:

[tex]\[ mg = \frac{mv^2}{r} \][/tex]

Solving for v, we get:

[tex]\[ v = \sqrt{gr} \][/tex]

Substituting the given radius [tex]\( r = \frac{33 \, \text{mm}}{2} = 0.0165 \, \text{m} \)[/tex] and the acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex], we find:

[tex]\[ v = \sqrt{(9.8 \, \text{m/s}^2)(0.0165 \, \text{m})} \approx 0.41 \, \text{m/s} \][/tex]

Therefore, the car's speed at the top of the loop is approximately 0.41 m/s.

With negligible air resistance and low speed, the only significant net force on a 0.7 kg ball is gravity, affecting the ball's y component of momentum. The x component remains constant, and z component changes are not discussed without additional forces.

When a ball of mass 0.7 kg flies through the air at low speed with air resistance negligible, the net force acting on the ball while it is in motion is primarily due to gravity, which will be impacting the y component of the ball's momentum. The x component of the ball's momentum remains unchanged because no horizontal force is applied, while the y component changes due to gravity, and the z component would only change if there were forces acting in a direction out of the horizontal plane, which are not mentioned in the scenario. As for the Earth-ball system, momentum is conserved in the vertical direction because the system experiences no net external vertical force.

Answer:

False

Explanation:

There is less pressure higher in the atmosphere, which means that air will expand, and thus cool

Answer:

[tex]K=\frac{GmM}{2R}[/tex]

Explanation:

The kinetic energy is defined as:

[tex]K=\frac{mv^2}{2}(1)[/tex]

Here, m is the object's mass and v its speed. In this case the speed of the satellite is the orbital speed, which is given by:

[tex]v_{orb}=\sqrt\frac{GM}{R}(2)[/tex]

Here, G is the gravitational constant, M is the mass of the object that the satellite is orbiting and R is the radius of its circular orbit. Replacing (2) in (1):

[tex]K=\frac{mv_{orb}^2}{2}\\K=\frac{m(\sqrt\frac{GM}{R})^2}{2}\\K=\frac{GmM}{2R}[/tex]

The kinetic energy of a satellite in a circular orbit can be found using the equation K=GMm/2r, involving the gravitational constant, mass of the satellite, mass of the body being orbited, and the radius of the orbit. This kinetic energy is half the potential energy and equal to the total energy of the satellite.

The kinetic energy K of a satellite with mass m in a circular orbit with radius R can be defined using the equation of the kinetic energy K = 1/2 * mv², where m is the mass of the satellite and v is its speed. However, in the context of gravitational forces, we must consider that the gravitational force provides the centripetal force necessary for the satellite to maintain its orbit with speed v. Therefore, we have GMm/r² = mv²/r where G is the gravitational constant and M is the mass of the body the satellite is orbiting.

When we solve for the speed v, we find that v=sqrt(GM/R), which leads us to an equation for the kinetic energy of K = GMm/2r. These equations demonstrate that the kinetic energy of the satellite is dependent not only on its own mass and speed, but also on the mass of the body it is orbiting and the radius of its orbit.

The concept that the kinetic energy of a satellite in circular orbit is half the magnitude of the potential energy, and the same as the magnitude of the total energy, is an important aspect of understanding these calculations. We also note that the gravitational constant G is by far the least well determined of all fundamental constants in physics.

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Answer:

e = 16.67 km/L

Explanation:

given,

fuel efficiency = 39.2 mi/gal

we need to convert fuel efficiency in Km/l

1 Km = 0.621 mile

1 gal = 3.786 L

now, efficiency

[tex]e = 39.2 \dfrac{mi}{gal}\times \dfrac{1\ km}{0.621\ mile}\times\dfrac{1\ gal}{3.786\ L}[/tex]

e = 16.67 km/L

hence, the efficiency of the car in km/L is equal to 16.67 km/L

Answer:

r = 3721.04 m          

Explanation:

Given that,

Weight of human, F = 650 N

Charge on two humans, [tex]q_1=q_2=1\ C[/tex]

We need to find the distance between charges if the electric attraction between them to equal their 650 N weight. It is given by :

[tex]F=\dfrac{kq^2}{r^2}[/tex]

[tex]r=\sqrt{\dfrac{kq^2}{F}}[/tex]

[tex]r=\sqrt{\dfrac{9\times 10^9\times 1^2}{650}}[/tex]

r = 3721.04 m

So, the distance between charges is 3721.04 m if the electric attraction between them to equal their 650 N weight. Hence, this is the required solution.

Answer:

The flea will move high to a height of 0.05 meters.

