A transparent. dielectric coating is applied to glass (εr = 4.μr=1, σ= 0) to eliminate the reflection of red light (wavelength in air of 750 nm).

a. What is the required dielectric constant and minimum thickness of the coating?
b. If violet light (wavelength in air of 420 nm) is shone onto this glass coating (6-0). what percentage of the incident power will be reflected?

Answers

Answer 1

Answer:

a) Dielectric constant ( λ ) = 750 * 10^-9 m

   minimum thickness of coating ( d )  = 187.5 nm

b) 3.6%

Explanation:

Given data:

wavelength of red light in air = 750 nm

εr = 4

μr = 1,  σ = 0

a) Determine the required dielectric constant and min thickness of coating used

Refractive index of coating ( n ) = √εr * μr =  √4*1 = 2

the refractive index of glass( ng)  = 1.5 which is < 2

λ = 750 * 10^-9 m

Dielectric constant ( λ ) = 750 * 10^-9 m

To determine the minimum thickness we will apply the formula below

d = m λ/2n ;  where  m = 1

∴ d = 750 nm / 2 ( 2 )

      = 187.5 nm

minimum thickness of coating ( d )  = 187.5 nm

b) Determine the percentage of the incident power that will be reflected

R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2

   = [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2

   = 0.03628 =  3.6%


Related Questions

Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?

Answers

I don’t agree what if it blows up?

Xác định ứng lực trong các thanh BC,
CF và FE của hệ giàn và chỉ rõ các thanh
chịu kéo hay nén. Cho P1=P2=600 lb,
P3=800 lb

Answers

Answer:

............................................

Explanation:

Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.

Answers

Answer:

c

Explanation:

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.

What is the matter?

Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.

These different states of matter have different characteristics according to which they vary their volume and shape.

It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same,  therefore the correct answer is C.

To earn more about the matter here,refer to the link;

brainly.com/question/9402776

#SPJ2

A physics student likes to study while listening to loud music. If electricity costs 12.00$/kWh (kilowatt-hour), how much would it cost the student to run a 220 W stereo system 8.0 hours per day for 10 days of studying?

Answers

Answer:

the cost of running the stereo is $211.2

Explanation:

Given;

cost of electricity, = 12.00$/kWh

power consumed by the stereo system, P = 220 W

duration of the power consumption, t = 8 hours

number of days, = 10 days

total time of the power consumption = 8 hours x 10 = 80 hours

power consumed in kW = 220 W / 1000 = 0.22 kW

Energy consumed = 0.22 kW x 80 h = 17.6 kWh

The cost of using 17.6 kWh

= 17.6 x $12

= $211.2

Therefore, the cost of running the stereo is $211.2

An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J

Answers

Answer:

b) - 3.3 J

Explanation:

Given;

mass, m = 2 kg

initial extension of the spring, x = 6 cm = 0.06 m

The weight of the mass on the spring;

W = mg

where;

g is acceleration due to gravity = 9.81 m/s²

W = 2 x 9.81

W = 19.62 N

The spring constant is calculated as;

W = kx

k = W/x

k = 19.62 / 0.06

k = 327 N/m

The work done by the spring when it is extended to an additional 10 cm;

work done = force x distance

distance = extension, x =  10 cm = 0.1 m

The work done by the spring opposes the applied force by acting in opposite direction to the force.

W = - Fx

W = - (kx) x

W = - kx²

W = - (327) x (0.1)²

W = - 3.27 J

W ≅ - 3.3 J

Therefore, the work done by the spring by opposing the applied force is -3.3 J

A car can go from 0 to 60 m/s in 5 seconds. What is the acceleration?

A) 50 m/s^2

B) 6 m/s^2

C) 12 m/s^2

D) 300 m/s^2

I think it’s 12 because I did the difference divided by 5 but some places said there was no acceleration

Answers

Answer:

12

Explanation:

i agree the answer is 12 because the acceleration is given by the difference in the velocity divided by the time taken

a=v-u/t

60-0/5

=12m/s²

I hope this helps

A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage of 0.0320 V is applied across the ends of the wire. From this information calculate the resistance of the wire.

Answers

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

We know that from Ohm's law

V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

How can I solve the following statement?

What is the magnitude of the electric field at a point midway between a −8.3μC and a +7.8μC charge 9.2cm apart? Assume no other charges are nearby.

