Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 ∘ C and 721 mmHg ?

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:

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]

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

Using momentum conserved

mb * (0) + mp * vp = ( mb + mp ) * vf

vf = ( mp / mp + mb) * vp

¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²

Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97  m /s) ²  / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]

Δx = 0.02599 m  ⇒ 2.59 cm

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:

Explanation:

The body can be split into two parts

1 ) cylinder part of mass 7/8 x 64 = 56 kg of radius 20 cm

2 ) In the form of rod attached with cylinder having length 40 cm and mass

of 1/8 x 64 = 8kg

moment of inertia of cylinder

= 1/2 mr²

= .5 x 56 x ( 20 x 10⁻²)²

= 1.12 kg m²

moment of inertia of rods

= 1/3 ml²

= 1/3 x 8 x ( 40 x 10⁻²)²

= .4267 kg m²

Total MI = 1.5467 Kg m²

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:

0.91125 km/s

Explanation:

[tex]v_1[/tex] = Velocity of planet initially = 54 km/s

[tex]r_1[/tex] = Distance from star = 0.54 AU

[tex]v_2[/tex] = Final velocity of planet

[tex]r_2[/tex] = Final distance from star = 32 AU

As the angular momentum of the system is conserved

[tex]mv_1r_1=mv_2r_2\\\Rightarrow v_1r_1=v_2r_2\\\Rightarrow v_2=\dfrac{v_1r_1}{r_2}\\\Rightarrow v_2=\dfrac{54\times 0.54}{32}\\\Rightarrow v_2=0.91125\ km/s[/tex]

When the exoplanet is at its farthest distance from the star the speed is 0.91125 km/s

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

[tex]2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}[/tex]

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

[tex]2t = 2\lambda_{film}[/tex]

[tex]t = \lambda_{film}[/tex]

The index of refraction of soap is given, then

[tex]\lambda_{film} = \frac{\lambda_{vacuum}}{n}[/tex]

Combining the results of all steps we get

[tex]t = \frac{\lambda_{vacuum}}{n}[/tex]

Rearranging, we find

[tex]\lambda_{vacuum} = tn[/tex]

[tex]\lambda_{vacuum} = (278)(1.33)[/tex]

[tex]\lambda_{vacuum} = 369.74nm[/tex]

Answer:

(a) maximum shear stress t=22.5mm

(b)maximum distortion energy t=19.48mm

Explanation:

Ф₁(sigma)=pd/2t

Ф₁=(5×10⁶×1.5)/2t

Ф₁=3.75×10⁶/t

Ф₂=pd/4t

Ф₂=1.875×10⁶/t

(a) According to maximum shear stress theory

|Ф₁|=ФY/1.5

3.75×10⁶/t=250×10⁶/1.5

t=22.5 mm

(b)According to distortion energy theory

Ф₁²+Ф₂²-(Ф₁Ф₂)=(ФY/1.5)²

(3.75×10⁶/t)²+(1.875×10⁶/t)-{(3.75×10⁶/t)(1.875×10⁶/t)}=(250×10⁶/1.5)²

t=19.48mm

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:

option (c)

Explanation:

mass of iron = 0.10 kg

mass of copper = 0.16 kg

rise in temperature, ΔT = 35°C

specific heat of iron = 450 J/kg°C

specific heat of copper = 390 J/kg°C

Heat by iron (H1) = mass of iron x specific heat of iron x ΔT

H1 = 0.10 x 450 x 35 = 1575 J

Heat by copper (H2) = mass of copper x specific heat of copper x ΔT

H1 = 0.16 x 390 x 35 = 2184 J

Total heat H = H1 + H2

H = 1575 + 2184 = 3759 J

by rounding off

H = 4000 J

Answer:b

Explanation:

Given

mass of iron Gear [tex]m_{iron}=0.1 kg[/tex]

mass of copper [tex]m_{cu}=0.16 kg[/tex]

specific heat of iron [tex]c_{iron}=450 J/kg-^{\circ}C[/tex]

specific heat of copper [tex]c_{cu}=390 J/kg-^{\circ}C[/tex]

heat gain by iron gear [tex]=m_{iron}c_{iron}\Delta T[/tex]

[tex]Q_1=0.1\times 450\times 35=1575 J[/tex]

heat gain by iron gear [tex]=m_{cu}c_{cu}\Delta T[/tex]

[tex]Q_2=0.16\times 390\times 35=2184 J[/tex]

Total heat [tex]Q_{net}=Q_1+Q_2[/tex]

[tex]Q_{net}=1575+2184=3759 J\approx 3800 J[/tex]

Answer:

Frequency = 658 Hz

Explanation:

The third harmonics of A is,

[tex]f_{A}=(3)(440Hz)=1320Hz[/tex]

The second harmonics of E is,

[tex]f_{E}=(2)(659Hz)=1318Hz[/tex]

The difference in the two frequencies is,

delta_f = 1320 Hz - 1318 Hz = 2 Hz

The beat frequency between the third harmonic of A and the second harmonic of E is,

delta_f = [tex]3f_{A}-2f_{E}[/tex]

[tex]f_{E}=\frac{3f_{A}-delta_f}{2}[/tex]

We have calculate the frequency of the E string when she hears four beats per second, then

delat_f = 4 Hz

[tex]f_{E}=\frac{3(440Hz)-4Hz}{2}[/tex]

[tex]f_{E}=658Hz[/tex]

Hope this helps!