Explanation:

Given that,

Acceleration of the flea, [tex]a=1000\ m/s^2[/tex]

Distance, d = 0.5 mm = 0.0005 m

Let u and v are the initial and final velocity of the flea. Using equation of motion as :

[tex]v^2-u^2=2ad[/tex]

[tex]v^2-u^2=2\times 1000\times 0.0005[/tex]

[tex]v^2-u^2=1[/tex]..........(1)

Using conservation of energy, we get :

[tex]\dfrac{1}{2}mu^2=\dfrac{1}{2}mv^2+mgh[/tex]

[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+(-g)h[/tex]

[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2-gh[/tex]

[tex]\dfrac{1}{2g}(u^2-v^2)=-h[/tex]

[tex]h=\dfrac{1}{2g}[/tex]

[tex]h=\dfrac{1}{2\times 9.8}[/tex]

h = 0.05 meters

So, the flea will move high to a height of 0.05 meters. Hence, this is the required solution.

Answer:

0.4rad/s²

Explanation:

Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)

w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m

a = ∆w/t

a = (w2 - w1)/t

a = (2 - 4)/5 = -2/5 =

a = - 0.4rad/s²

The -ve sign indicates a deceleration in the motion

Good luck

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

Answer:

Force, [tex]F=7.04\times 10^{-5}\ N[/tex]

Explanation:

Given that,

Distance between 1.70 A from the plutonium nucleus, [tex]d=1.7\times 10^{-10}\ m[/tex]

The number of electron in plutonium is 94.

To find,

The magnitude of the force on an electron.

Solution,

Total charge in the plutonium nucleus is, [tex]q=94\times 1.6\times 10^{-19}=1.504\times 10^{-17}\ C[/tex]. The electric force between charges is given by :

[tex]F=\dfrac{kq^2}{d^2}[/tex]

[tex]F=\dfrac{9\times 10^9\times (1.504\times 10^{-17})^2}{(1.7\times 10^{-10})^2}[/tex]

[tex]F=7.04\times 10^{-5}\ N[/tex]

So, the magnitude of the force on an electron is [tex]7.04\times 10^{-5}\ N[/tex]. Hence, this is the required solution.

Answer:

2.24×10⁻⁸ C

Explanation:

From coulomb's law,

F = kq1q2/r² ............................ Equation 1

Where F = Repulsive force between the two charges, q1 = charge on the first sphere, q2 = charge on the second sphere, r = distance between the sphere, k = proportionality constant.

Note: q1 = q2 = q,

Then, we can rewrite equation 1 as,

F = kq²/r²

making q the subject of the equation

q = √(Fr²/k)................................. Equation 2

Given: F = 21 mN = 0.021 N, r = 15 mm = 0.015 m

Constant: k = 9.0×10⁹ Nm²/C²

Substituting these values into equation 2

q = √(0.021×0.015²/9.0×10⁹)

q = √(5×10⁻¹⁶)

q = 2.24×10⁻⁸ C.

Hence the charge on each sphere = 2.24×10⁻⁸ C

The magnitude of the charge on each sphere is [tex]2.29*10^{-8}C[/tex]

The force between two charges is given by coulombs law,

                  [tex]F=k\frac{Q_{1}*Q_{2}}{r^{2} }[/tex]

Where r is distance between charges.

And k is constant,  [tex]k=9*10^{9} Nm^{2}/C^{2}[/tex]

Given that, [tex]Q_{1}=Q_{2},r=15mm=15*10^{-3}m,F=21mN=21*10^{-3}N[/tex]

Substitute above values in above formula.

       [tex]21*10^{-3}=9*10^{9}*\frac{Q^{2} }{(15*10^{-3} )^{2} } \\\\Q^{2}=\frac{21*10^{-3}*225*10^{-6} }{9*10^{9} } \\\\Q^{2}=5.25*10^{-16} \\\\Q=\sqrt{5.25*10^{-16}} =2.29*10^{-8}C[/tex]

Hence, the magnitude of the charge on each sphere is [tex]2.29*10^{-8}C[/tex]

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Answer:

0.06

Explanation:

The thickness of a human hair is about 60 micrometers. When converted to millimeters, this measures to 0.06 millimeters.

The thickness of a human hair is about 60 micrometers (um). To convert this to millimeters (mm), you need to understand that 1 millimeter is equal to 1000 micrometers. Therefore, you can convert 60 um to mm by dividing 60 by 1000.

This gives an answer of 0.06 mm. So, the thickness of a human hair is 0.06 millimeters.

