Ifthe amount of ice on the planet increased to the amount that was present during the last glacial maximum, how muchwould it cause global sea level to fall? If you do not remember the approximate size of this rise, the volume of additionalice is 51000000 km3 and the surface area of the world's ocean is 361,000,000 km;A. less than 10 centimetersB. 10 centimeters to 1 meterC. 1 meter to 10 metersD. 10 meters to 50 metersE. more than 50 meters

Answer:

.

Explanation:

Answer:

l = 3.21 mm

Explanation:

The efficiency of a fiber-reinforced is the following:

[tex] \eta = \frac{l - 2x}{l} [/tex]

Where:

η: is the efficiency

l: is the fiber length

x: is the length of the fiber at each end that doesn't contribute to the load transfer

So the length required for a 0.62 efficiency of reinforcement is:

[tex] l = \frac{2x}{1- \eta} = \frac{2 \cdot 0.61 mm}{1- 0.62} = 3.21 mm [/tex]

I hope it helps you!  

The length required for a 0.62 efficiency of reinforcement in a fiber-reinforced composite can be calculated using the equation η = (l-x)/l, where l represents the total length of the fiber and x represents the length at each end that does not contribute to load transfer. By rearranging the equation, we can solve for l.

To calculate the length required for a 0.62 efficiency of reinforcement in a fiber-reinforced composite, we can use the equation η = (l-x)/l. Here, l represents the total length of the fiber, and x represents the length at each end that does not contribute to load transfer. We are given that x is 0.61 mm. We can rearrange the equation to solve for l:

η = (l-x)/l

0.62 = (l-0.61)/l

0.62l = l - 0.61

0.62l - l = -0.61

0.62l = -0.61

l = -0.61/0.62

l = -0.983

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Answer:

B. Marginal cost equals long-run average total cost.

Explanation:

The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.

Answer:

Explanation:

Given

after 5.5 revolution wheel comes to stop

i.e. radian turned before stopping

[tex]\theta =2\pi \times 5.5 rad[/tex]

initial angular velocity [tex]\omega _0=3.5 rad/s[/tex]

[tex]\omega ^2-\omega _0^2=2\cdot \alpha \cdot \theta [/tex]

where  [tex]\alpha =angular\ acceleration\ or\ deceleration[/tex]                                                                                                                                                                                                                                                                                                    

[tex]0-(3.5)^2=2(\alpha )(2\pi \cdot 5.5)[/tex]

[tex]\alpha =-0.1772 rad/s^2[/tex]

angle turned when final angular velocity is [tex]2 rad/s[/tex]

[tex]2^2-3.5^2=2\cdot (-0.1772)(\theta )[/tex]

[tex]\theta =23.27\ radians[/tex]

Answer:

560.06714 Nm

Explanation:

[tex]\omega_f[/tex] = Final angular velocity

[tex]\omega_i[/tex] = Initial angular velocity = 0

[tex]\alpha[/tex] = Angular acceleration

[tex]\theta[/tex] = Angle of rotation = [tex]\pi[/tex] (Half rotation)

v = Velocity of bat = 29.8 m/s

M = Mass of bat = 11.3 kg

m = Mass of ball = 0.196 kg

R = Radius of swing = 0.984 m

[tex]\omega_f=\dfrac{v}{r}\\\Rightarrow \omega_f=\dfrac{29.8}{0.984}\\\Rightarrow \omega_f=30.28455\ rad/s[/tex]

From equation of rotatational motion

[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{30.28455^2-0^2}{2\times \pi}\\\Rightarrow \alpha=145.96958\ rad/s^2[/tex]

Moment of inertia is given by

[tex]I=\dfrac{1}{3}MR^2+mR^2\\\Rightarrow I=\dfrac{1}{3}11.3\times 0.984^2+0.196\times 0.984^2\\\Rightarrow I=3.83687577\ kgm^2[/tex]

Torque is given by

[tex]\tau=I\alpha\\\Rightarrow \tau=3.836875776\times 145.96958\\\Rightarrow \tau=560.06714\ Nm[/tex]

The torque the pitcher applies is 560.06714 Nm

The average torque this pitcher must apply to the softball is 560.64 Newton.

