A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total elapsed time of 8 years for the trip. An identical clock on Earth shows that the total elapsed time for the trip was 10 years. What was the speed of the spaceship relative to the Earth?

Taking moments about B

N x L'' = W x L'

Here the moment is = force x perpendicular distance between the axis of rotation and the point of applied force .

Here L' = L/3 and L'' = 9

Thus from figure

471 x 9 = W x [tex]\frac{L}{3}[/tex]

But L'' = [tex]\frac{1}{2}[/tex]( L - [tex]\frac{L}{3}[/tex] ) = [tex]\frac{L}{3}[/tex] = 9

Thus W = 471 N

The value of the weight ( W ) is ; 471 N

First step : take moments about B

N * L" = W * L'

Where : L' = L/3,  L" = 9

From the figure

471 * 9 = W * [tex]\frac{L}{3}[/tex]

also:

L" = 1/2 ( L - L/3 ) = L/3 = 9

Hence ; W = 471 N

Hence we can conclude that The value of the weight ( W ) is ; 471 N

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Answer:

a

The total friction drag for the long side of the plate is 107 N

b

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image  

Where the U is the initial velocity = 6 m/s

    So the equation we will be working with is

               [tex]F = \frac{1}{2} \rho C_fAU^2[/tex]

    Where [tex]\rho[/tex] is the density of SAE 10W = [tex]870\ kg/m^3[/tex] This is obtained from the table of density at 20° C

                [tex]C_f[/tex] is the friction drag coefficient

   This coefficient is dependent on the Reynolds number if the Reynolds number is less than [tex]5*10^5[/tex] then the flow is of laminar type and

          [tex]C_f[/tex]  = [tex]\frac{1.328}{\sqrt{Re} }[/tex]

But if the Reynolds number is greater than [tex]5*10^5[/tex] the flow would be of Turbulent type and

         [tex]C_f = \frac{0.074}{Re_E^{0.2}}[/tex]

Where Re is the Reynolds number

   To obtain the  Reynolds number  

                                      [tex]Re = \frac{\rho UL}{\mu}[/tex]

          where L is the length of the long side = 110 cm = 1.1 m

 and [tex]\mu[/tex] is the Dynamic viscosity of SAE 10W oil [tex]= 1.04*10^{-1} kg /m.s[/tex]

  This is gotten from the table of Dynamic viscosity of oil

  So        

                    [tex]Re = \frac{870 *6*1.1}{1.04*10^{-1}}[/tex]

                          [tex]= 55211.54[/tex]

Since            55211.54 < [tex]5.0*10^5[/tex]

Hence

                    [tex]C_f = \frac{1.328}{\sqrt{55211.54} }[/tex]

                          [tex]= 0.00565[/tex]

                 [tex]A[/tex] is the area of the plate  = [tex]\frac{ (110cm)(55cm)}{10000}[/tex] =[tex]0.55m^2[/tex]

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

                [tex]F = \frac{1}{2} \rho C_f(2A)U^2[/tex]

                [tex]F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2[/tex]

                 [tex]F = 107N[/tex]

Considering the short side

            To obtain the Reynolds number

                      [tex]Re = \frac{\rho U b}{\mu}[/tex]

Here b is the short side

                        [tex]Re =\frac{870*6*0,55}{1.04*10^{-1}}[/tex]

                              [tex]=27606[/tex]

Since the value obtained is not greater than [tex]5*10^5[/tex] then the flow is laminar

   And

              [tex]C_f = \frac{1.328}{\sqrt{Re} }[/tex]

                    [tex]= \frac{1.328}{\sqrt{27606} }[/tex]

                   [tex]= 0.00799[/tex]

The next thing to do is to obtain the total friction drag

             [tex]F = \frac{1}{2} \rho C_f(2A)U^2[/tex]

      Substituting values

           [tex]F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2[/tex]

                [tex]= 151.4 N[/tex]

Final answer:

To calculate the total friction drag on the plate immersed in the given fluid stream, use the drag force formula. Compute the drag for both the long and short sides by multiplying the relevant dimensions with the velocity and viscosity.

Explanation:

The drag force per unit area on the plate is given by Fdrag = μSA, where μ is the viscosity of the fluid and A is the area of the plate. Here, the total friction drag can be calculated by multiplying the drag force per unit area by the total area of the plate.

