There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long (in s) after the asteroid hit the Moon, which is 3.77 ✕ 105 km away, would the light first arrive on Earth?

Answer:

F ’= 1/32 F

We see that the value of the force is the initial force over 32

Explanation:

In this problem the sphere that is touching the others is connected to ground, after each touch,

Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half

½ (Q / 2) = ¼ Q

For the red dial repeat the same scheme

with the first touch the charge is reduced to Q / 2

with the second touch e reduce to ½ (Q / 2) = ¼ Q

with the third toce it is reduced to ½ (¼ Q) = ⅛ Q

Now let's analyze what happens to the electric force

if the force is F for when the charge of each sphere is Q

        F = k Q Q / r²

with the remaining charge strength is

        F ’= k (¼ Q) (⅛ Q) / r²

        F ’= 1/32 k Q Q / r²

        F ’= 1/32 F

We see that the value of the force is the initial force over 32

Answer:

it is the physics that explains how everything works. The best description we have of the. nature of the particles that make up matters and the forces with which they interact. It underlines how atoms work, and so why chemistry and biology work as they do

The position of a particle is r(t)= (4.0t²i+ 2.4j- 5.6tk) m. (Express your answers in vector form.)

(a) Determine its velocity (in m/s) and acceleration (in m/s²) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.)

v(t)= ________m/s

a(t)= ________m/s²

(b) What are its velocity (in m/s) and acceleration (in m/s²) at time t 0?

v(0) =_______ m/s

a(0)=_______ m/s²

(a)

v(t)= [tex]8ti - 5.6k[/tex] m/s

a(t)= 8i m/s²

(b)

v(0) = -5.6k m/s

a(0)= 8i m/s²

From the question, the position of the particle is given by;

r(t)= (4.0t²i+ 2.4j- 5.6tk)        -----------------(i)

(a)

(i)To get the velocity, v(t), of the particle, we'll take the first derivative of the position of the particle (given by equation (i)) with respect to time, t, as follows;

v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]\frac{d(4.0t^2i + 2.4j - 5.6tk)}{dt}[/tex]

v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]8ti +0j - 5.6k[/tex]

v(t) = [tex]8ti - 5.6k[/tex]          --------------------(ii)

(ii) To get the acceleration, a(t), of the particle, we'll take the first derivative of the velocity of the particle (given by equation (ii)) with respect to time, t, as follows;

a(t) = [tex]\frac{dv(t)}{dt}[/tex]  = [tex]\frac{d(8ti - 5.6k)}{dt}[/tex]

a(t) = 8i                    --------------------(iii)

(b)

(i) To get the velocity of the particle at time t = 0, substitute the value of t = 0 into equation (ii) as follows;

v(t) = [tex]8ti - 5.6k[/tex]  

v(0) = 8(0)i - 5.6k

v(0) = 0 - 5.6k

v(0) = -5.6k

(ii) To get the acceleration of the particle at time t = 0, substitute the value of t = 0 into equation (iii) as follows;

a(t) = 8i

a(0) = 8i

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

Answer:

Roche limit = 1.89 of earth radius

Explanation:

We know that,

Mass of earth = 5.972 × 10²⁷ g

New radius = 1.5(old radius) = 1.5(6.371 × 10⁸) = 9.5565 × 10⁸

Density of earth = 5.5132 g/cm³

New density of earth = Mass of earth / (4/3)πr³

New density of earth = 5.972 × 10²⁷ kg / (4/3)(22/7)( 9.5565 × 10⁸)³

New density of earth = 1.634 g/cm³

Roche limit = [2(Density of earth)/(New density of earth)]¹/³r

Roche limit = 1.89 of earth radius

Answer:

r = 0.405m = 40.5cm

Explanation:

In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]       (1)

Fc: centripetal acceleration (tension force on the string) = 12N

m: mass of the ball = 60g = 0.06kg

r: length of the string = ?

v: linear speed of the ball = 9.0m/s

You solve for r in the equation (1) and replace the values of the other parameters:

[tex]r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m[/tex]

The length of the string between Wanda and the ball is 0.405m = 40.5cm

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.