Given the following data:

To determine the average torque this pitcher must apply to the ball:

First of all, we would determine the final angular speed of the ball by using this formula:

[tex]\omega_f =\frac{v}{r} \\\\\omega_f =\frac{29.8}{0.984} \\\\\omega_f = 30.29\;rad/s[/tex]

Next, we would determine the constant angular acceleration by using the equation of circular motion:

[tex]\alpha =\frac{\omega^2_f - \omega^2_i}{2\theta} \\\\\alpha =\frac{30.29^2 - 0^2}{2\times \pi} \\\\\alpha =\frac{917.48}{2\times 3.142} \\\\\alpha =\frac{917.48}{6.284}\\\\\alpha = 146\;rad/s^2[/tex]

For the moment of inertia:

For a point mass, moment of inertia is given by this formula:

[tex]I =\frac{1}{3} Mr^2 + mr^2\\\\I =\frac{1}{3} \times 11.3 \times 0.984^2 + 0.196 \times 0.984^2\\\\I = 3.84 \;Kgm^2[/tex]

Now, we can determine the average torque:

[tex]Torque = I\alpha \\\\Torque =3.84 \times 146[/tex]

Torque = 560.64 Newton.

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Answer:

a)  ΔV = - 4.346 10⁻² , b)   ρ’= 1.082 10³ kg / m³

Explanation:

The volume module is defined as the ratio of the pressure and the unit deformation, with a negative sign, for the module to be positive

       B = - P / (ΔV/V)

a) The ΔV volume change

     ΔV/V = -P / B

     ΔV = - P V / B

     ΔV = - 1.13 10⁸ 0.9 /2.34 10⁹

     ΔV = - 4.346 10⁻²

b) Density at the bottom of the sea

On the surface

     ρ = m / V

      m = ρ V

     m = 1.03 10³ 0.9

     m = 0.927 10³ kg

Body mass does not change with depth

Deep down

    ρ’= m / V’

    ΔV = 4.346 10⁻²

    [tex]V_{f}[/tex]- V₀ = 4,346 10⁻²

    [tex]V_{f}[/tex] = 0.0436 + Vo

    [tex]V_{f}[/tex]= -0.04346 + 0.9

    [tex]V_{f}[/tex] = 0.85654 m³

    ρ’= 0.927 10³ / 0.85654

    ρ’= 1.082 10³ kg / m³

Using the formula for volume change under pressure and density calculations, it can be determined that the change in volume of 0.9 m³ seawater when taken to the Mariana Trench is -0.043 m³ and its density at the bottom is approximately 1072 kg/m³.

The pressure in the ocean increases with depth due to the weight of the overlying water. This high pressure can compress the volume of the water at such depths, altering its density. As the question provides, the bulk modulus of seawater is given as 2.34 x 10⁹ N/m² and the pressure in the Mariana Trench is 1.13 x 10⁸ N/m².

(a) To find the change in volume, we can use the formula ΔV = -(PΔV/B), where P is the pressure, ΔV is the change in volume, and B is the bulk modulus. Inserting the given values, we get the change in volume to be -0.043 m³.

(b) The density of a substance is its mass divided by its volume. At the bottom of the Mariana Trench, the volume of water has decreased due to the high pressure, but its mass remains the same. Therefore, as the volume decreases, the density increases. The new density, ρ', is calculated using the formula ρ' = ρ/(1-ΔV/V), where ρ is the initial density and V is the initial volume. Substituting into this equation, we get ρ' = 1.03 x 10³ kg/m³ / (1 -(-0.043), which gives us a density at the bottom of the Mariana Trench of approximately 1072 kg/m³.

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From the definition we have that the Torque corresponds to the Multiplication between the Force (or its respective component) and the radius of distance of the Force to the inertial turning point.

Mathematically this can be expressed,

[tex]\tau = F \times d[/tex]

Where,

F = Perpendicular component of force

d = distance from pivot point

The total sum of the torques would be equivalent to

[tex]\tau_{net} = \tau_1 +\tau_2[/tex]

According to the values given, torque 1 and 2 would be given by

[tex]\tau_1 = 6*1.2 = 7.2N\cdot m (+)[/tex]

[tex]\tau_2 = -5.2sin(30) = -7.8N\cdot m (-)[/tex]

Therefore the net Torque is

[tex]\tau_{net} = \tau_1+\tau_2[/tex]

[tex]\tau_{net} = 7.2-7.8[/tex]

[tex]\tau_{net} = -0.6N\cdot m[/tex]

Therefore the net torque about the pivot is -0.6Nm

Answer:

33 g.

