The wave number (k) and the angular frequency of a wave are 6.2rad/m and 12rad/s respectively. Which of the followings could be the equation of the wave?
a. 8sin(12x-6.2t)
b. 8sin(6.2x-12t)
c. 12 sin (6.2x-8t)
d. all of the above

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:

C is the best answer for this question

Answer:

It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction

Answer:

There could be electric and magnetic fields, oriented in opposite directions

Explanation:

Lorentz force, is the force that may be exerted on a body of a specified magnitude of charge q, moving with a velocity v, in a magnetic field B and in an electric field of intensity E. This Lorentz force is given by; F= qE+qvBsin ϕ

However, if the motion of the particle is opposite to the magnetic field such a that ϕ = 0, then there is no net magnetic force on the charge and it moves freely, with a constant velocity and in a straight line. Hence, there is no magnetic field in the region.

The charge moves with constant speed due to same direction of magnetic and electric field.

There could be electric and magnetic fields that is oriented in the same direction or the other reason is that there is no magnetic field and electric field in that region where the charge moves. If the electric and magnetic field are present at the same direction then it means that it applies no force on the charge.

This is due to more distance from the charge as well as the charge travels away from the field occupies by the magnetic and electric field so we can conclude that the charge moves with constant speed due to same direction of magnetic and electric field.

Learn more: https://brainly.com/question/17132472

Answer:

D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity

Explanation:

Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.

Answer:

Option A

Explanation:

Given that

Distance = Speed / Time

So, they are in inverse relation.

Such that when the time passes, the distance from the reacing point will become less and vice versa.

So, Yes! The more time that passes, the farther away a person riding a bike will be.

Answer:

118 minutes( 2 hours approximately )

Explanation:

Here, we are interested in calculating the orbital period of the satellite

Please check attachment for complete solution

Answer:

T = 7101 s = 118.35 mins = 1.9725 hrs

Explanation:

To solve the question, we apply the formula for gravitational acceleration

a = GM/r², where

a = acceleration due to gravity

G = gravitational constant

M = mass of the earth

r = distance between the satellite and center of the earth

Now, if we make r, subject of formula, we have

r = √(GM/a)

Recall also, that

a = v²/r, making v subject of formula

v = √ar

If we substitute the equation of r into it, we have

v =√a * √r

v =√a * √[√(GM/a)]

v = (GM/a)^¼

Again, remember that period,

T = 2πr/v, we already have v and r, allow have to do is substitute them in

T = 2π * √(GM/a) * [1 / (GM/a)^¼]

T = 2π * (GM/a³)^¼

T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼

T = 6.284 * [(3.982*10^14) / 244.140]^¼

T = 6.284 * (1.63*10^12)^¼

T = 6.284 * 1130

T = 7101 s

T = 118.35 mins

T = 1.9725 hrs

Answer:

The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].

Explanation:

Let be an uniform rod of length L whose origin is located at one one end and axis is perpendicular to the rod, such that:

[tex]\lambda = \frac{dm}{dr}[/tex]

Where:

[tex]\lambda[/tex] - Linear density, measured in kilograms per meter.

[tex]m[/tex] - Mass of the rod, measured in kilograms.

[tex]r[/tex] - Distance of a point of the rod with respect to origin.

Mass differential can translated as:

[tex]dm = \lambda \cdot dr[/tex]

The equation of the moment of inertia is represented by the integral below:

[tex]I = \int\limits^{L}_{0} {r^{2}} \, dm[/tex]

[tex]I = \lambda \int\limits^{L}_{0} {r^{2}} \, dr[/tex]

[tex]I = \lambda \cdot \left(\frac{1}{3}\cdot L^{3} \right)[/tex]

[tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex] (as [tex]m = \lambda \cdot L[/tex])

The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

∴ βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

Answer:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.pls see attached file for explanations

Answer:

2.5×10¹⁹

Explanation:

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

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:

(a) f

(b) r

(c) s

Explanation:

There are two forces on the sphere: weight and buoyancy.

Sum of forces in the y direction:

∑F = ma

B − mg = 0

B = mg

Buoyancy is equal to the weight of the displaced fluid, or ρVg, where ρ is the density of the fluid and V is the displaced volume.

ρVg = mg

ρV = m

V = m/ρ

(a) The mass decreases, so the displaced volume decreases.

(b) The sphere's density is constant and its radius increases, which means its mass increases, so the displaced volume increases.

(c) The mass stays the same, so the displaced volume is the same.