For the long side: Total friction drag = μ(6)(0.55)(1.1)

For the short side: Total friction drag = μ(6)(0.11)(1.1)

To augment the rectifier circuit with a capacitor, connect it in parallel with the load resistor. Calculate the capacitor value to achieve the desired ripple voltage. Determine the average output voltage, fraction of cycle during diode conduction, average diode current, and peak diode current using the ripple voltage and peak output voltage.

To augment the rectifier circuit with a capacitor, we need to connect the capacitor in parallel with the load resistor. The peak-to-peak ripple voltage depends on the value of the capacitor. To calculate the average output voltage, we need to consider the ripple voltage and the peak output voltage. The fraction of the cycle during which the diode conducts, average diode current, and peak diode current can also be determined using the ripple voltage and peak output voltage.

(a) To achieve a peak-to-peak ripple voltage of 10% of the peak output, we can calculate the capacitor value using the equation Vr = (1/2πfc) * (Id / Ic) * Vp, where Vr is the ripple voltage, Id is the diode current, Ic is the capacitor current, Vp is the peak output voltage, f is the frequency, and c is the capacitor value. With the calculated capacitor value, we can find the average output voltage.

(b) The fraction of the cycle during which the diode conducts can be calculated by dividing the time during which the diode conducts by the total time of the cycle.

(c) The average diode current can be calculated by dividing the total charge passing through the diode by the total time period of the cycle.

(d) The peak diode current can be calculated by multiplying the average current by the diode conduction time.

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Answer:

d. There is no way of knowing the phase composition without more information.

Explanation:

There is no way to ascertain the phases present in the container when equilibrium is established because we are not furnished with enough information.

Answer:

The original energy that is left over after the friction does work to remove some is 33.724 J

Explanation:

The original energy that is left in the system can be obtained by removing the energy loss in the system.

Given the mass m = 2.9 kg

        the height h = 2.2 m

        the distance d = 5 m

        coefficient of friction μ = 0.3

         θ = 36.87°

         g = 9.8 m/[tex]s^{2}[/tex]

Since the block is at rest the initial energy can be expressed as;

[tex]E_{i} = mgh[/tex]

   = 2.9 kg x 9.8 m/[tex]s^{2}[/tex] x 2.2 m

   = 62.524 J

The energy loss in the system can be obtained with the expression below;

[tex]E_{loss}[/tex] = (μmgcosθ) x d

The parameters have listed above;

[tex]E_{loss}[/tex]  = 0.3 x 2.9 kg x 9.8 m/[tex]s^{2}[/tex] x cos 36.87° x 5 m

[tex]E_{loss}[/tex] = 28.8 J

The original energy that is left over after the friction does work to remove some can be express as;

[tex]E = E_{i} -E_{loss}[/tex]

E = 62.524 J - 28.8 J

E = 33.724 J

Therefore the original energy that is left over after the friction does work to remove some is 33.724 J

To calculate the remaining kinetic energy of a block sliding down an incline, subtract the work done by friction from the initial potential energy, considering the mass of the block, the height of the incline, the distance slid, the coefficient of friction, and the angle of the incline.

The question concerns the calculation of the kinetic energy (KE) of a block sliding down an incline, taking into account the work done by friction and the conservation of energy. The block starts with potential energy (PE) due to its elevation h and ends with kinetic energy at the bottom of the incline. The force of friction, which depends on the coefficient of friction μ, the gravitational acceleration, and the normal force, does work along the distance d that removes some of this energy.

Initially, the block's total mechanical energy is all potential: PE = mgh. As it slides down the ramp, work done by friction (which is a non-conservative force) is given by Wfriction = μmgcos(θ)d. The final kinetic energy of the block when it reaches the bottom of the incline is calculated by subtracting the work done by friction from the initial potential energy: KE = PE - Wfriction. Substituting the given values and doing the math will provide us with the amount of energy that remains as kinetic when the block reaches the ground.

Answer:

0.44999 m

Explanation:

f = Actual wavelength = 696 Hz

v = Speed of sound in air = 343 m/s

[tex]v_o[/tex] = Velocity of observer = 33.4 m/s

[tex]v_s[/tex] = Velocity of source = 60.3 m/s

From Doppler's effect we have

[tex]f_o=f\left(\dfrac{v-v_o}{v-v_s}\right)\\\Rightarrow f_o=696\left(\dfrac{343-33.4}{343-60.3}\right)\\\Rightarrow f_o=762.22709\ Hz[/tex]

Wavelength is given by

[tex]\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{762.22709}\\\Rightarrow \lambda=0.44999\ m[/tex]

The wavelength at any position in front of the ambulance for the sound from the ambulance’s siren is 0.44999 m.

Answer:

[tex]u_x=55.208\ m.s^{-1}[/tex]

Explanation:

Given:

horizontal distance form the point of shooting where the arrow hits ground, [tex]s=107\ ft[/tex] [tex]=32.614\ m[/tex]

angle below the horizontal form the point of release of arrow where it hits ground, [tex]\theta=3^{\circ}[/tex]

So the height above the ground from where the arrow was shot:

[tex]\tan3^{\circ}=\frac{h}{107}[/tex]

[tex]h=5.6076\ ft=1.71\ m[/tex]

[tex]v_y=\sqrt{2g.h}[/tex]

[tex]v_y=\sqrt{2\times 9.8\times 1.71}[/tex]

[tex]v_y=5.789\ m.s^{-1}[/tex]

Using equation of motion:

[tex]v_y=u_y+g.t[/tex]

where:

t = time taken

[tex]5.789=0+9.8\times t[/tex]

[tex]t=0.591\ s[/tex]

[tex]u_x=\frac{s}{t}[/tex]

[tex]u_x=\frac{32.614}{0.591}[/tex]

[tex]u_x=55.208\ m.s^{-1}[/tex]

The principle of mechanical energy conservation describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops.

The most simplified form of the law of conservation of energy that describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops is the principle of mechanical energy conservation. This principle states that the total mechanical energy of a system remains constant as long as no external forces do work on the system. In this case, as the block slides, the potential energy it loses due to its change in height is transformed into kinetic energy until the block stops and all of its energy is in the form of kinetic energy.

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v=6ft/sec

To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block. First, let's find the acceleration of the block using the force applied. Next, we need to consider the frictional force acting on the block. Finally, we can calculate the net force and use the equation of motion to determine the velocity of the block.

To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block.

First, let's find the acceleration of the block using the force applied. The force applied is given by F = 8t^2 lb, where t is the time in seconds. So, at t = 0, the force is F = 0 lb, and at t = 1, the force is F = 8 lb.

We can now use the equations of motion to find the acceleration. Since the mass of the block is not given, we can assume it to be 10 lb (as mentioned in the question). From the second law of motion, F = ma, where F is the net force, m is the mass, and a is the acceleration. Therefore, 8 = 10a. Solving for a, we get a = 0.8 ft/s^2.

Next, we need to consider the frictional force acting on the block. The coefficient of kinetic friction is given as μs = 0.2. The frictional force, Ff, can be calculated using Ff = μs * N, where N is the normal force. The normal force is equal to the weight of the block, N = mg, where g is the acceleration due to gravity, approximately 32 ft/s^2. So, N = 10 * 32 = 320 lb. Substituting the values, we get Ff = 0.2 * 320 = 64 lb.

Now, we can calculate the net force acting on the block. The net force, Fn, is given as Fn = F - Ff, where F is the force applied. Substituting the values, Fn = 8t^2 - 64 lb.

Finally, we can use the equation of motion, v^2 = u^2 + 2as, to determine the velocity of the block when it moves a distance of 30 ft. Here, v is the final velocity, u is the initial velocity (which is 0 ft/s because the block starts from rest), a is the acceleration, and s is the distance traveled. Substituting the values, v^2 = 0 + 2 * 0.8 * 30. Solving for v, we find

v = 6 ft/s.

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Answer:

a) (vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429

where vᵣₘₛ₁ represents the vᵣₘₛ for 235-UF6 and vᵣₘₛ₂ represents the vᵣₘₛ for 238-UF6.

b) T = 767.34 K

c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.

Explanation:

The vᵣₘₛ for an atom or molecule is given by

vᵣₘₛ = √(3RT/M)

where R = molar gas constant = J/mol.K

T = absolute temperature in Kelvin

M = Molar mass of the molecules.