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Answer:

The maximum no. of electrons- [tex]2.25\times 10^{22}[/tex]

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

[tex]I = \frac{Q}{t}[/tex]

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

[tex]Q = 1.0\times 3600 = 3600\ C[/tex]

Maximum number of electrons, n is given by:

[tex]n = \frac{Q}{e}[/tex]

where

e = charge on an electron = [tex]1.6\times 10^{- 19}\ C[/tex]

Thus

[tex]n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}[/tex]

The void ratio of the soil is 1.2045, the specific gravity of solids is 2.6329, and the saturated unit weight is 82.7586 lb/ft3.

The void ratio of the soil can be determined using the formula:

e = (1 + w) / (1 - w)

where e is the void ratio and w is the moisture content. Plugging in the values, we get:

e = (1 + 0.17) / (1 - 0.17) = 1.2045

To calculate the specific gravity of solids, we can use the formula:

Gs = (1 + e) * (1 - S) / (1 - e * S)

where Gs is the specific gravity of solids and S is the degree of saturation. Substituting the given values:

Gs = (1 + 1.2045) * (1 - 0.6) / (1 - 1.2045 * 0.6) = 2.6329

The saturated unit weight can be found using the equation:

Gamma_sat = Gamma_dry / (1 + w)

where Gamma_sat is the saturated unit weight and Gamma_dry is the dry unit weight. Given that the unit weight of the soil is 96 lb/ft3, we have:

Gamma_sat = 96 / (1 + 0.17) = 82.7586 lb/ft3

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Answer:

10.84406 Nm/rad

0.068625 kgm²

2.00066 rad/s

0.49983 s

Explanation:

F = Force = 4.29 N

R = Radius = [tex]\dfrac{30}{2}=15\ cm[/tex]

[tex]\theta[/tex] = Angle = [tex]3.4\ ^{\circ}[/tex]

m = Mass of disk = 6.1 kg

Torsional constant is given by

[tex]J=\dfrac{\tau}{\theta}\\\Rightarrow J=\dfrac{FR}{\theta}\\\Rightarrow J=\dfrac{4.29\times 0.15}{3.4\times \dfrac{\pi}{180}}\\\Rightarrow J=10.84406\ Nm/rad[/tex]

The torsion constant is 10.84406 Nm/rad

Moment of inertia is given by

[tex]I=\dfrac{1}{2}mr^2\\\Rightarrow I=\dfrac{1}{2}6.1\times 0.15^2\\\Rightarrow I=0.068625\ kgm^2[/tex]

The moment of inertia is 0.068625 kgm²

Frequency is given by

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{J}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{10.84406}{0.068625}}\\\Rightarrow f=2.00066\ rad/s[/tex]

The frequency is 2.00066 rad/s

Time period is given by

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{2.00066}\\\Rightarrow T=0.49983\ s[/tex]

The time period is 0.49983 s

Answer:

a. its constitution and mixed government.

Explanation:

Polybius tell us about the Roman mixed constitution as a fundamental in the Roman victory and the Carthaginian defeat in the Punic war, conceiving the mixed constitution as the best, since this constitution was at its peak, which implies that among the elements of the constitution, the aristocratic component was the dominant one.

Answer:

The right leg (oil on top) is higher

Explanation:

Given:

The mass density of oil is lesser than the mass density of water.

This is justified by the equation:

[tex]P=\rho.g.h[/tex]

where:

[tex]\rho =[/tex] density of the liquid

[tex]g=[/tex] acceleration due to gravity

[tex]h=[/tex] height of the liquid column

The extension in the rubber band for the restoring force of 3 newtons is 1.5 cm.

The force according to Hooke's law is given as:

[tex]F=kx[/tex]

Here F is the restoring force, k is the proportionality constant and x is the stretch or compression.

Given:

Restoring force, [tex]F= 2\ N[/tex]

Stretch in the band, [tex]x=1\ cm[/tex]

The value of the k is computed as:

[tex]F=kx\\2=k \times 1cm\\k= 2\ N/cm[/tex]

The stretch in the band for a restoring force of 3 N is computed as:

[tex]F=kx\\x=\frac{F}{k}\\x= \fract{3}{2}\\x = 1.5 cm[/tex]

Therefore, the extension in the rubber band for the restoring force of 3 newtons is 1.5 cm.

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According to Hooke's Law, a force of 3 newtons will stretch the rubber band 1.5 centimeters.

To answer this question, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, we are given that when a rubber band is stretched 1 cm beyond its natural length, it exerts a restoring force of 2 newtons. This means that the force constant of the rubber band is 2 newtons per centimeter (N/cm).