Explanation:

Assuming no heat transfer can be possible except for heat exchange between water and steel, we can say that the heat lost by the knife, must be equal to the heat gained by the water.

As we have a limit for the maximum temperature of both elements (once reached a final thermal equilibrium), of 100ºC, which means that the maximum allowable change in temperature will be of 300º C for the knife, and of 80º C for the water.

Empirically , it has been showed that for a heat exchange process using only conduction, the heat needed to raise the temperature of a body, is proportional to the mass, being the proportionality constant a factor that depends on the material, called specific heat.

So, we can write the following equation:

cs*mk*Δtk = cw*mw*Δtw

Replacing by the givens of the question, we have:

0.11 cal/gºC * 80 g * 300ºC = 1 cal/gºC*mw*80ºC

Solving for mw = 2,640 cal / 80 cal/g =33 g.

Answer:

0.00389 m

Explanation:

Archimedes principle states that the buoyant force when a body is immersed in a liquid equals the weight of the liquid displaced

B = weight of the liquid displaced

density is defined as the mass divided by the volume of the substance and the units is  in kg/m³

Density of glycerin = mass / volume

ρg × volume = mass of glycerin displaced

since the object was floating, the upthrust from the liquid equals the weight of the liquid

ρg × volume × g = mg

divide both side by g

ρg × volume = 0.180  where ρg (density of glycerin) = 1260 kg / m³

1260 × volume of glycerin displaced = 0.18 / 1260 = 0.000143 m³

volume of glycerin displaced = πr²h₁ where h₁ = of liquid displaced

πr²h₁ = 0.000143

h₁ = 0.000143 / ( 3.142 × 0.055² ) = 0.01505 m

when the ice completely melted, it will displaces liquid equal to it own volume

density of water = mass of water /volume

1000 = 0.18 / v

v = 0.18 / 1000 = 0.00018 m³

volume = πr²h₂ where h₂ = height of the melted water

πr²h₂ = 0.00018

h₂ = 0.00018 / ( 3.142 × 0.055²) = 0.01894 m

change in height of the liquid = h₂ - h₁ =  0.01894 m - 0.01505 m = 0.00389 m

When the 0.180 kg ice cube melts, it will form water that spreads out in the glycerin-filled cylinder, raising the level by approximately 1.89 cm.

To solve this problem, we first need to understand that when the ice cube melts, it will not change the overall volume of liquid in the cylinder. In other words, the total volume before the ice melts (volume of glycerin + volume of ice) is equal to the total volume after the ice melts (volume of glycerin +volume of water).

The volume of the water, resulting from the melted ice, can be calculated using the mass of the ice cube and the density of water. The volume (V) is equal to the mass (m) divided by the density (ρ). For ice, m = 0.180 kg and ρ = 1000 kg/m³ (the density of water), which gives V = 0.00018 m³ or 180 cm³.

This water spreads out in the cylinder, raising the level. The height (h) to which it raises can be found by dividing the volume of the water (V) with the cross-sectional area (A) of the cylinder (h = V/A). The area of the cylinder is πr², where r is the radius, so A = π * (5.5 cm)² = 95.03 cm². Thus, h = 180 cm³ / 95.03 cm² = 1.89 cm. Consequently, when the ice cube melts, the glycerin's height will increase by approximately 1.89 cm.

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The normal force of the slide on the child is 217.3 N and the kinetic frictional force from the slide is 166.8 N when the child slides down the slide at a constant speed.

The normal force on a slope, which is always perpendicular to the surface, is equal to the weight component of the object that is perpendicular to the slope. As the child slides down the slide at a constant speed, the net force on the child is zero. In this scenario, let's denote mass (m) as 28 kg, inclination angle (θ) as 38 degrees, and g as gravitational acceleration which is 9.8 m/s². So, the normal force (N), which is equal to m*g*cosθ, can be calculated as: 28 kg * 9.8 m/s² * cos(38) = 217.3 N.