Answer:

   R = r_protón / 2

Explanation:

The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement

      F = m a

Since the magnetic force is perpendicular, the acceleration is centripetal.

       a = v² / R

the magnetic force is

       F = q v x B = q v B sin θ

the field and the speed are perpendicular so the sin 90 = 1

we substitute

          qv B = m v² / R

          R = q v B / m v²

in the exercise they indicate

the charge  q = 2 e

the mass     m = 4 m_protón

        R = 2e v B / 4m_protón v²

we refer the result to the movement of the proton

         R = (e v B / m_proton) 1/2

the data in parentheses correspond to the radius of the proton's orbit

         R = r_protón / 2

Answer:

Explanation:

he moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is:  m L²/ 3  where m is mass and L is length of rod

If the axis of rotation passes through the center of the rod, then the moment of inertia is:   m L² / 12

So for the former case , moment of inertia is higher that that in the later case .

In the former case , the axis is at one extreme end . Hence range of distance of any point on the rod from axis is from zero to L .

In the second case , as axis passes through middle point , this range of distance of any point on the rod from axis is from zero to L / 2 .

Since range of distance from axis is less , moment of inertia too will be less because

Moment of inertia = Σ m r² where r is distance of mass m from axis .

Answer:

μ = 0.03

Explanation:

In order for the trunk not to slide the frictional force between the turntable and the trunk must be equal to the unbalanced force applied on the trunk by the motion of the turntable. Therefore,

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma (Newton's second law of motion)

Frictional Force = μN = μW = μmg

Therefore,

ma = μmg

a = μg

μ = a/g

where,

μ = coefficient of static friction between the trunk and the turntable = ?

a = tangential acceleration of trunk = 0.3 m/s²

g = 9.8 m/s²

Therefore,

μ = (0.3 m/s²)/(9.8 m/s²)

μ = 0.03

Answer:

There are 10 cubes in the box

Explanation:

Let the total number of cubes be x and the total number of pyramids be y

Since there are 17 total objects;

Then;

x + y = 17 •••••••••(i)

Also, the total number of sides of cubes is 6 * x = 6x ( a single cube has 6 sides)

For the pyramid we have 5 * y = 5y

adding both gives the total number of sides

6x + 5y = 95 •••••• (ii)

From i, we cabs say y = 17-x

plug this in ii

6x + 5(17-x) = 95

6x + 85 -5x = 95

6x-5x = 95-85

x = 10

Answer:

x = A sin w t           displacement in SHM

v = A w cos w t      velocity in SHM

PE = 1/2 k x^2 = 1/2 k A^2 sin^2 w t

KE = 1/2 m v^2 = 1/2 m w^2 A^2 cos^2 w t

If KE = PE then

k sin^2 w t = m w^2 cos^2 w t

sin^2 wt / cos^2 w t = tan^2 w t = m w^2 / k

but k / m = w^2

So tan^2 w t = 1  and tan w t = 1   or w t = pi / 4 or theta = 45 deg

Then  x = r sin w t  = r sin 45 = .707 r

Answer:

The impulse is 2145 kg-m/s

The final velocity is -8.34 m/s or 8.34 m/s in he opposite direction.

Explanation:

Force on the rail = 3900 N

Elapsed time of impact = 0.55 s

Impulse is the product of force and the time elapsed on impact

I = Ft

I is the impulse

F is force

t is time

For this case,

Impulse = 3900 x 0.55 = 2145 kg-m/s

If the initial velocity was 2.95 m/s

and mass of car plus driver is 190 kg

neglecting friction, the initial momentum of the car is given as

P = mv1

where P is the momentum

m is the mass of the car and driver

v1 is the initial velocity of the car

initial momentum of the car P = 2.95 x 190 = 560.5 kg-m/s

We know that impulse is equal to the change of momentum, and

change of momentum is initial momentum minus final momentum.

The final momentum = mv2

where v2 is the final momentum of the car.

The problem translates into the equation below

I = mv1 - mv2

imputing values, we have

2145 = 560.5 - 190v2

solving, we have

2145 - 560.5 = -190v2

1584.5 = -190v2

v2 = -1584.5/190 = -8.34 m/s

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Answer:

a) true.

b) True

c) False. In the equation above the mass does not appear

d) True

e) False. Mass does not appear in the equation

f) False. The load even when distributed in the space can be considered concentrated in the center

Explanation:

1. The electric force is given by the relation

           F = k Q e / r2

where k is the Coulomb constant, Q the charge used, e the charge of the electron and r the distance between the two.