Answer:

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Explanation:

The power dissipation by an electrical circuit is given by the following formula:

Power Dissipation = (Voltage)(Current)

P = VI

but, from Ohm's Law, we know that:

Voltage = (Current)(Resistance)

V = IR

Substituting this in formula of power:

P = (IR)(I)

P = I²R   ---------------- equation 1

Now, if we double the current , then the power dissipated by that circuit will be:

P' = I'²R

where,

I' = 2 I

Therefore,

P' = (2 I)²R

P' = 4 I²R

using equation 1

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Complete question:

Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasing four hours later.

Answer:

The rate at which the distance between the cars is increasing four hours later is 35 mi/h.

Explanation:

Given;

speed of one car, dx/dt = 28 mi/h South

speed of the second car, dy/dt = 21 mi/h West

The distance between the cars is the line joining west to south, which forms a right angled triangle with the two positions.

Apply Pythagoras theorem to evaluate this distance;

let the distance between the cars = z

x² + y² = z² -------- equation (1)

Differentiate with respect to time (t)

[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex] ----- equation (2)

Since the speed of the cars is constant, after 4 hours their different distance will be;

x: 28(4) = 112 mi

y: 21(4) = 84 mi

[tex]z = \sqrt{x^2 + y^2} \\\\z = \sqrt{112^2 + 84^2} \\\\z = 140 \ mi[/tex]

Substitute in the value of x, y, z, dx/dt, dy /dt into equation (2) and solve for dz/dt

[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\\\2(112)(28) + 2(84)(21) = 2(140)\frac{dz}{dt} \\\\9800 = 280\frac{dz}{dt} \\\\\frac{dz}{dt} = \frac{9800}{280} \\\\\frac{dz}{dt} = 35 \ mi/h[/tex]

Therefore, the rate at which the distance between the cars is increasing four hours later is 35 mi/h

Answer:

If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.

Explanation:

The frequency of a vibrating string is primarily based on three factors:

The sounding length (longer is lower, shorter is higher)

The tension on the string (more tension is higher, less is lower)

The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)

To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.

Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.

And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.

Many things can influence the amplitude.

What is producing the sound?

How far are you from the source of the sound? The farther away the smaller the amplitude.

Intervening material. Sound does not travel through walls as well as air.

Depends on what is detecting the wave sound. Ear vs. microphone.

Answer:

The frequency will determine the pitch

the amplitude will determine the loudness

Explanation:

The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.

The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

The magnitude of the electric field.

Magnitude is the size of the object in properties that is determines the size of the object. It also displays the result of the order of class of the object. The direction of the electric field tells us about the position of the field in four different directions. As per the question, the answer is 1.4x10^6N/C.

Learn more about the uniformly charged.

brainly.com/question/12088419.

For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.

When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.

For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.

Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.

Learn more: https://brainly.com/question/5019199

Answer:

answer is D

Explanation:

horizontally, CCW

Answer:

A) his observation is of little importance ,

B) This observation is very important since the movement of the point of light depends on the relationship between the already magnetic electric force

C)   see that in this second case it is 4 times less  

D) the force of gravity is of the order of 10⁻⁴⁰

therefore it is 10²⁸ times less than the electric force,

Explanation:

A) This observation is of little importance since the cacodylate ray tube always emits electrons, regardless of the material of which it is made.

B) This observation is very important since the movement of the point of light depends on the relationship between the already magnetic electric force

C) the elect's load is 1.6 10⁻¹⁹ C its mass is 9.1 10⁻³¹ kg, let's look for its relation

        e / m = 1.6 10⁻¹⁹ / 9.1 10⁻³¹

         e / m = 0.1 758 10 10¹² N

look for this in the case of an atom, let's use the lightest atom hydrogen

the homogenize have an electron of charge 1.6 10⁻¹⁹ C

and a mass of 1.6735575 10⁻²⁷ ka

        e / M = 1.6 10⁻⁻¹⁹ / 1.67 10⁻²⁷

       e / M = 0.96 10⁸ N

We see that in this second case it is 4 times less

D) the force of gravity is of the order of 10⁻⁴⁰

therefore it is 10²⁸ times less than the electric force, therefore it should not contribute to the movement of the light beam

Answer:

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

Explanation:

- To find the direction of the conventional current in the wire you use the following formula:

[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex]       (1)

i: current in the wire = ?