The frictional force from the slide acts in the opposite direction to the motion. When the sliding speed is constant, this kinetic frictional force equals the component of the child's weight that is parallel to the slope (m*g*sinθ). Hence, the kinetic frictional force would be: 28 kg * 9.8 m/s² * sin(38) = 166.8 N.

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The normal force of the slide on the child can be found by multiplying the child's weight by the cosine of the angle of inclination. The kinetic frictional force from the slide is equal to the horizontal force exerted by the rope.

To find the normal force of the slide on the child, we need to determine the component of the child's weight perpendicular to the slide. Since the slide is inclined at 38∘ above horizontal, the normal force is equal in magnitude to the component of the child's weight perpendicular to the slide, which is given by:

Normal force = weight * cos(38∘)

Next, to find the kinetic frictional force from the slide, we need to use the horizontal force exerted by the rope. Since the child slides with a constant speed, the kinetic frictional force must be equal in magnitude to the horizontal force exerted by the rope, which is given as 30 N.

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To solve this problem we will use the concepts related to thermal expansion in a body for which the initial length, the coefficient of thermal expansion and the temperature change are related:

[tex]\Delta L = L0\alpha\Delta T[/tex]

Where,

[tex]\Delta L[/tex] = Change in Length

[tex]\alpha[/tex] = Coefficient of linear expansion

[tex]\Delta T[/tex] = Change in temperature

[tex]L_0[/tex] = Initial Length

Our values are:

[tex]L_0 = 3.45m[/tex]

[tex]\alpha = 5.5*10^{-7} \°C^{-1}[/tex]

[tex]\Delta T = 235-20 = 215\°C[/tex]

Replacing we have,

[tex]\Delta L = (3.49) (5.5*10^{-7}) [(215)[/tex]

[tex]\Delta L = 0.0004126m[/tex]

[tex]\Delta L = 0.4126mm[/tex]

Therefore the change in milimiters was 0.4126mm

Answer:

β max = 2.86 x 10 ⁻¹² T

β max = 1.432 x 10 ⁻¹² T

β max = 5.726 x 10 ⁻¹² T

β max = 8.589 x 10 ⁻¹² T

Explanation:

Given:

R = 35.0 mm , d = 4.1 mm , f = 60 Hz , V = 160v

And knowing

μ₀ = 4 π x 10 ⁻⁷ T * m / A ,  ε₀ = 8.85 x 10⁻¹² C² / N * m²

To find β max can use the equation

β max = [ π * f * μ₀ * ε₀ * r * V ] / d

r = R

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.035 m * 160 ] / (4.1 x 10⁻³ )

β max = 2.86 x 10 ⁻¹² T

r = 17.5 mm

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.0175 m * 160 ] / (4.1 x 10⁻³ )

β max = 1.432 x 10 ⁻¹² T

r = 70 mm

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.070 m * 160 ] / (4.1 x 10⁻³ )

β max = 5.726 x 10 ⁻¹² T

r = 105.0 mm

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.105 m * 160 ] / (4.1 x 10⁻³ )

β max = 8.589 x 10 ⁻¹² T

Answer:

Part A. 5.36x10²³ molecules of air

Part B. 6.9kg  

Explanation:

Part A.

To calculate the number of molecules of the air we need first find the number of moles of air using the equation of ideal gas law:

[tex] PV = nRT [/tex]    (1)

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

[tex] n = \frac{PV}{RT} = \frac{735torr \cdot 1atm/760torr \cdot 2.35L}{0.082 Latm/Kmol \cdot (37 + 273)K} = 0.089 moles [/tex]

Now by using the Avogadro's number we can find the number of molecules of air:

[tex] number of molecules = \frac{6.022 \dot 10^{23}}{1mol} \cdot 0.089moles = 5.36 \cdot 10^{22} molecules [/tex]

Part B.

Similarly, to calculate the mass of air first we need to detemine the number of moles using equation (1):

[tex] n = \frac{PV}{RT} = \frac{1.07atm \cdot 5.0\cdot 10^{3}L}{0.082 Latm/Kmol \cdot (0.5 + 273)K} = 238.55 moles [/tex]

So, the mass of air is:

[tex] m = moles \cdot M [/tex]

where M: is the average molar mass of air

[tex] m = 238.55moles \cdot 28.98g/mol = 6.9 kg [/tex]

I hope it helps you!  