 The strength depends on:

a) true.

b) True

c) False. In the equation above the mass does not appear

d) True

e) False. Mass does not appear in the equation

f) False. The load even when distributed in the space can be considered concentrated in the center

two.

a) True

b) Treu

c) Fail

f) false

For a single electron located at a distance from a positive charge, we have:

1. The force on the electron depends on the distance between it and the positive charge (option a) and the charge of both particles (option b and d).      

2. The electric field at the electron's position depends on the distance between the positive charge and it (option a) and the charge of the positive particle (option d).    

The force on a single electron at a distance from the point charge is given by Coulomb's law:

[tex] F = \frac{Kq_{1}q_{2}}{r^{2}} [/tex]    (1)

Where:

As we can see in equation (1), the force on the electron by the positive charge depends on both charges q₁ and q₂, and the distance, so the correct options are:

a. The distance between the positive charge and the electron

b. The charge on the electron

d. The charge of the positive charge

The other options (c, e, f, and g) are incorrect because the electric force does not depend on the particles' masses or their radii.

The electric field (E) at a distance "r" from a point charge is given by:

[tex] E = \frac{Kq_{1}}{r^{2}} [/tex]   (2)

From equation (2), we can see that the electric field is directly proportional to the charge and inversely proportional to the distance of interest (r).  

The electric field at the electron's position is given by the one produced by the positive charge, so the correct options are:

a. The distance between the positive charge and the electron

d. The charge of the positive charge

The other options (b, c, e, f, and g) are incorrect because the electric field is independent of the mass of the charges involved and their radii.

Therefore, the correct options for part 1 are a, b, and d and for part 2 are a and d.

Learn more about the electric field here:

brainly.com/question/13308086

I hope it helps you!

Answer:

C) 8 m/s²

Explanation:

Given:

v₀ = 40 m/s

v = 0 m/s

Δy = 100 m

Find: a

v² = v₀² + 2aΔy

(0 m/s)² = (40 m/s)² + 2a (100 m)

a = -8 m/s²

Answer:

a) 94.26 g/s

b) 4.713 cm/s

c) 2356.5 cm^2

Explanation:

a) velocity of blood through the aorta = 30 cm/s

radius of aorta = 1 cm

density of blood = 1 g/cm^3

Area of the aorta = [tex]\pi r^{2}[/tex] = 3.142 x [tex]1^{2}[/tex] = 3.142 cm^2

Flow rate through the aorta Q = AV

where A is the area of aorta

V is the velocity of blood through the aorta

Q = 3.142 x 30 = 94.26 cm^3/s

Current of blood through aorta [tex]I[/tex] = Qρ

where ρ is the density of blood

[tex]I[/tex] = 94.26 x 1 = 94.26 g/s

b) Velocity of blood in the major aorta = 30 cm/s

Area of the aorta = 3.142 cm^2

Velocity of blood in the major arteries = ?

Area of major arteries = 20 cm^2

From continuity equation

[tex]A_{ao} V_{ao} = A_{ar} V_{ar}[/tex]

where

[tex]V_{ao}[/tex] = velocity of blood in the major arteries

[tex]A_{ao}[/tex] = Area of the aorta

[tex]V_{ar}[/tex] = velocity of blood in the major arteries

[tex]A_{ar}[/tex] = Area of major arteries

substituting values, we have

3.142 x 30 = 20[tex]V_{ar}[/tex]

94.26 = 20[tex]V_{ar}[/tex]

[tex]V_{ar}[/tex]  = 94.26/20 = 4.713 cm/s

c) From continuity equation

[tex]A_{ar} V_{ar} = A_{c} V_{c}[/tex]

where

[tex]A_{ar}[/tex] = Area of major arteries = 20 cm/s

[tex]V_{ar}[/tex] = velocity of blood in the major arteries = 4.713 cm/s

[tex]A_{c}[/tex] = Area of the capillary system = ?

[tex]V_{c}[/tex] = velocity of blood in the capillary system = 0.04 cm/s

substituting values, we have

20 x 4.713 = [tex]A_{c}[/tex]  x 0.04

94.26 = 0.04[tex]A_{c}[/tex]

[tex]A_{c}[/tex]  = 94.26/0.04 = 2356.5 cm^2

This question involves the concepts of volumetric flow rate, continuity equation, and flow velocity.

a) Total current of the blood passing through the aorta is "94.2 g/s".

b) The velocity of blood in major arteries is "4.71 cm/s".

c) The cross-sectional area of the capillary system is "2356.2 cm²".

a)

First, we will find the volumetric flow rate of the blood, using the continuity equation's formula:

[tex]Q=Av[/tex]

where,

Q = volumetric flow rate = ?

