greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.

Answer:

Pressure is  a scalar quantity defined as per unit area.

Density is the objects ,times its  the acceleration due to gravity.

Explanation:

Pressure is the alternative object increases the area of contact decrease .

Pressure is the force component  to the surface used to calculate pressure.

pressure is that collisions of the gas to container as the per unit time .

pressure is an physical important quantity to play the solid and  fluid .

Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.

Density is the  objects, times, volume of the object that times acceleration objects.

Density is the used to the system complex objects and materials.

Density  force is the weight of a region or objects static fluid.

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

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

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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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:

Maximum acceleration is 800m/s^2

Explanation:

See attached file

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  [tex]f= -75 \ cm = - 0.75 \ m[/tex]

b

  [tex]P = -1.33 \ diopters[/tex]

Explanation:

From the question we are told that

    The  image distance is  [tex]d_i = -75 cm[/tex]

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  [tex]d_o = \infty[/tex]

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

              [tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]

substituting values

             [tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]

=>         [tex]f= -75 \ cm = - 0.75 \ m[/tex]

Generally the power of the corrective lens is  mathematically represented as

        [tex]P = \frac{1}{f}[/tex]

substituting values

       [tex]P = \frac{1}{-0.75}[/tex]

        [tex]P = -1.33 \ diopters[/tex]

Answer:

9.6×10⁹ A

Explanation:

From the question above,

P = VI.................... Equation 1

Where P = Electric power, V = Voltage, I = current.

make I the subject of the equation

I = P/V............. Equation 2

Given: P = 200 kW = 200×10³ W, V = 48000 V.

Substitute these vales into equation 2

I = 200×10³×48000

I = 9.6×10⁹ A.

Hence the current in the line is 9.6×10⁹ A.

Answer:

Q = 832 kJ

Explanation:

It is given that,

Mass of the water, m = 2.5 kg

The latent heat of fusion, L = 333 kJ/kg

We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :

Q = mL

[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]

or

Q = 832 kJ

So, the heat needed to melt the water is 832 kJ.

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:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m  

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I =  3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

Answer:

(b) Increases

Explanation:

The potential energy between two point charges is given as;

[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]

Where;

k is the coulomb's constant

q₁  ans q₂ are the two point charges

r is the distance between the two point charges

Since the two charges have opposite sign;

let q₁ be negative and q₂ be positive

Substitute in these charges we will have

[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]

The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.

Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.

Answer:

B=uonI/2

Explanation:

See attached file

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:

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) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a          --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r          [v = linear velocity of particle, r = radius of circular path]

Therefore, equation (i) becomes;

F = m v²/ r             --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ          -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

Combine equations (ii) and (iii) as follows;

m (v² / r) = qvBsinθ         [divide both side by v]

m v / r = qBsinθ              [make r subject of the formula]

r = (m v) / (qBsinθ)              ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v          --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

Answer:

7.68×10^25kg

Explanation:

The formula for energy used per year is calculated as

Energy used per year =12 x Energy used per month

By substituting Energy used per month in the above formula, we get

Energy used per year =12 x 1600kWh

= 19200kWh

Conversion:

From kWh to J:

1 kWh=3.6 x 10^6 J

Therefore, it is converted to J as

19200 kWh =19200 x 3.6 x 10^6 J

= 6.912×10^10 J

Hence, energy used per year is 6.912×10^10 J

To find the mass that is converted to energy per year.

E = MC^2 ............1

E is the energy used per year

C is the speed of light = 3.0× 10^8m/s

Where E= 6.912×10^10 J

Substituting the values into equation 1

6.912×10^10 J = M × 3.0× 10^8m/s

M = 6.912×10^10 J / (3.0× 10^8m/s)^2

M = 6.912×10^10 J/9×10^16

M = 7.68×10^25kg

Hence the mass to be converted is

7.68×10^25kg

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:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

A) The speed of the scooter at which the driver will feel weightlessness is;

v = 68.586 m/s

B) The apparent weight of both the driver and the scooter at the top of the hill is;

F_net = 779.1 N

We are given;

Mass; m = 159 kg

Radius; r = 480 m

A) Since it's motion about a circular hill, it means we are dealing with centripetal force.