The student needs to use the ideal gas law and Avogadro's number to convert the volume of air to moles and then to molecules. For the blue whale, use the ideal gas law to convert volume to moles, then multiply by the molar mass of air to get mass.

To begin with, we first convert the volume into molecules. Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature. The number of moles, n, can be found by rearranging the equation to n=PV/RT.

Substituting the given values (converting pressure to atm and volume to L, and temperature to Kelvin), we will compute n. After finding the no of moles, the number of molecules is calculated by multiplying the number of moles by Avogadro's number.

In the case of the blue whale, we again use the ideal gas law to find the number of moles of air in the lungs, then multiply by the molar mass of air (28.98 g/mol) to find the mass in grams. Finally, we then convert from grams to kilograms.

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Speed can be found through the application of concepts related to potential energy and kinetic energy, for which you have

[tex]KE = PE[/tex]

[tex]\frac{1}{2}mv^2 = mgh[/tex]

Where,

m = mass

v = Velocity

g = Gravitational acceleration

h = Height

Re-arrange to find the velocity we have,

[tex]v^2 = 2gh[/tex]

[tex]v = \sqrt{2gh}[/tex]

[tex]v = \sqrt{2(9.8)(0.54)}[/tex]

[tex]v = 3.253m/s[/tex]

Therefore the speed at which water squirts out of the hole is .3253m/s

Answer: 5.5km

Explanation:

Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).

Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.

Pressure decreasing with higher altitudes also means that  air pressure decreases rapidly at lowerevels but more slowly at higher levels.

It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).

Answer:

B) along the -z-axis.

Explanation:

Answer:

option (b)

Explanation:

direction of velocity is along + X axis

direction of force is along - Y axis

So, by use of the

[tex]\overrightarrow{F}=q\left ( \overrightarrow{v} \times \overrightarrow{B}\right )[/tex]

Where, B be the magnetic field

As charge of electron is negative

so the direction of magnetic field is along negative z axis .

Answer:

The series A test tube has some left amount of glucose left in it.

Explanation:

Let's assume that a fixed amount of glucose is synthesized, for the fixed quantity the bacteria produced in A and B be x and y respectively,

Therefore, the condition on x and y is,    y > x  as the no. of bacteria present in B is greater.

As a result B would require a greater amount of energy for its functioning, these energy would be derived from the already fixed amount of glucose present.

A test tube would also require the energy for its x number of bacteria, but it is less than that of B.

Therefore, there would be some unused glucose left in Test Tube Series A which has unused energy.

Answer:

Explanation:

Given

mass of object [tex]m=2 kg[/tex]

inclination [tex]\theta =50^{\circ}[/tex]

[tex]\mu _k=0.3[/tex]

[tex]\mu _s=0.4[/tex]

initial velocity [tex]u=3 m/s[/tex]

acceleration of block during upward motion

[tex]a=g\sin \theta -\mu _kg\cos \theta [/tex]

[tex]a=g(\sin 50-0.3\cos 50)[/tex]

[tex]a=5.617 m/s^2[/tex]

using relation

[tex]v^2-u^2=2a\cdot s[/tex]

where [tex]s=distance\ moved [/tex]

[tex]v=final\ velocity[/tex]

v=0 because block stopped after moving distance s

[tex]0-(3)^2=2\cdot (-5.617)\cdot s[/tex]

[tex]s=\frac{4.5}{5.617}[/tex]

[tex]s=0.801[/tex]

If block stopped after s m then force acting on block is

[tex]F=mg\sin \theta =[/tex]friction force [tex]f_r=\mu mg\cos \theta [/tex]

[tex]F>f_r[/tex] therefore block will slide back down to the bottom            

Answer:

B) the amount of magnetic field that passes through a surface

Explanation:

The magnetic flux is defined as the total magnetic field times the area normal to the magnetic field lines passing through it.

Mathematically:

[tex]\phi_B=\vec{B}.\vec{A}[/tex]

where:

[tex]\vec{A}=[/tex] area vector directed normal to the surface

[tex]\vec{B}=[/tex] magnetic field vector

It is a quantity defined for the convenience of use in Faraday's Law.

Final answer:

The correct phrase for describing 'magnetic flux' is 'the amount of magnetic field that passes through a surface,' represented by the formula Φ = BA cos θ. The unit of magnetic flux is the weber (Wb) or volt second (Vs).