₁₂

Let the vᵣₘₛ of 235-UF6 be vᵣₘₛ₁

And its molar mass = M₁ = 349.0 g/mol

vᵣₘₛ₁ = √(3RT/M₁)

√(3RT) = vᵣₘₛ₁ × √M₁

For the 238-UF6

Let its vᵣₘₛ be vᵣₘₛ₂

And its molar mass = M₂ = 352.0 g/mol

√(3RT) = vᵣₘₛ₂ × √M₂

Since √(3RT) = √(3RT)

vᵣₘₛ₁ × √M₁ = vᵣₘₛ₂ × √M₂

(vᵣₘₛ₁/vᵣₘₛ₂) = (√M₂/√M₁) = [√(352)/√(349)]

(vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429

b) Recall

vᵣₘₛ₁ = √(3RT/M₁)

vᵣₘₛ₂ = √(3RT/M₂)

(vᵣₘₛ₁ - vᵣₘₛ₂) = 1 m/s

[√(3RT/M₁)] - [√(3RT/M₂)] = 1

R = 8.314 J/mol.K, M₁ = 349.0 g/mol = 0.349 kg/mol, M₂ = 352.0 g/mol = 0.352 kg/mol, T = ?

√T [√(3 × 8.314/0.349) - √(3 × 8.314/0.352) = 1

√T (8.4538 - 8.4177) = 1

√T = 1/0.0361

√T = 27.7

T = 27.7²

T = 767.34 K

c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.

Explanation:

Below is an attachment containing the solution.

Answer:

the question is incomplete, below is the complete question

"Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance."

a. [tex]47KW[/tex]

b. 90.1KW

c. -10.5KW

Explanation:

Data given

mass m=1150kg,

length,l=100m

angle, =30

time=12s

Note that the power is define as the rate change of the energy, note we consider the potiential and the

Hence

[tex]Power,P=\frac{energy}{time} \\P=\frac{mgh+1/2mv^{2}}{t}[/tex]

to determine the height, we draw the diagram of the hill as shown in the attached diagram

a. at constant speed, the kinetic energy is zero.Hence the power is calculated as

[tex]p=\frac{mgh}{t} \\p=\frac{1150*9.81*100sin30}{12}\\p=47*10^{3}W\\[/tex]

b.

for a change in velocity of 30m/s

we have the power to be

[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*30^2}{12}\\ P=47kw +43.1kw\\P=90.1KW[/tex]

c. when i decelerate, we have the power to be

[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*\\((5^2)-(35^2))}{12}\\ P=47kw -57.5kw\\P=-10.5KW[/tex]

Answer:

The energy of an individual photon that comes out of the laser pointer is 1.91 eV

Explanation:

The energy of a photon can be obtained using the expression below

E = hc/λ

where E is the energy;

h is the Planck's constant  = 6.626 x 10-34 Js;

c is the speed of light = 3.00 x [tex]10^{8}[/tex] m/s (speed of light);

λ is the wavelength = 650 nm =650 x [tex]10^{-9}[/tex] m.

E = (6.626 x 10-34 Js) x (3.00 x [tex]10^{8}[/tex] m/s) /650 x [tex]10^{-9}[/tex] m

E = 3.058 x [tex]10^{-19}[/tex] J

1 joule = 6.242 x [tex]10^{18}[/tex] eV

3.058 x [tex]10^{-19}[/tex] J = 3.058 x [tex]10^{-19}[/tex] J x 6.242 x [tex]10^{18}[/tex] eV = 1.91 eV

Therefore the energy of an individual photon that comes out of the laser pointer is 1.91 eV

Answer:

Distance traveled in the next 5.30 seconds is 24.454 meters.

Explanation:

Considering the block moves with a constant acceleration when the force is applied, the average speed would be half of the speed at the end of 5.20 seconds of acceleration.