Formula for centripetal force is given as;

F = mv²/r

Now, we want to find the speed of the scooter if the driver feels weightlessness.

This means that the centripetal force would be equal to the gravitational force.

Thus;

mg = mv²/r

m will cancel out to give;

v²/r = g

v² = gr

v = √(gr)

v = √(9.8 × 480)

v = √4704

v = 68.586 m/s

B) Now, he is travelling with speed of;

v = 68.586 m/s

And the radius is 2r

Let's first find the centripetal acceleration from the formula; α = v²/r

Thus; α = 4704/(2 × 480)

α = 4.9 m/s²

Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;

F_app = m(g - α)

F_net = 159(9.8 - 4.9)

F_net = 779.1 N

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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:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

Answer:

450°C

Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .

η = 1 - T2 / T1

η = 1/6 initially

when T2 is reduced by 65°C then η becomes 1/3

Solution

η = 1/6

1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)

When η = 1/3 :

η = 1 - ( T2 - 75 ) / T1

1/3 = 1 - (T2 - 75)/T1.........................(2)

T2 - T1 = -75 [ because T2 is reduced by 75°C ]

T2 = T1 - 75...........................(3)

Put this in (2) :

> 1/3 = 1 - ( T1 - 75 - 75 ) / T1

> 1/3 = 1 - (T1 - 150 ) /T1

> (T1 - 150) / T1 = 1 - 1/3

> ( T1 -150 ) / T1 = 2/3

> 3 ( T1 - 150 ) = 2 T1

> 3 T1 - 450 = 2 T1

Collecting the like terms

3 T1- 2 T1 = 450

T1 = 450

The temperature initially was 450°C

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K  , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

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:

2.5×10¹⁹

Explanation:

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

Answer:

a) Height reached before coming to rest is 42.86 m

b) Energy lost to friction is 17902.45 J

Explanation:

mass of the motorcycle = 185 kg

speed of the towards the hill = 29 m/s

The wheels weigh 12 kg each

Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m

a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height

the kinetic energy of the motorcycle is given as

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

where m is the mass of the motorcycle

v is the velocity of the motorcycle

[tex]KE[/tex]  = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J

This will be converted to potential energy

The potential energy up the hill will be

[tex]PE[/tex] = mgh

where m is the mass

g is acceleration due to gravity 9.81 m/s^2

h is the height reached before coming to rest

[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h

equating the  kinetic energy to the potential energy for energy conservation, we'll have

77792.5 = 1814.85h

height reached before coming to rest  = 77792.5/1814.85 = 42.86 m

b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is

[tex]PE[/tex] = mgh

[tex]PE[/tex]  = 185 x 9.81 x 33 = 59890.05 J

The energy lost to friction will be the kinetic energy minus this potential energy.

energy lost = 77792.5 - 59890.05 = 17902.45 J

A) The motorcycle can coast up the hill by ; 42.86m  

B) The amount of energy lost to friction :  17902.45 J

A) Determine how high the motorcycle can coast up the hill when friction is neglected

apply the formula for kinetic and potential energies

K.E = 1/2 mv²  ---- ( 1 )

P.E = mgH  ---- ( 2 )

As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.

∴ K.E = P.E

1/2 * mv² = mgH

∴ H = ( 1/2 * mv² ) / mg  ---- ( 3 )

where ; m = 185 kg ,  v = 29 m/s ,  g = 9.81

Insert values into equation ( 3 )

H ( height travelled by motorcycle neglecting friction ) =  42.86m  

B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest  

P.E = mgh = 185 * 9.81 * 33  = 59890.05 J

where : h = 33 m , g = 9.81

K.E = 1/2 * mv² = 77792.5 J   ( question A )

∴ Energy lost ( ΔE ) =  ( 77792.5  - 59890.05 ) = 17902.45 J

Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction :  17902.45 J.

Learn more : https://brainly.com/question/3586510

Answer:

Explanation:

a = F / m

where a is acceleration , F is thrust and m is mass

taking log and differentiating

da / a = dF / F - dm / m

(da / a)x 100 = (dF / F)x100 - (dm / m) x100

percentage increase in a = percentage increase in F - percentage increase in m

= percentage increase in acceleration a   = 39 - 13 = 26 %

required increase = 26 %.

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %