A system loses 510 J of potential energy. In the process, it does 430 J of work on the environment and the thermal energy increases by 100 J.

Required:
Find the change in kinetic energy.

Answer:Centripetal force that acts an object keep it along a moving circular path.

Explanation:Centripetal force along a path circular of radius(r) with velocity(V) acceleration the center of the path.

a=v/r

object will along moving continue a straight path unless by the external force.External force is the centripetal force.

Centripetal force is to moving in horizontal circle,Centripetal force is not a fundamental force.Gravitational force satellite and orbit of centripetal force.

Centripetal acceleration and centripetal force are used to calculate the motion of objects in circular motion. The main answer to the question is given below:The centripetal force is given by:F = mv²/rwhere m is the mass of the object, v is the speed of the object and r is the radius of the circle. The unit of centripetal force is Newtons (N).The centripetal acceleration is given by:a = v²/rThe unit of centripetal acceleration is meters per second squared

(m/s²).Explanation:When an object moves in a circular motion, there is a force that acts upon it. This force is called the centripetal force. This force always points towards the center of the circle. It is responsible for keeping the object moving in a circular motion.The centripetal force is related to the centripetal acceleration.

The centripetal acceleration is the acceleration of an object moving in a circle. It is always directed towards the center of the circle.The magnitude of the centripetal force is given by:F = mv²/rwhere F is the force, m is the mass of the object, v is the speed of the object and r is the radius of the circle.The magnitude of the centripetal acceleration is given by:a = v²/rwhere a is the acceleration, v is the speed of the object and r is the radius of the circle.

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Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s

final velocity of bullet [tex]v_{2}[/tex] = 140 m/s

loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]

==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

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

70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]

566.28 = [tex]v^{2}[/tex]

[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

Answer:

The tension in the string at that point is 90.75 N

Explanation:

Given;

mass of the object, m = 3 kg

length of string, r = 0.25 m

the angular velocity, ω = 11 rad/s

The tension on string can be equated to the centrifugal force on the object;

T = mω²r

Where;

T is the tension in the string

m is mass of the object

ω is the angular velocity

r is the radius of the circular path

T = 3 x (11)² x 0.25

T = 90.75 N

Therefore, the tension in the string at that point is 90.75 N

Answer:

Tension, T = 0.556 N

Explanation:

It is given that,

Length of vocal cords, l = 22 mm = 0.022 m

Mass per unit length, [tex]\mu=0.0042\ kg/m[/tex]

We need to find the tension is required in the vocal cords in order to produce a tone of middle C of frequency 261.62 Hz. The frequency in terms if tension is given by :

[tex]f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/tex]

T = tension in the vocal cords

[tex]f^2=\dfrac{1}{4l^2}\times \dfrac{T}{\mu}\\\\T=4l^2\mu f^2\\\\T=4\times (0.022)^2\times 0.0042 \times (261.62 )^2\\\\T=0.556\ N[/tex]

So, the tension in the vocal cords is 0.556 N.

Answer:

a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c)  λ = 0.486 m , d)     λ = 0.35 m

Explanation:

a) The speed of a wave on a string is

          v = √T /μ

also all the waves fulfill the relationship

          v = λ f

they indicate that the fundamental frequency is f = 980 Hz.

The wavelength that is fixed at its ends and has a maximum in the center

          L = λ / 2

          λ = 2L

we substitute

           v = 2 L f

let's calculate

           v = 2  0.243  980

           v = 476.28 m / s

b) The tension of the rope

             T = v² μ

the density of the string is

            μ = m / L

            T = v² m / L

            T = 476.28²   0.717 / 0.243

            T = 6.69 10⁵ N

c)          λ = 2L

            λ = 2  0.243

            λ = 0.486 m

d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air

          v = λ f

          λ= v / f

          λ = 343/980

          λ = 0.35 m

Answer:

Radius of gyration = a/2.

Explanation:

So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;

"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."

(NB: I changed the "rho" word to its symbol).

Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.

Therefore, the Radius of gyration = a/2.

Answer:

a) 6738.27 J

b) 61.908 J

c)  [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

[tex]I[/tex] =  [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] =  [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

[tex](I_{1} +I_{2} )w[/tex]

where the subscripts 1 and 2 indicates the values first and second  flywheels

[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

where m is the mass of the car

[tex]v_{car}[/tex] is the velocity of the car

Equating the energy

2246.09 =  [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

making m the subject of the formula

mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Answer:

a

Explanation:

The correct answer would be that the apparent depth of the object is less than the real depth.

The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.

Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.

The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.

Correct option: a

Answer:

I would say its a deep ocean trench

Explanation:

This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim

Answer:

64 times

Explanation:

if increase of 1 gives you 32

then increase of 2 will give you its double

64

If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.

An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.

The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.

The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.

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Answer: D(t) = [tex]8.e^{-0.4t}.cos(\frac{\pi }{6}.t )[/tex]

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = [tex]a.sin(\omega.t)[/tex] or y = [tex]a.cos(\omega.t)[/tex]

where:

|a| is initil displacement

[tex]\frac{2.\pi}{\omega}[/tex] is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

[tex]y=a.e^{-ct}.cos(\omega.t)[/tex] or [tex]y=a.e^{-ct}.sin(\omega.t)[/tex]

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

[tex]y=a.e^{-ct}.cos(\omega.t)[/tex]

period = [tex]\frac{2.\pi}{\omega}[/tex]

12 = [tex]\frac{2.\pi}{\omega}[/tex]

ω = [tex]\frac{\pi}{6}[/tex]

Replacing values:

[tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex]

The equation of displacement, D(t), of a spring with damping factor is [tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex].

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

Answer:

its paramagnetic

Explanation:

i took this quiz

Answer:

the water that remain in the tank in one hour will be 200 lb

Explanation:

Initial mass of water in the tank = 2000 lb

hot water is delivered through the first inlet pipe at a rate of = 0.8 lb/s

cold water is delivered through the second inlet pipe at a rate of = 1.2 lb/s

exit pipe flow rate = 2.5 lb/s

amount of water in the tank after one hour = ?

In one hour, there are 60 x 60 seconds = 3600 sec, therefore

the water through the first inlet pipe in one hour = 0.8 x 3600 = 2880 lb

the water through the second inlet pipe in one hour = 1.2 x 3600 = 4320 lb

the water through the exit in one hour = 2.5 x 3600 = 9000 lb

The total amount of water in the tank = 2000 + 2880 + 4320 = 9200 lb

The total amount of water that leaves the tank = 9000 lb

therefore, in one hour, the water that remain in the tank will be

==> 9200 lb - 9000 lb = 200 lb

Answer:

The y-value  is  z = 0.759 m

Explanation:

From the question we are told that

     The position of the first y-axis is  [tex]y_1 = 0.300 \ m[/tex]

     The current on the first wire is  [tex]I_ 1 = 26.0 \ A[/tex]

      The force per unit length on each wire is  [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]

Generally the force per unit length on first wire is mathematically represented as

                [tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]

Where  [tex]\mu _o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

    substituting values

                    [tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]

                [tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]

                 [tex]I_2 = 17.0 \ A[/tex]

Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis  so

             [tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]

          [tex]I_2 (y_1 - z) = I_1 * y_1[/tex]

substituting values

         [tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]

         z = 0.759 m

Answer:

The current is  [tex]I = 2042\ A[/tex]

Explanation:

From the question we are told that

    The length of the solenoid is  [tex]l = 2.2 \ m[/tex]

    The  radius is  [tex]r_i = 30 \ cm = 0.30 \ m[/tex]

    The number of turn is [tex]N = 1200 \ turns[/tex]

    The  magnetic field is  [tex]B = 1.4 \ T[/tex]

The  magnetic field produced  is mathematically represented as

         [tex]B = \frac{\mu_o * N * I }{l }[/tex]

making [tex]I[/tex] the subject

       [tex]I = \frac{B * l}{\mu_o * N }[/tex]

Where  [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]

 substituting values

        [tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]

        [tex]I = 2042\ A[/tex]

Answer:

a) P = 1028.6 W = 1.03 KW

b) Cost = $0.25

Explanation:

a)

The rate of electrical energy transfer or power of the heater is given as:

P = VI

where,

P = Rate of Electrical Energy Transferred = ?

V = Potential Difference = 120 V

I = Current

but, from Ohm's Law:

V = IR

I = V/R

Therefore,

P = V²/R

where,

R = Resistance = 14 Ω

Therefore,

P = 120²/14

P = 1028.6 W = 1.03 KW

b)

First we find energy used:

Energy = E = Pt

where,

t = time = 5 h

Therefore,

E = (1.03 KW)(5 h)

E = 5.14 KWh

Now, the cost is given as:

Cost = (E)(Unit Price)

Cost = (5.14 KWh)($0.05/KWh)

Cost = $0.25

Answer:

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = 14691.456 J   is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

[tex]P = \frac{w}{t}[/tex]

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds

Answer:

A: 28°

B. 1x10^-3M

Explanation:

See attached file

Answer:

t = 27.5

Explanation:

[tex]3 + 5 -220t = 0[/tex]

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

Answer: 70.5 cm

Explanation:

The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.

You will use the moment techniques.

That is,

Sum of the clockwise moment = sum of anticlockwise moments

Please find the attached file for the remaining explanation and solution.

Answer:

Explanation:

know that there is no external force on skater and the stone so the total momentum of the system will remains constant

so we will have

here we have

so the skater will move back with above speed

now the deceleration of the skater is due to friction given as

Answer:

(a) 3.553 m/s

(b) 21.46 m

Explanation:

(a) Applying the law of of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u'  = mv+m'v'.................. Equation 1

Where m and m' are the mass of skater and stone respectively,  u and u' are the initial velocity of skater and stone respectively, v and v' are the final velocity of the skater and the stone respectively.