This means:

Average Speed = Total distance / Time

Average Speed = 12 / 5.20 = 2.307 m/s

Speed at end of 5.20 seconds = 2 * Average Speed

Speed at end of 5.20 seconds = 2 * 2.307

Speed at end of 5.20 seconds = 4.614 m/s

Now that we have the speed at which the block is traveling, and there is no friction to reduce or change the speed, we can solve for the distance covered in the next 5.30 seconds in the following way:

Distance traveled in the next 5.30 seconds:

Distance = Speed * Time

Distance = 4.614 * 5.30

Distance = 24.454 meters

Answer:46.05 mm

Explanation:

Given

[tex]Power=260\ hp\approx 260\times 746=193.96\ KW[/tex]

speed [tex]N=550\ rev/min[/tex]

allowable shear stress [tex](\tau )_{max}=15\ kpsi\approx 103.421\ MPa[/tex]

Power is given by

[tex]P=\frac{2\pi NT}{60}[/tex]

[tex]193.96=\frac{2\pi 550\times T}{60}[/tex]

[tex]T=3367.6\ N-m[/tex]

From Torsion Formula

[tex]\frac{T}{J}=\frac{\tau }{r}-----1[/tex]

where J=Polar section modulus

T=Torque

[tex]\tau [/tex]=shear stress

For square cross section

[tex]r=\frac{a}{2}[/tex]

where a=side of square

[tex]J=\frac{a^4}{6}[/tex]

Substituting the values in equation 1

[tex]\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}[/tex]

[tex]a=0.04605\ m[/tex]

[tex]a=46.05\ mm[/tex]

A square cross-section solid steel shaft of approximately 5 inches on each side can transmit 260 hp at 550 rev/min without exceeding a maximum allowable shear stress of 15 kpsi.

The size of the steel shaft can be calculated by using the power-transmitting capacity of a shaft equation, which states:

P = (16πNT) / (60 * 33000), where P is power in hp, N is rotational speed in rev/min, and T is torque in lb-ft.

We also know that the shear stress (τ) is given by the equation: τ = (16T) / (πd^3), where d is the diameter in inches.

To find the correct torque, we start by rearranging the power equation for T: T = (P * 60 * 33000) / (16πN).

Substituting in the given power and rotational speed, we find that the torque is approximately 13400 lb-ft.

We then substitute this value and the allowable shear stress into the shear stress equation, solving for d to get d ≈ 5 inches.

So, a square cross-section solid steel shaft about 5 inches on each side should be able to transmit the given power at the stated speed without exceeding the maximum allowable shear stress.

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Explanation:

The given data is as follows.

     v = [tex]6.00 \times 10^{5} m/s[/tex]

     B = 0.0525 T,    q = [tex]1.60 \times 10^{-19}[/tex]

     m = [tex]3.34 \times 10^{-27} kg[/tex]

It is known that relation between mass and magnetic field is as follows.

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

or,       r = [tex]\frac{mv^{2}}{Bvq}[/tex]

So, putting the given values into the above formula and we will calculate the radius as follows.

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

               = [tex]\frac{3.34 \times 10^{-27} kg \times 6.00 \times 10^{5} m/s}{0.0525 T \times 1.60 \times 10^{-19}}[/tex]

               = [tex]\frac{20.04 \times 10^{-22}}{0.084 \times 10^{-19}}[/tex]

               = 0.238 m

Thus, we can conclude that radius of the circular path is 0.238 m.

Answer:

The radius is [tex]238.57\times10^{-3}\ m[/tex]

Explanation:

Given that,

Speed [tex]v=6.00\times10^{5}\ m/s[/tex]

Magnetic field = 0.0525 T

We need to calculate the radius

Using relation of centripetal force and magnetic force

[tex]F=qvB[/tex]

[tex]\dfrac{mv^2}{r}=qvB[/tex]

[tex]r=\dfrac{mv^2}{qvB}[/tex]

[tex]r=\dfrac{mv}{qB}[/tex]

Put the value into the formula

[tex]r=\dfrac{3.34\times10^{-27}\times6.00\times10^{5}}{1.60\times10^{-19}\times0.0525}[/tex]

[tex]r=0.238\ m[/tex]

[tex]r=238.57\times10^{-3}\ m[/tex]

Hence, The radius is [tex]238.57\times10^{-3}\ m[/tex]

Answer:

the tractor's power output = 1318.47 watts

Explanation:

Weight of bale = (mg) = (150 kg * 9.81 m/s²) = 1471.5 newtons

Resolve that weight force into its components at right angles to the ramp and parallel to the slope.