Answers

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d =  9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

[tex]E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C[/tex] leftwards

The electric field due to the charge q' at midpoint is

[tex]E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C[/tex]

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density​

Answers

Answer:

velocity is the answer of this question.

Answer:

Velocity is the right answer ok

A generator uses a coil that has 270 turns and a 0.48-T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

Answers

Answer:

The total length of wire is 0.24 m.

Explanation:

Number of turns, N = 270

magnetic field, B = 0.48 T

frequency, f = 60 Hz

rms value of emf = 120 V

maximum value of emf, Vo = 1.414 x 120 = 169.68 V

let the area of square is A and the side is L.

The maximum emf is given by

Vo = N B A w

169.68 = 270 x 0.48 x A x 2 x 3.14 x 60

A = 3.5 x 10^-3 m^2

So,

L = 0.0589 m

Total length of wire, P = 4 L = 4 x 0.0589 = 0.24 m

If Light is travels from air into pure water with an incident angle of 30°. What is the angle of refraction?

Answers

Explanation:

For air, n1 = 1.00003; for water, n2 = 1.3330

Given: θ2 = 30 degrees, then

θ1 = arcsin [(n2/n1) sin θ2]

= arcsin [(1.3330/1.0003) sin (40)]

= 58.93 degrees

Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.

A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels

Answers

Answer: [tex]53\ rev/min[/tex]

Explanation:

Given

angular speed of wheel is [tex]\omega_1 =530\ rev/min[/tex]

Another wheel of 9 times the rotational inertia is coupled with initial wheel

Suppose the initial wheel has moment of inertia as I

Coupled disc has [tex]9I[/tex] as rotational inertia

Conserving angular momentum,

[tex]\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min[/tex]

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?

Answers

Answer:

The charges are + 74.3 μC and - 74.3 μC

Explanation:

Let the charges be q and q'.

Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by

F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m

When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.

They now repel each other.

So, the magnitude of the force of repulsion is given by

F' = k[(q + q')/2][(q + q')/2]/r²

F' = k[(q + q')²/4r²

Since the magnitude of the force of attraction and repulsion are the same, we have that

F = F'

kqq'/r² = k[(q + q')²/4r²

qq' = (q + q')²/4

(q + q')² = 4qq'

q² + 2qq' + q'² = 4qq'

q² + 2qq' - 4qq' + q'² = 0

q² - 2qq' + q'² = 0

(q - q')² = 0

q - q' = 0

q = q'

Substituting q = q' into F, we have

F = kqq'/r²

F = kq²/r²

making q subject of the formula, we have

q² = Fr²/k

q = √(Fr²/k)

q = r√(F/k)

Substituting the values of the variables into the equation, we have

q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)

q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)

q = 0.189 m(0.3923 × 10⁻³ C/m)

q = 0.0743 × 10⁻³ C

q = 74.3 × 10⁻³ × 10⁻³ C

q = 74.3 × 10⁻⁶ C

q = 74.3 μC

Since q and q' initially attract, it implies that they initially had opposite charges.

So, q = 74.3 μC and q' = -74.3 μC

So, the charges are + 74.3 μC and - 74.3 μC

Which of the following statements about magnetism is TRUE?
a) The direction of the magnetic force on a current-carrying wire is parallel to the wire.
b) Magnetic poles always occur in pairs (N and S).
c) Magnetic field lines begin at south poles and end on north poles.
d) Moving charges do not experience a force in magnetic fields.

Answers

(B) option magnetic poles always occur in pairs (N andS ).

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision

Answers

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

 

a) the initial impulse is

         p₀ = m v₁₀ + 0

         p₀ = 0.6 2

         p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

        p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

 

we write the final moment for each axis

X axis

         p₀ₓ = 1.2 kg m / s

         p_{fx} = m v1f cos 20 + m v2f cos θ

         p₀ = p_f

        1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

         1.2482 = v_{2f} cos θ

Y axis  

        p_{oy} = 0

        p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

        0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

        0.2736 = v_{2f} sin θ

we write our system of equations

         0.2736 = v_{2f} sin θ

         1.2482 = v_{2f} cos θ

divide to solve

         0.219 = tan θ

          θ = tan⁻¹ 0.21919

          θ = 12.36

let's look for speed

            0.2736 = v_{2f} sin θ

             v_{2f} = 0.2736 / sin 12.36

            v_{2f} = 1.278 m / s

Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.

Answers

Answer:

An object in either state of equilibrium must have no net force acting on it.