F: magnitude of the magnetic force on the wire = 15.1N

B: magnitude of the magnetic field = 6.1T

l: length of the wire that is affected by the magnetic field = 0.45m

The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).

The direction of the current must be in the +y direction (+^j). In fact, you have:

^j X ^k = ^i

The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):

[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

. . . is carrying more energy than a wave in the same medium with a lower amplitude.

Explanation:

Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR

Answer:

E=7453.99 V/m

Explanation:

The electric field on the charged is given by

E= Kqx/(r^2 +x^2)^3/2

Where;

K= constant of Coulomb's law

q= magnitude of charge= 30.0×10^-9 C

r= radius of the rings= 5 cm or 0.05m

x= distance between the rings = 18cm = 0.18 m

Substituting values;

E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2

E= 48.6/(2.5×10^-3 + 0.0324)^3/2

E= 48.6/(0.0025 + 0.0324)^3/2

E= 48.6/6.52×10^-3

E=7453.99 V/m

Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.

Explanation:

The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:

F = BqVSinØ

If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field

According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.

Answer:

1.06 atm

Explanation:

On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water

The pressure due to a height of water = ρgh

where ρ is the density of water = 1000 kg/m^3

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

h is the height of the water column

The height of water column on the open end = 100 cm = 1 m

pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa

Atmospheric pressure = 101325 Pa

The total pressure on the open end =  101325 Pa + 9810 Pa = 111135 Pa

The pressure due to the water column on the closed end = ρgh

The height of the water in the closed end = 40 cm = 0.4 m

The pressure due to this column of water = 1000 x 9.81 x 0.4 = 3924 Pa

The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa = 107211 Pa

In atm unit, this pressure = 107211/101325 = 1.06 atm

Answer:

The test charge will take the south-west direction indicated in option 6.

Explanation:

The image is shown below.

Since all the charges are positively charged, they will all repel each other. If we consider the force on +q due to +Q and +Q, then we can proceed as follows

The +Q particle at the top left corner of the cube will exert a vertical downward force on +q in the -ve y-axis.

The +Q particle at the bottom right corner of the cube will exert a force on +q towards the horizontal left on the -ve x-axis.

Both of these forces will act at angle of 90°, and therefore, the resultant force will act at an angle of 45° to horizontal and vertical forces.

The result is that the +q charge will move in a south-west direction of the cube.

Question:

Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.652 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer in Joules (J)

Answer:

-263J

Explanation:

Though its difficult and infact impossible to measure the internal energy of a system, the change in internal energy ΔE, can however be determined. This change when it is accompanied by work(W) and transfer of heat(Q) in or out of the system, can be calculated as follows;

ΔE = Q + W       ----------------(i)

Q is negative if heat is lost. It is positive otherwise

W is negative if work is done by the system. It is positive otherwise.

From the question;

Q = -0.652kJ = -652J    {the negative sign shows heat loss}

W = +389J                      {the positive sign shows work done on the system(balloon)}

Substitute these values into equation (i) as follows;

ΔE = -652 + 389

ΔE = -263J

Therefore the change in internal energy is -263J

PS: The negative sign shows that the process is exothermic. This means that the system (balloon) lost some energy to the environment.

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is [tex]\frac{1}{6}[/tex]

So, the equation is

[tex]\frac{1}{6} = 1 - \frac{T_2}{T_1}[/tex]

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is [tex]\frac{1}{3}[/tex]

Therefore the next equation is

[tex]\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}[/tex]

Now solve both the equations

solve equations (1) and (2)

[tex]2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)[/tex]

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

Answer:

Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects (e.g. water striders), usually denser than water, to float and slide on a water surface.

Explanation:

Answer:

It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.