A = cross-sectional area of aorta

A =  [tex]\pi(r)^2=\pi(1\ cm)^2= 3.14\ cm^2[/tex]

v = flow velocity = 30 cm/s

Therefore,

[tex]Q=(3.14\ cm^2)(30\ cm/s)[/tex]

Q = 94.25 cm³/s

Now, the blood current will be given as:

I = Qρ

where,

I = current = ?

ρ = blood density = 1 g/cm³

Therefore,

I = (94.2 cm³/s)(1 g/cm³)

I = 94.2 g/s

b)

Now, this volumetric flow rate will be constant in major arteries:

[tex]Q = A_r v_r\\\\v_r=\frac{Q}{A_r}[/tex]

where,

Ar = cross-section area of major arteries = 20 cm²

vr = flow velocity of blood in major arteries = ?

Therefore,

[tex]v_r=\frac{94.25\ cm^3/s}{20\ cm^2}[/tex]

vr = 4.71 cm/s

c)

Now, this volumetric flow rate will be constant in capillaries:

[tex]Q = A_c v_c\\\\A_c=\frac{Q}{v_c}[/tex]

where,

Ac = cross-section area of capillaries = ?

vc = flow velocity of blood in capillaries = 0.04 cm/s

Therefore,

[tex]A_c=\frac{94.25\ cm^3/s}{0.04\ cm/s}[/tex]

Ac = 2356.2 cm²

Learn more about the continuity equation here:

https://brainly.com/question/24905814?referrer=searchResults

Answer:

(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.

Explanation:

The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"

(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:

[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]

Where:

[tex]U_{g,o}[/tex], [tex]U_{g,f}[/tex] - Initial and final gravitational potential energies, measured in joules.

[tex]m[/tex], [tex]M[/tex] - Masses of the rocket and planet Earth, measured in kilograms.

[tex]G[/tex] - Universal gravitation constant, measured in newton-square meters per square kilogram.

[tex]r_{o}[/tex], [tex]r_{f}[/tex] - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.

The initial distance and rocket mass are converted to meters and kilograms, respectively:

[tex]r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)[/tex]

[tex]r_{o} = 6,437,360\,m[/tex]

[tex]m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)[/tex]

[tex]m = 7000\,kg[/tex]

Given that [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]r_{f} \rightarrow +\infty[/tex], the work equation is reduced to this form:

[tex]U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}[/tex]

[tex]U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}[/tex]

[tex]U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J[/tex]

4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.

(b) The needed change in gravitational potential energy is:

[tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex]

The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:

[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]

[tex]\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}[/tex]

[tex]\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}[/tex]

[tex]r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}[/tex]

If [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex]  and [tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex], then:

[tex]r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}[/tex]

[tex]r_{f} \approx 12,874,502.49\,m[/tex]

The final distance with respect to the center of the Earth in miles is:

[tex]r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)[/tex]

[tex]r_{f} = 7999.865\,mi[/tex]

The distance travelled by the rocket is: ([tex]r_{f} = 7999.865\,mi[/tex], [tex]r_{o} = 4000\,mi[/tex])

[tex]\Delta r = r_{f}-r_{o}[/tex]

[tex]\Delta r = 7999.865\,mi - 4000\,mi[/tex]

[tex]\Delta r = 3999.865\,mi[/tex]

The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.

Answer:

A. 2.97x 10^-6C

B. 1.48x10^ -6 C

Explanation:

Pls see attached file

Answer:

1) +9.4 x 10^-7 C

2) +4.72 x 10^-7 C  and  +1.9 x 10^-6 C

Explanation:

The two positive charges will repel each other

Repulsive force on charges = 0.200 N

distance apart = 20.0 cm = 0.2 m

charge on each sphere = ?