Note, u = 0 m/s, u' = 0 m/s

Therefore,

0 = mv+m'v'

-mv = m'v'................ Equation 2

make v the subject of the equation

v = -m'v'/m............. Equation 3

Given: m = 45 kg, m' = 7.65 kg, v' = 20.9 m/s

Substitute into equation 3

v = 7.65(20.9)/45

v = -3.553 m/s

Hence the speed of the skater = 3.553 m/s

(b) F = mgμ..............Equation 4

But F = ma

Therefore,

ma = mgμ

a = gμ............... Equation 5

Where a = acceleration of the skater, g = acceleration due to gravity, μ = coefficient of kinetic friction

Given: μ = 0.03, g = 9.8 m/s²

Substitute into equation 5

a = 0.03(9.8)

a = 0.294 m/s²

Using the equation of motion,

v² = u²+2as............. Equation 6

Where s = distance moved by the skater.

note that u = 0 m/s.

therefore,

v² = 2as

s = v²/2a................ Equation 7

Given: v = 3.553 m/s, a = 0.294

Substitute into equation 7

s = 3.553²/(2×0.294)

s = 12.62/0.588

s = 21.46 m

Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.

The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The wavelength is   [tex]\lambda= \frac{2 \pi }{k}[/tex]

Explanation:

From the question we are told that  

      The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]

       The magnetic field is  [tex]\= B = B_0 sin (kx -wt) \r k[/tex]

From the above equation

and  k is the wave number which is mathematically represented as

        [tex]k = \frac{2 \pi }{\lambda }[/tex]

=>     [tex]\lambda= \frac{2 \pi }{k}[/tex]

Where [tex]\lambda[/tex] is the wavelength

Answer:

the expression is False

Explanation:

From the kinematics equations we can find the speed of a body in a clean fall

        v = v₀ - g t

         v² = V₀² - 2 g y

If the body starts from rest, the initial speed is zero (vo = 0)

            v= √ (2g y)

In the first equation it gives us the relationship between speed and time.

With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.

Therefore the expression is False

Answer:

the direction of angular momentum = EAST

Explanation:

given

Direction of position = r = north

Direction of velocity = v = up

angular momentum = L = m(r x v)

where m is the mass, r is the radius, v is the velocity

utilizing the right hand rule, the right finger heading towards the course of position vector and curl them toward direction of velocity, at that point stretch thumb will show the bearing of the angular momentum.

then L = north x up = East

Complete  Question

In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?

Answer:

a

  [tex]\theta = 0.3819^o[/tex]

b

  [tex]y = 0.00386 \ m[/tex]

Explanation:

From the question we are told that

    The slit separation is  [tex]d = 150 \lambda[/tex]

    The  distance from the screen is  [tex]D = 57.9 \ cm = 0.579 \ m[/tex]

Generally the condition for constructive interference is mathematically represented as

            [tex]dsin (\theta ) = n * \lambda[/tex]

=>        [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]

where  n is the order of the maxima  and value is 1 because we are considering the central maximum and an adjacent maximum

     and  [tex]\lambda[/tex] is the wavelength of the light

So

       [tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]

       [tex]\theta = 0.3819^o[/tex]

Generally the distance between the maxima is mathematically represented as

       [tex]y = D tan (\theta )[/tex]

=>    [tex]y = 0.579 tan (0.3819 )[/tex]

=>    [tex]y = 0.00386 \ m[/tex]

Answer:

b. Rodger work = 1300 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of the force.

From the question,

Work is done by Rodger using a force of 100 N  in pushing the skateboard through a distance of 13.0 m.

W = F×d............. Equation 1

Where W = work done, F = force, d = distance.

Given: F = 100 N, d = 13 m

Substitute these values into equation 1

W = 100(13)

W = 1300 J.

Hence the right option is b. Rodger work = 1300 J

Answer:

2.5×10¹⁹

Explanation:

4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second

Answer:

a) 6.57 m/s

b) 53.75 J

c) 6.37 m/s

d) -98.297 J

Explanation:

mass of player = [tex]m_{p}[/tex] = 117.5 kg

speed of player = [tex]v_{p}[/tex] = 6.5 m/s

mass of ball = [tex]m_{b}[/tex] = 0.43 kg

velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] =  2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = 6.57 m/s

b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = 53.75 J  gained

c) if they travel in opposite direction, equation becomes

[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 = 6.37 m/s

d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex]  = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is 98.297 J  lost

Answer:

E = 12.25 x 10³ N/C = 12.25 KN/C

Explanation:

In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,

Electrostatic Force = Weight

E q = mg

where,

E = Electric Field = ?

m = Mass of the Charge = 5 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C

Therefore,

E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)

E = 0.049 N/4 x 10⁻⁶ C

E = 12.25 x 10³ N/C = 12.25 KN/C