Answer:

The  potential is  [tex]V = 153.659 \ V[/tex]

Explanation:

From the question we are told that

     The diameter of the plastic sphere is  [tex]d = 0.410 \ cm = 0.0041 \ m[/tex]

      The magnitude of the charge is  [tex]q = 35.0 pC = 35.0 *10^{-12} \ C[/tex]

The radius of the plastic sphere is  mathematically evaluated as

          [tex]r = \frac{d}{2}[/tex]

=>     [tex]r = \frac{0.0041}{2}[/tex]

       [tex]r = 0.00205 \ m[/tex]

The  potential near the surface is mathematically represented as

         [tex]V = \frac{k * q}{r }[/tex]

Where k is the Coulombs constant with value [tex]9 *10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

substituting values  

       [tex]V = \frac{9*10^9 * 35 *10^{-12}}{0.00205}[/tex]

       [tex]V = 153.659 \ V[/tex]

Answer:

Yes the monkey will get hit and it will not avoid the dart.

Explanation:

Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.

No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.

What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.

28.29m/s

In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;

([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v            -----------------(i)

Where;

[tex]m_{C}[/tex] = mass of car

[tex]u_{C}[/tex] = initial velocity of car before collision

[tex]m_{D}[/tex] = mass of deer

[tex]u_{D}[/tex] = initial velocity of the deer before collision

v = common velocity with which the car and the deer move after collision

From the question;

[tex]m_{C}[/tex] = 900kg

[tex]u_{C}[/tex] = +30.0m/s    (direction of the motion of the car taken positive)

[tex]m_{D}[/tex] = 150kg

[tex]u_{D}[/tex] = +18.0m/s    (relative to the direction of the car, the velocity of the deer is also positive )

Substitute these values into equation (i) as follows;

(900 x 30.0) + (150 x 18.0) = (900 + 150)v

27000 + 2700 = 1050v

29700 = 1050v

v = [tex]\frac{29700}{1050}[/tex]

v = 28.29m/s

Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.

Answer:

Explanation:

The distance of middle point from centres of spheres will be as follows

From each of 2 cm diameter sphere

R  = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m

Expression for electric field = Q / 4πε R²

Electric field due to positive charge

E₁ = 70  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 33.3 x 10⁴ N/C

Electric field due to negative  charge

E₂ = 40  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 19.02 x 10⁴ N/C

E₁ and E₂ act in the same direction so

Total field = (33.3 + 19.02 ) x 10⁴

= 52.32 x 10⁴ N/C .

Answer:

P= 2414.9 Pa

Explanation:

given

density of air , p = 1.29 kg/m³

speed of air over the upper surface , v₁ = 125 m/s

speed of air over the lower surface , v₂ = 109 m/s

the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)

P = 0.5 × 1.29 × ( 125² - 109²)

P= 0.645(3744)

P = 2414.9 Pa

the pressure difference on an airplane wing is 2414.9 Pa

Answer:

The  acceleration is [tex]a = 5 \ m/s^2[/tex]  

Explanation:

From the  question we are told that

      The  coefficient of kinetic friction is  [tex]\mu_k = 0.24[/tex]

       The coefficient of static friction is  [tex]\mu_s = 0.75[/tex]

       The horizontal force is [tex]F_h = 93 \ N[/tex]

Generally the static frictional force is  mathematically represented as

         [tex]F_F = \mu_s * (m * g )[/tex]

The  static frictional force is the equivalent to the maximum possible force for which the crate does not begin to slide So

       [tex]F_h = F_F = \mu_s * (m * g )[/tex]

=>      [tex]93 = \mu_s * (m * g )[/tex]

=>        [tex]m = \frac{93}{\mu_s * g }[/tex]

substituting values  

          [tex]m = \frac{93}{0.75 * 9.8 }[/tex]

        [tex]m = 12.65 \ kg[/tex]

When the crate is already sliding the frictional force is

      [tex]F_s = \mu_k *(m * g )[/tex]

substituting values  

     [tex]F_s = 0.24 * 12.65 * 9.8[/tex]

     [tex]F_s = 29.82 \ N[/tex]

Now the net force when the horizontal force is applied during sliding is  

      [tex]F_{net} = F_h - F_s[/tex]

substituting values  

     [tex]F_{net} = 93 - 29.8[/tex]

     [tex]F_{net} = 63.2 \ N[/tex]

This  net force is mathematically represented as

     [tex]F_{net } = m * a[/tex]

Where a is the acceleration of the crate

So  

      [tex]a = \frac{F_{net}}{m }[/tex]

      [tex]a = \frac{ 63.2}{12.65 }[/tex]

      [tex]a = 5 \ m/s^2[/tex]