Explanation:

The phrase that best describes the term magnetic flux is 'B) the amount of magnetic field that passes through a surface.' This is due to the fact that magnetic flux is a measure of the total number of magnetic field lines that penetrate a given surface area. The formula for calculating magnetic flux (symbolized by the Greek letter phi, Φ) is Φ = BA cos θ, where B represents the magnetic field strength, A is the area through which the field lines pass, and θ is the angle between the magnetic field and the perpendicular to the surface.

The unit of magnetic flux is the weber (Wb), which can also be expressed in terms of magnetic field per unit area, or tesla per square meter (T/m²), and as a volt second (Vs). Understanding magnetic flux is essential in applying Faraday's Law and in the study of electromagnetic induction, where changes in magnetic flux through a conductor generate an electromotive force (emf).

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2, the minimum tractive effort  = 220.3 lb.

To find the minimum tractive effort,

vehicle weight  is 2500 lb

drag coefficient is 0.35

frontal area 22 ft^2

vehicle speed is 60 mi/hr = 70 ft/sec

gradient - 5%

air density = 0.002045 slugs/ft^3

The force which exerted on a solid body moving in some relation to a fluid by the fluid's movement is known as a drag force.

Drag force is given as,

Fd = [tex]\frac{1}{2}[/tex]CdρAV²

= [tex]\frac{1}{2}[/tex] × 0.35 × 0.002045× 20× 88²

= 55.4 lb.

Force due to vehicle weight,

Fw = 0.01 ( 1+ υ/147)W

= 0.01 ( 1 + 88/147) 2500

= 39.965 lb.

Force due to gradient

Fg = W × g

= 2500 × 0.05

= 125 lb

Minimum tractive effort

F = Fd + Fw +Fg

 = 55.4 + 39.965 + 125

 = 220.365 lb

Thus, minimum tractive force can be found as 220.365 lb.

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Answer:10000N

Explanation:formuler for calculating force is given by F=Ma

M(mass)=500kg

a(acceleration)=20m/s^2

Therefore by substitution we have F=500*20

F=10000N

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

(a) Yes, this situation is possible. No, It is not possible in more than one way.

(b) The position of the wire is at a distance of 16.3 cm, in the direction left of the wire

(c) The magnitude of the current in wire 3 (I3) is 3.267 A and in a direction pointing in the downward direction.

(a)

The diagrammatic illustration of the information provided in the question can be seen in the image attached below.

From the image: suppose the forces on the wire as a result of the two other wires are equal and opposite, then the third wire will experience no net force. Hence, we can say Yes,  the situation is possible.

Given that:

It will not be possible to have more than one way, this is because the force acting on the other two wires will have opposite directions but they would not have the same equal magnitude.

The force per unit length existing in-between two current-carrying parallel wires can be determined by using the formula:

[tex]\mathbf{\dfrac{F}{l} = \dfrac{\mu_o I_1I_2}{2 \pi r} }[/tex]

where:

SInce F1 = F2, Then:

[tex]\mathbf{\dfrac{\mu_o\times I_1\times I_3\times l}{2 \pi r}= \dfrac{\mu_o\times I_2\times I_3\times l}{2 \pi (r+0.2)} }[/tex]

[tex]\mathbf{\dfrac{I_1}{r}= \dfrac{ I_2}{ (r+0.2)} }[/tex]

[tex]\mathbf{\dfrac{1.80}{r_1}= \dfrac{ 4.00}{ (r_1+0.2)} }[/tex]

1.80(r₁ + 0.2) = 4.00r₁

1.80r₁ +0.36 = 4.00 r₁

0.36 = 2.2r₁

r₁ = 0.36/2.2

r₁ = 16.3 cm in the direction left of the wire.

Again, we are to determine the magnitude of the current and its direction in wire 3 (I₃).