Down the slope we have 1471.5sin(15°) = 380.85 N

At right angles to the ramp we have 1471.5cos(15°) = 1421.35 N

OK, now we need the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface.  In this case Fn is equal in magnitude to the component of weight force at right angles to the surface.}

Ff = μ * Fn = (0.40 * 1421.35) = 568.54 N

The bale is not accelerating, so the "pull force" up the incline = component of weight + friction force down

= (380 N + 568.54 N)

= 948.54N

We need the speed in m/s not km/h

5.0 km/h = 5000 m/h

= (5000/3600)

= 1.39 m/s

Power = (948.54 N * 1.39m/s)

= 1318.47 N.m/s

= 1318.47 watts

The power output of the tractor is 1318.47 W

According to the question the weight of bale

mg = 150 × 9.81 = 1471.5 N

Resolving the weight we get:

The normal reaction perpendicular to the ramp is given by:

N = 1471.5cos(15°) = 1421.35 N

The force of friction is given by:

[tex]f = \mu N = 0.40 \times 1421.35\\\\f = 568.54 N[/tex]

The bale is not accelerating, so the force up the incline = component of weight + friction force down

F = mgsin(15°) + f

F = 380 N + 568.54 N

F = 948.54N

Now, the engine speed is:

v = 5.0 km/h = 5000 m/h

v = (5000/3600)

v = 1.39 m/s

The power is defined as the product of force and speed, so:

Power P = Fv

P = 948.54 × 1.39

P = 1318.47 W

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Answer: potential to kinetic and then potential energies

Explanation:

Answer:

KE₁ = PE₂

Explanation:

From the principle of conservation of energy we know that

KE₁ + PE₁ = KE₂ + PE₂

where KE is the kinetic energy and PE is potential energy  

KE = ½mv²

PE = mgd

Initially, you gave a push to crate and it started moving with some velocity

KE₁ = ½mv₁²

Initially, the potential energy is zero since it didnt cover any distance

PE₁ = mgd₁ = 0

At the final state, the crate stopped and its finally velocity become zero, therefore, the kinetic energy is zero.

KE₂ = ½mv₂² = 0

At the final state, the crate covered some distance and the final potential energy is

PE₂ = mgd₂

Therefore, the energy relation becomes

KE₁ + PE₁ = KE₂ + PE₂

KE₁ + 0 = 0 + PE₂

KE₁ = PE₂

Hence, the initial kinetic energy of the system is converted to final potential energy of the system.

Answer:

The maximum height is [tex]h_{max} = \frac{v^2}{2g}[/tex]

Explanation:

From the question we are given that

      [tex]K_{\rm i}+U_{\rm i}=K_{\rm f}+U_{\rm f}\;\;\;\;,[/tex]

     and [tex]K=(1/2)mv^2[/tex]

     while [tex]U=mgh[/tex]

 Now at the minimum height the kinetic energy is maximum and the potential energy is 0

     [tex]K_i \ is \ max[/tex]

     [tex]U_i = 0[/tex]

At maximum height   [tex]h_{max}[/tex]

           The  kinetic energy is 0 and  kinetic energy is

Hence the above equation

                   [tex]K_i = U_f[/tex]

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

Making  [tex]h_{max}[/tex] the subject of the formula we have

              [tex]h_{max} = \frac{v^2}{2g}[/tex]

Answer:

[tex]\epsilon_{max} =12.064\ V[/tex]

Explanation:

Given,

Number of turns, N = 64

angular velocity, ω = 377 rad/s

Magnetic field, B = 0.050 T

Area, a = 0.01 m²

maximum output voltage = ?

We know,

[tex]\epsilon_{max} = NBA\omega[/tex]

[tex]\epsilon_{max} = 64\times 0.05\times 0.01\times 377[/tex]

[tex]\epsilon_{max} =12.064\ V[/tex]

The maximum output voltage is equal to 12.064 V.

Answer:

59.55 cm

Explanation:

Note: A meter stick has a length of 100 cm, and it is balanced at 50 cm.

From the principle of moment,

Sum of clockwise moment = sum of anti clockwise moment

Taking moment about the center

mg(50-x) = m'(y-50)g.................. Equation 1

Where m = first mass, m' = second mass, x = position of the first mass, y = position of the second mass, g = acceleration

make y the subject of the equation

y = (m(50-x)/m')+50.................... Equation 2

y = (0.11(50-17)/0.38)+50

y = (0.11(33)/0.38)+50

y = 9.55+50

y = 59.55 cm

Answer:

a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. Explanation:

The coefficient of volume expansion is [tex]0.000810 \, \text{per} \, ^\circ\text{C}.[/tex]

The coefficient of volume expansion (\(\beta\)) can be calculated using the formula:

[tex]\[\Delta V = V_0 \beta \Delta T\][/tex]

where:

- [tex]\(\Delta V\)[/tex] is the change in volume,

- [tex]\(V_0\)[/tex] is the initial volume,

- [tex]\(\beta\)[/tex] is the coefficient of volume expansion,

- [tex]\(\Delta T\)[/tex] is the change in temperature.

We need to solve for [tex]\(\beta\):[/tex]

[tex]\[\beta = \frac{\Delta V}{V_0 \Delta T}\][/tex]

Substituting the given values:

[tex]\[\beta = \frac{0.0920 \, \text{m}^3}{2.35 \, \text{m}^3 \times 48.5 \, ^\circ\text{C}}\][/tex]

First, calculate the denominator:

[tex]\[2.35 \, \text{m}^3 \times 48.5 \, ^\circ\text{C} = 113.575 \, \text{m}^3 \cdot ^\circ\text{C}\][/tex]

Now, calculate [tex]\(\beta\)[/tex]:

[tex]\[\beta = \frac{0.0920 \, \text{m}^3}{113.575 \, \text{m}^3 \cdot ^\circ\text{C}}\][/tex]

[tex]\[\beta \approx 0.000810 \, \text{per} \, ^\circ\text{C}\][/tex]

Therefore, the coefficient of volume expansion is approximately [tex]\[\beta \approx 8.10 \times 10^{-4} \, ^\circ\text{C}^{-1}\][/tex].

Answer:

The value of  the maximum length of a surface flaw that is possible without fracture a = 1.02 × [tex]10^{-11}[/tex] mm

Explanation:

Given data

Tensile stress [tex]\sigma[/tex] = 50 [tex]\frac{N}{mm^{2} }[/tex]

Specific surface energy [tex]\gamma_{s}[/tex] = 0.5 [tex]\frac{J}{m^{2} }[/tex] = 0.5 × [tex]10^{-6}[/tex] [tex]\frac{J}{mm^{2} }[/tex]

Modulus of elasticity E = 80 × [tex]10^{3}[/tex] [tex]\frac{N}{mm^{2} }[/tex]

The critical stress is given by [tex]\sigma_{c}^{2}[/tex] = [tex]\frac{2 E \gamma_{s} }{\pi a}[/tex] ----- (1)

In the limiting case [tex]\sigma[/tex] = [tex]\sigma_{c}[/tex]

⇒ [tex]\sigma^{2}[/tex] = [tex]\frac{2 E \gamma_{s} }{\pi a}[/tex] ------ (2)

Put all the values in above formula we get,

⇒ [tex]50^{2}[/tex] = 2 × 80 × [tex]10^{3}[/tex] × 0.5 × [tex]10^{-6}[/tex] × [tex]\frac{1}{3.14}[/tex] × [tex]\frac{1}{a}[/tex]

⇒ a = 1.02 × [tex]10^{-11}[/tex] mm

This is the value of  the maximum length of a surface flaw that is possible without fracture.

Answer:

32 W.

Explanation:

Power: This can be defined as the ratio of energy to time. The S.I unit of power is watt(W). The formula for power is given as,

P = W/t.................... Equation 1

Where P = power, W = work done, t = time.

Given: W = 400 J, t = 10 s.

Substitute into equation 1

P = 400/10

P = 40 W.

If the motors that is choose is 80% efficient,

P' = P(0.8)

Where P' = minimum power

P' = 40(0.8)

P' = 32 W.