To find out how far a force of 3 newtons will stretch the rubber band, we can use the formula for Hooke's Law: F = kx, where F is the force, k is the force constant, and x is the displacement. Rearranging the formula, we have x = F/k. Plugging in the values, we get:

x = 3 N / (2 N/cm) = 1.5 cm

Therefore, a force of 3 newtons will stretch the rubber band 1.5 centimeters.

Answer:

a) 88.1 rad/s

b) 0.286

Explanation:

given information:

diameter, d = 8 cm = 0.08 m

sphere's mass, m = 400 g = 0.4 kg

the distance from rest to the tip, h = 2.1 m

incline angle, θ = 25°

a) What is the sphere's angular velocity at the bottom of the incline?

mg(h sinθ) = 1/2 Iω² + 1/2mv²

I of solid sphere = 2/5 mr², thus

mg(h sinθ) = 1/2 (2/5 mr²) ω² + 1/2 mv², now we can remove the mass

g h sin θ = 1/5 r² ω² + 1/2 v²

ω = v/r, v = ωr

so,

g h sin θ = 1/5 r² ω² + 1/2 (ωr)²

g h sin θ = (7/10) r² ω²

ω² = 10 g h sin θ/7 r²

ω = √10 g h sin θ/7 r²

   = √10 (9.8) (2.1) sin 25° / 7 (0.04)²

   = 88.1 rad/s

b) What fraction of its kinetic energy(KE) is rotational?

fraction of its kinetic energy = rotational KE / total KE

total KE = total potential energy

             = m g h sin θ

             = 0.4 x 9.8 x 2.1 sin 25°

             = 3.48 J

rotational KE = 1/2 Iω²

                      = 1/5 mr²ω²

                      = 1/5 0.4 (0.04)²(88.1)²

                      = 0.99

fraction of its KE = 0.99/3.48

                            = 0.286

A) The sphere's angular velocity at the bottom of the incline is; ω = 88.1 rad/s

B) Fraction of its kinetic energy that is rotational is; 0.286

We are given;

Diameter; d = 8 cm = 0.08 m

Mass of sphere; m = 400 g = 0.4 kg

Distance from rest to the tip; h = 2.1 m

Angle of inclination; θ = 25°

a) To get the sphere's angular velocity at the bottom of the incline, we will use the expression;

mg(h*sinθ) = ¹/₂Iω² + ¹/₂mv²

where;

I of solid sphere = ²/₅mr²

Thus;

mg(h*sinθ) = ¹/₂(²/₅mr²)ω² + ¹/₂mv²

The mass m will cancel out to give;

gh*sin θ = ¹/₅r²ω² + ¹/₂v²

where v = ωr

Thus;

gh*sin θ = ¹/₅r²ω² + ¹/₂r²ω²

gh*sin θ = ⁷/₁₀r²ω²

ω = √[(¹⁰/₇)*g*h*(sin θ)/r²]

ω = √[(¹⁰/₇)*9.8*2.1*(sin 25)/(0.04)²]

ω = 88.1 rad/s

b) Fraction of its kinetic energy that is rotational = rotational KE/total KE

But, total KE = total potential energy

Thus;

KE_tot = mgh*sin θ

KE_tot = 0.4 * 9.8 * 2.1 sin 25°

KE_tot = 3.48 J

KE_rot = ¹/₂Iω²

I of solid sphere = ²/₅mr². Thus;

KE_rot = ¹/₅mr²ω²

KE_rot = ¹/₅ * 0.4 * 0.04² * 88.1²

KE_rot = 0.99 J

Fraction of its kinetic energy that is rotational = 0.99/3.48 = 0.286

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Answer:

(a) [tex]E=2.42eV[/tex]

(b) [tex]E=1.45*10^{-5}eV[/tex]

(c) [tex]E=1.66*10^{-7}eV[/tex]

Explanation:

The Planck-Einstein relation allows us to know the energy (E) of a photon, knowing its frequency (f). According to this relation, the energy of the photon is defined as:

[tex]E=hf[/tex]

Here h is the Planck constant.

(a)

[tex]E=(4.14*10^{-15}eV\cdot s)(585*10^{12}Hz)\\E=2.42eV[/tex]

(b)

[tex]E=(4.14*10^{-15}eV\cdot s)(3.50*10^{9}Hz)\\E=1.45*10^{-5}eV[/tex]

(c)

[tex]E=(4.14*10^{-15}eV\cdot s)(40.0*10^{6}Hz)\\E=1.66*10^{-7}eV[/tex]