From the image, the forces of F21 = F23

∴

[tex]\mathbf{\dfrac{\mu_o \times I_1 \times I_2 \times l }{2 \pi r} = \dfrac{\mu_o \times I_2 \times I_3 \times l }{2 \pi r}}[/tex]

[tex]\mathbf{\dfrac{\mu_o \times 1.8 \times 4.0 \times l }{2 \pi \times 0.2} = \dfrac{\mu_o \times 4 \times I_3 \times l }{2 \pi \times 0.363}}[/tex]

Making I₃ the subject by equating both equations together, we have:

[tex]\mathbf{I_3 = \dfrac{1.8 \times 0.363}{0.2}}[/tex]

I₃ = 3.267 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

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Final answer:

Calculations from the provided wave equation involve finding the period (T), horizontal wave travel, wave number and frequency, wave crest speed, and the maximum vertical displacement speed. These reveal wave motion characteristics including time intervals and speeds for complete wave patterns and individual points within a wave.

Explanation:

The student's question about a wave on a lake is one that involves understanding the characteristics and behavior of waves, which is a key concept in physics. Specifically, the student is asked to find the time for a complete wave pattern to pass a stationary observer, the horizontal distance traveled by a wave crest in that time, the wave number and frequency, the wave crest speed, and the maximum speed of a vertical displacement caused by the wave. Here's how we can calculate each of these:

Time for one complete wave pattern: The time taken for one complete wave to pass is the reciprocal of the frequency, known as the period (T). We use the angular frequency (5.05 s-1) provided in the equation y(x,t) = (3.30 cm) cos(0.400 cm-1x + 5.05 s-1t). We find T by taking 2π divided by the angular frequency.

Horizontal distance traveled: To find this, multiply the wave speed by the period (T).

Wave number and frequency: The wave number (k) is already provided as 0.400 cm-1, which can be converted to meters if needed. The frequency (f) is the angular frequency divided by 2π.

Wave crest speed: This can be found by dividing the angular frequency by the wave number.

Maximum speed of vertical displacement (cork floater): This is the amplitude times the angular frequency, which gives the maximum speed of any vertical movement due to the wave.

These calculations help understand the dynamics of wave motion and how waves interact with objects in their path like a fisherman's cork floater.

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

The current amplitude in the circuit is 1.51A. The impedance of the circuit is 168Ω. The reactance of the inductor can have two values, which can be calculated using the equation XL = 2&pivL.

The current amplitude in the circuit can be calculated using the equation:

II = Vo/R = 116V / 77.0Ω = 1.51A.

The impedance of the circuit can be calculated using the equation:

ZZ = √(R2 + (XL - XC)2) = √(77.0Ω2 + (473Ω - 364Ω)2) = 168Ω.

The reactance of the inductor can have two values. From the equation XL = 2&pivL, we can set XL = XC = 473Ω and solve for L, which gives L = XC / (2&piv) = 473Ω / (2πf) where f is the frequency of the AC voltage source.

To find the values for which the angular frequency is less than the resonance angular frequency, we need to compare the values of XL obtained for different frequencies. If the frequency is such that XL > XC, then the angular frequency is less than the resonance angular frequency.

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Answer:

option A

Explanation:

the initial moment of inertia of system , I₀ = 13 kg.m²

the final moment of inertia of the system , I = 2.6 kg.m²

time = t = 2 s

mass = 5 Kg

the initial angular speed ,

[tex]\omega = \dfrac{2\pi}{T}[/tex]

[tex]\omega = \dfrac{2\pi}{2}[/tex]

ω = 3.14 rad/s

[tex]\omega = \dfrac{3.14}{2\pi}[/tex]

[tex]\omega_0= 0.5\ rev/s[/tex]

let the final angular speed be ω₀

using conservation of angular momentum

I₀ x ω₀= I x  ω

13 x ω₀= I x  ω

13 x  0.5 = 2.6 x ω

ω = 2.5 rev/s

hence, the correct answer is option A

Answer:

Collisions are basically two types: Elastic, and inelastic collision. Elastic collision is defined as the colliding objects return quickly without undergoing any heat generation. Inelastic collision is defined as the where heat is generated, and colliding objects are distorted.

In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.

Final answer:

Elastic collisions conserve both momentum and kinetic energy with objects bouncing off each other, while inelastic collisions conserve momentum but not kinetic energy, often resulting in the objects sticking together or showing permanent deformations. Mathematically analyzing an elastic collision is generally easier due to the conservation of both momentum and kinetic energy.

Explanation:

The primary physical difference between elastic and inelastic collisions lies in how kinetic energy is conserved in the collision process. In an elastic collision, both momentum and kinetic energy are conserved. This means that after the collision, the total kinetic energy of the two objects remains the same as before they collided. The objects bounce off each other and do not stick together. As a helpful trick, remember that 'elastic' implies the ability to return to the original shape, just like objects in an elastic collision bounce off one another and return to their separate ways.

On the other hand, an inelastic collision is one where the kinetic energy is not conserved, although the momentum is still conserved. The objects may stick together after the collision, which can be referred to as a perfectly inelastic collision. Since kinetic energy is lost in inelastic collisions, the objects typically do not separate after colliding; rather, they may move as a single entity or with less kinetic energy than they had initially. An inelastic collision involves a permanent deformation of the colliding bodies or the generation of heat.

To your other questions: Analyzing an elastic collision might be easier, as both momentum and kinetic energy conservation can be used to solve for final velocities, while in an inelastic collision, only momentum is conserved. Verification can be done both mathematically and graphically, but mathematical methods often provide a direct computation of the final velocities. When considering the recoil in two objects thrown by two people, if the objects have the same velocity but different masses, question 15 from 8.3 suggests that the person who threw the heavier object will gain more velocity upon recoil. However, in reality, since momentum is conserved, both individuals would experience the same amount of recoil velocity, opposite in direction to the thrown object.

Answer:

22.1 m

Explanation:

[tex]v_{o}[/tex] = initial speed of ball = 14.3 m/s

[tex]\theta[/tex] = Angle of launch = 27°

Consider the motion of the ball  along the vertical direction.

[tex]v_{oy}[/tex] = initial speed of ball = [tex]v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}[/tex]

[tex]a_{y}[/tex] = acceleration due to gravity = - 9.8 ms⁻²

[tex]t[/tex]  = time of travel

[tex]y[/tex]  = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

[tex]y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s[/tex]

Consider the motion of the ball along the horizontal direction.

[tex]v_{ox}[/tex] = initial speed of ball = [tex]v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}[/tex]

[tex]X[/tex]  = Horizontal distance traveled

[tex]t[/tex]  = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

[tex]X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m[/tex]

Potential energy is the mechanical energy associated with the location of a body within a force field (gravitational, electrostatic, etc.) or the existence of a force field inside a body (elastic energy)

The gravitational potential force is subject to the relationship

[tex]PE = mgh[/tex]

Where,

m = mass

g = Gravitational Energy

h = Height

If our height is 3 meters, the mass is 0.8Kg and the earth has a gravitational acceleration of 9.8m / s, we will have to

PE = (9.8)(3)(0.8)

PE = 23.52J

The closest answer is D.

Answer:

5 m

80 m

320 m

Explanation:

[tex]v_{o}[/tex] = Initial speed of the car = 10 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d[/tex] = Stopping distance of the car = 20 m

[tex]a[/tex] = acceleration of the car

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 10^{2} + 2 (20) a\\a = - 2.5 ms^{-2}[/tex]

[tex]v_{o}[/tex] = Initial speed of the car = 5 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d'[/tex] = Stopping distance of the car

[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'\\0^{2} = 5^{2} + 2 (- 2.5) d'\\d' = 5 m[/tex]

[tex]v_{o}[/tex] = Initial speed of the car = 20 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d''[/tex] = Stopping distance of the car

[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d''\\0^{2} = 20^{2} + 2 (- 2.5) d''\\d'' = 80 m[/tex]

[tex]v_{o}[/tex] = Initial speed of the car = 40 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d'''[/tex] = Stopping distance of the car

[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'''\\0^{2} = 40^{2} + 2 (- 2.5) d'''\\d''' = 320 m[/tex]

To solve this problem we need to apply the concepts related to the average electromagnetic energy density. Which is given as

[tex]U = \frac{1}{2}\epsilon_0 E^2[/tex]

Where,

\epsilon_0 = Permettivity of free space constant

E = Electric Field amplitude

Since the average electromagnetic energy density is directly proportional to the amplitude of the magnetic field then we have to

[tex]E = \frac{1}{2} (8.85*10^{-12}C^2/N\cdot m^2)(0.9V/m)^2[/tex]

[tex]E = 3.6*10^{-12}J/m^3[/tex]

[tex]E = 3.6pJ/m^3[/tex]

Therefore the correct answer is C.