Answer:102.84 N

Explanation:

Given

Mass of board [tex]m=12.8\ kg[/tex]

Length of board [tex]l=6.1\ m[/tex]

First support is at one end and second support is at a distance of 2.38 m from the other end

Suppose [tex]R_1[/tex] and [tex]R_2[/tex] are the reactions at the two support

Taking moment about the first end we can write

[tex]mg\times \dfrac{6.1}{2}-R_2(6.1-2.38)=0[/tex]

[tex]R_2=\dfrac{12.8\times 9.8 \times 3.05}{3.72}[/tex]

[tex]R_2=102.84\ N[/tex]    

Answer:

[tex]\boxed {q_1=-4q_2}[/tex]

Explanation:

Using the attached figure

Considering that the distance of separation is 2d then  

[tex]F_1=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}[/tex]

Also, considering that distance of separation between  and  is d then

[tex]F_2=\frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}[/tex]

The net force acting on  is

[tex]F=F_1+F_2=0\\F=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}+ \frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}=0\\F=\frac {q_3}{4\pi \epsilon_o d^{2}}(q_2+0.25q_1)=0\\F=0.25q_1+q_2=0[/tex]

Therefore

[tex]\boxed {q_1=-4q_2}[/tex]

The question pertains to a classical physics problem dealing with the equilibrium of three charged particles along a line. It is solved using Coulomb's law to balance the forces acting on the 3rd charge, leading to a condition that determines the values of the charges.

This situation falls under the domain of Physics, specifically the study of electromagnetism. When the charges are in equilibrium, it means the net electrostatic force acting on the third charge, q3, is zero. This equilibrium condition allows us to create an equation. The electrostatic force F between two charges q1 and q2 separated by distance d is described by Coulomb's law: F = k*q1*q2/d^2, where k is Coulomb's constant. It follows then that for q3 to be in equilibrium, the forces from q1 and q2 must balance out. That is, the force of attraction or repulsion between q1 and q3 must equal the force between q2 and q3.

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Explanation:

The given data is as follows.

         Work done by the force, (W) = 79.0 J

         Compression in length (x) = 0.190 m

So, formula for parallel combination of springs  equivalent is as follows.

           [tex]k_{eq} = K_{1} + K_{2}[/tex]

                    = 2k

Hence, work done is as follows.

             W = [tex]\frac{1}{2}k_{eq} \times x^{2}[/tex]

           [tex]k_{eq} = \frac{2W}{x^{2}}[/tex]

                       = [tex]\frac{2 \times 79}{(0.19)^{2}}[/tex]

                       = [tex]\frac{158}{0.0361}[/tex]

                       = 4376.73 N/m

Hence, magnitude of force required to hold the platform is as follows.

               F = [tex]k_{eq}x[/tex]

                  = [tex]4376.73 N/m \times 0.19 m[/tex]

                  = 831.58 N

Thus, we can conclude that magnitude of force you must apply to hold the platform in this position is 831.58 N.

Answer:

2.

Explanation:

Final answer:

To compute Vout with Vin = 0.5 V: a) Using the equation VO = G (V+-V-) with G = 10⁶ would theoretically give 500,000 V, but this is practically limited by the op-amp's supply voltage. b) Using the Golden Rules, assuming an ideal op-amp, Vout would be equal to Vin, thus Vout = 0.5 V.

Explanation:

To compute Vout when Vin = 0.5 V, we can use two different methods:

a) Using the equation VO = G (V+-V-) with G = 10⁶, we assume that for an ideal operational amplifier (op-amp), the voltage at the non-inverting input (V+) is equal to the voltage at the inverting input (V-). Since the question suggests that V- is at circuit common and therefore is 0 V, we have VO = G * (V+ - V-). Substituting values gives VO = 10⁶ * (0.5 V - 0 V) = 500,000 V. However, because real op-amps cannot output such high voltages, Vout would typically be limited by the supply voltage of the op-amp.

b) Using the Golden Rules of op-amps, which state that no current flows into the input terminals of an op-amp and the voltage at the inverting terminal is equal to the voltage at the non-inverting terminal, we can infer that because Vin = 0.5 V is applied to the non-inverting terminal and no current flows into the op-amp, then the voltage difference across the inputs is zero. Thus, Vout must also be 0.5 V to maintain this condition. Therefore, Vout = Vin = 0.5 V under the ideal op-amp approximation.

To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

This leads to the direct conclusion that the resulting energy is 20J.

The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.

Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.

Based on the given question, we can infer that a. the kinetic energy of object increases, mechanical energy decreases.

This refers to the form of energy which makes use of the motion of an object to move.

Hence, we can see that based on the principles of energy conservation, we can see that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

With this in mind, we can see that the resulting energy is 20J.

Read more about conservation of energy here:

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Answer: (a) peak emf = 13.38V

(b) frequency = 591.78Hz

Explanation: Please see the attachments below