Answer

given,

mass of the car = 2510 Kg

angle of inclination = 10°

initial speed = v₁ = 20 m./s

skid length = 20 m

coefficient of friction = 0.5

Using conservation of energy

[tex]\Delta E = \Delta KE + \Delta U[/tex]

[tex]\Delta E = \dfrac{1}{2}mv^2 + m g h[/tex]

h = d sin θ

[tex]\Delta E = \dfrac{1}{2}mv^2- m g (dsin\theta)[/tex]

[tex]\Delta E = \dfrac{1}{2}\times 2510 \times 20^2- 2510\times 9.8 \times (20 sin10^0)[/tex]

[tex]\Delta E =416572.04\ J[/tex]

now, calculating the  magnitude of frictional force is equal to

E = F_f d

[tex]F_f = \dfrac{E}{d}[/tex]

[tex]F_f = \dfrac{416572.04}{20}[/tex]

[tex]F_f = 20828\ N[/tex]

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:

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:

D) resistance, resistivity

Explanation:

Resistance is a physical quantity that indicates the opposition of an object to conduct electricity, this quantity depends on different factors such as temperature, material, object length, among other things. The resistance of two objects of the same material may be different, because it depends on the specifications of the object.

On the other hand, resistivity is a more general quantity, since it is assigned to materials and depends only on the nature of the material and its temperature.

So the resistivity is related to the meterial rather than the object.

The answer is: Resistance is a property of an object while resistivity is a property of a material.

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:

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:

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:

.

Explanation:

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]

Answer:

Explanation:

1 . When the Celsius temperature doubles, the Fahrenheit temperature doubles  - FALSE

A change of temperature from 300 F to 600 F changes celsius degree from 148 to 315 °C

2. When heat is added to a system, the temperature must rise.

- FALSE

When we add heat to a system of mixture of ice and water , temperature does not rise until whole of ice melts.

3. In a bimetallic strip of aluminum and brass which curls when heated, the aluminum is on the inside of the curve

- FALSE

Coefficient of linear thermal expansion of aluminium is more so it will be on the outer side.

4. An iron plate has a hole cut in its center. When the plate is heated, the hole gets smaller.

- FALSE

hole will be bigger. There will be photogenic expansion .

5.When the pendulum of a grandfather clock is heated, the clock runs more slowly.

TRUE

Time period is directly proportional to square root of length of pendulum.

when time period increases , clock becomes slow.

6.Cold objects do not radiate heat energy

- FALSE

Object radiates energy  at all temperature .

6 .

The statements are (1)False, (2)False, (3)True, (4)False, (5)True, (6)False

Here are the answers to your statements:

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.

Read more on torque here: https://brainly.com/question/14839816

Answer:

a. 1.0 eV

Explanation:

Given that

Voltage difference ,ΔV = 1 V

From work power energy

Work =Change in the kinetic energy

We know that work on the charge W= q ΔV

For electron ,e= 1.6 x 10⁻¹⁹ C

q=e= 1 x 1.6 x 10⁻¹⁹ C

Change in the kinetic energy

ΔKE= q ΔV

Now by putting the values

ΔKE=  1 x 1.6 x 10⁻¹⁹   x 1   C.V

We can also say that

ΔKE=  1 e.V

Therefore the answer will be a.

a). 1.0 eV

Answer:

Part 1)

- 224.6 J

Part 2)

- 339.4 J

Explanation:

[tex]P[/tex] = Constant pressure acting on gas = 0.643 atm = (0.643) (1.013 x 10⁵) Pa = 0.651 x 10⁵ Pa

[tex]V_{i}[/tex] = initial volume = 5.46 L = 0.00546 m³

[tex]V_{f}[/tex] = final volume = 2.01 L = 0.00201 m³

Work done on the gas is given as

[tex]W = P (V_{f} - V_{i})\\W = (0.651\times10^{5}) ((0.00201) - (0.00546))\\W = - 224.6 J[/tex]

Part 2)

[tex]\Delta U[/tex] = Change in the internal energy

[tex]Q[/tex] = Heat energy escaped = - 564 J

Using First law of thermodynamics

[tex]Q = W + \Delta U\\- 564 = - 224.6 +  \Delta U\\ \Delta U = - 339.4 J[/tex]

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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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:

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

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

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The spring will stretch by 13.79 cm when 51.7 grams is hung on it.

To find the stretch of the spring when a weight of 51.7 grams is hung on it, we will use Hooke's Law. First, we need to convert the weight to Newtons. Since 1 g is equal to 0.0098 N, the weight in Newtons is 51.7 grams * 0.0098 N/g = 0.50546 N. Now we can rearrange Hooke's Law equation, F = -k * s, to solve for s, the stretch of the spring. Plugging in the values, we get 0.50546 N = -k * s. Rearranging further, we have s = 0.50546 N / -3.662 = -0.1379 m. Since the question asks for the answer in centimeters, we can convert -0.1379 m to centimeters by multiplying by 100. Therefore, the spring will stretch by 13.79 cm when 51.7 grams is hung on it.