Electrical force on charged spheres at a distance is given as

F = [tex]\frac{kQq}{r^{2} }[/tex]

where F is the force on the spheres

k is the Coulomb's constant = 8.98 x 10^9 kg⋅m³⋅s⁻²⋅C⁻²

Q is the charge on of the spheres

q is the charge on the other sphere

r is their distance apart

since the charges are equal, i.e Q = q, the equation becomes

F = [tex]\frac{kQ^{2} }{r^{2} }[/tex]

making Q the subject of the formula

==> Q = [tex]\sqrt{\frac{Fr^{2} }{k} }[/tex]

imputing values into the equation, we have

Q = [tex]\sqrt{\frac{0.2*0.2^{2} }{8.98*10^{9} } }[/tex] = +9.4 x 10^-7 C

If one charge has four times the charge on the other, then

charge on one sphere = q

charge on the other sphere = 4q

product of both charges = [tex]4q^{2}[/tex]

we then have

F = [tex]\frac{4kq^{2} }{r^{2} }[/tex]

making q the subject of the formula

==> q =  [tex]\sqrt{\frac{Fr^{2} }{4k} }[/tex]

imputing values into the equation, we have

q = [tex]\sqrt{\frac{0.2*0.2^{2} }{4*8.98*10^{9} } }[/tex] = +4.72 x 10^-7 C

charge on other sphere = 4q = 4 x 4.72 x 10^-7 = +1.9 x 10^-6 C

Answer:

a)    a = 17.1 m / s², b)    F = 3.04 N

Explanation:

This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities

* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components

* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components

We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is

                θ = 10 -30 = -20º

The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force

              sin (-20) = F_{y} / F

              cos (-20) = Fₓ / F

              F_{y} = F sin (-20)

              Fₓ = F cos (-20)

              F_y = 18 sin (-20) = -6.16 N

              Fₓ = 18 cos (-20) = 16.9 N

The decomposition of the weight is the customary

               sin 30 = Wₓ / W

               cos 30 = W_y / W

               Wₓ = W sin 30 = mg sin 30

                W_y = W cos 30 = m g cos 30

                Wₓ = 0.8 9.8 sin 30 = 3.92 N

                 W_y = 0.8 9.8 cos 30 = 6.79 N

Notice that in the case  the angle is measured with respect to the axis y perpendicular to the plane

Now we can write Newton's second law for each axis

X axis

      Fₓ - fr = m a

Y Axis  

      N - [tex]F_{y}[/tex] - Wy = 0

      N =F_{y} + Wy

      N = 6.16 + 6.79

They both go to the negative side of the axis and

      N = 12.95 N

The friction force has the formula

        fr = μ N

we substitute

        Fₓ - μ N = m a

        a = (Fₓ - μ N) / m

we calculate

       a = (16.9 - 0.25 12.95) / 0.8

       a = 17.1 m / s²

b) now the block slides down with constant speed, therefore the acceleration is zero

ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced

       Newton's law for the x axis

              Fₓ -fr = 0

              Fₓ = fr

              F cos 20 = μ N

              F = μ N / cos 20

we calculate

              F = 0.25 12.95 / cos 20

              F = 3.04 N

this is the force applied at an angle of 10º to the horizontal

Explanation:

proton number + neutron number = atomic mass

30 + 35 = 65

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:

velocity = 2.62m/s

Explanation:

950= (4 x A)/0.12

950 x 0.12 = 4 x A

114 = 4 x A

A = 114/4

A = 28.5m/s²

The maximum velocity of the mass is equal to 1.85,/s when the amplitude of oscillation is 0.12m.

The spring force will be acting on the spring when the spring is stretched or compressed, which opposes the load force. These springs are divided into many types based on how this load force is applied to them.

F = -kx

where k is the spring constant and x is the displacement of the spring attached with mass.

Given, the mass attached to the spring, m = 4.0 Kg

The value of spring constant, k = 950 N/m

The amplitude of oscillation, A = 0.12m

The maximum velocity can be calculated as:

[tex]\frac{mv_{max}^2}{2} =\frac{kA^2}{2}[/tex]

[tex]v_{max} =\sqrt{\frac{kA^2}{m} }[/tex]

Substitute the values of the m, k, and A in the above equation:

Vmax = [tex]=\frac{950N/m(0.12m)^2}{4Kg}[/tex]

Vmax = √3.42 m/s

Vmax = 1.85m/s

Therefore, the maximum velocity of the mass is equal to 1/85 m/s.

Learn more about spring force, here:

https://brainly.com/question/14655680

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Answer:

Explanation:

Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.

Mathematically, F = GMaMb/R² ... 1

G is the gravitational constant

Ma and Mb are the masses of the balls

R is the distance between the balls

If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;

F₂ = GMa(3Mb)/(3R)²

F₂ = 3GMaMb/9R² ... 2

Dividing equation 1 by 2 we will have;

F₂/F = (3GMaMb/9R²)/GMaMb/R²

F₂/F =  3GMaMb/9R² * GMaMb/R²

F₂/F = 3/9

F₂/F = 1/3

F₂ = 1/3 F

This shows that the magnitude of the new attractive force is one-third that of the initial attractive force