Explanation:

Answer: An object in either state of equilibrium must have no net force acting on it.

Explanation:

It is easy to produce a potential difference of several thousand volts between your body and the floor by scuffing your shoes across a nylon carpet. When you touch a metal doorknob, you get a mild shock. Yet contact with a power line of comparable voltage would probably be fatal. Why is there a difference?

Answers

Answer:

In sof the friction with the nylon is very small and the current with the line e is largeummary

Explanation:

When we rub the shoes against the carpet, static electricity is produced, when you touch the metal knob you close the circuit and the current can circulate to three of the body, the value of this current is of the order of micro volts, for which a small discharge, the power that circulates through the body is very small of the order of 0.005 A

When you touch the power line, the voltage may be small, but the amount of current that can generate them is of the order of tens of amps, the electric shock is much greater per location.

In general there is a rule that if the body resumes more than P = 4000W the discharge could be fatal.

In sof the friction with the nylon is very small and the current with the line e is largeummary, the difference is that the current at the stop , so the paper that passes through the body is large and can be dangerous.

In contact with metal doorknob, get a mild shock  while with power line of same voltage, fatal the body as the amount of current is more.

What is charging by friction?

When the two materials are rubbed each other, then the electric charged generated between them.

This charging of materials, due to the rubbing of two materials against each other, is called the charging by friction.

It is easy to produce a potential difference of several thousand volts between the body and the floor by scuffing your shoes across a nylon carpet.

In this case, the potential difference may be higher, but the value of current is very low. Thus, when the body touches a metal doorknob, it will get a mild shock.

Now, in another case, the contact with a power line of comparable voltage would probably be fatal. This is because in the power line the amount of current is much higher.

Hence, in contact with metal doorknob, get a mild shock  while with power line of same voltage, fatal the body as the amount of current is more.

Learn more about the charging by friction here;

https://brainly.com/question/8418256

A cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel. The bucket starts from rest and falls for 4.00 s.

Required:
a. What is the linear acceleration of the falling bucket?
b. How far does it drop?
c. What is the angular acceleration of the reel?

Answers

I think the answer is C

A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Answers

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

a method of reducing friction​

Answers

Answer:

Lubrication

Explanation:

People oil/lubricate bicycle chains because the chain turns around the cogs and rub together so this help with friction.

Hope this helps :)

Answer:

The method of reducing friction are :

i) In moving parts of machine friction can be reduced by using a ball bearing between the moving surfaces

ii) The bodies of aeroplane ,ship ,boat etc are made streamlined to reduce friction.

iii) Friction can be reduced by polishing rough surfaces. For example : carrom boards are highly polished to reduce friction.

I hope this help you:)

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.

Answers

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant speed. If the driver of the car hears the horn at a frequency of 246 Hz, then what is the speed of the car? Use 340 m/s for the speed of the sound

Answers

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn [tex]f_o=228\ Hz[/tex]

Apparent frequency [tex]f'=246\ Hz[/tex]

Speed of sound is [tex]V=340\ m/s[/tex]

Doppler frequency is

[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]

Where,

[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]

Insert values

[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]

Thus, the speed of the car is [tex]26.84\ m/s[/tex]

1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?

Answers

Explanation:

the car will be brought back

Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance:___________.
A) must be directed inward.
B) acts like it originated in a point charge Q at the center of the distribution.
C) must be directed outward.
D) must be greater than zero.
E) must be zero.

Answers

Answer:

Q at the center of the distribution.

Explanation:

The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.

what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...

Answers

(C)

Explanation:

[tex]E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})[/tex]

[tex]= 5.3×10^{-18}\:\text{J}[/tex]

Answer:

It's D

Explanation:

It's from alvs

Sunlight above the Earth's atmosphere has an intensity of 1.36 kW/m2. If this is reflected straight back from a mirror that has only a small recoil, the light's momentum is exactly reversed, giving the mirror twice the incident momentum. (a) Calculate the force per square meter of mirror (in N/m2). N/m2 (b) Very low mass mirrors can be constructed in the near weightlessness of space, and attached to a spaceship to sail it. Once done, the average mass per square meter of the spaceship is 0.170 kg. Find the acceleration (in m/s2) of the spaceship if all other forces are balanced. m/s2 (c) How fast (in m/s) is it moving 24 hours later

Answers

Answer:

a)  [tex]F=9.2*10^{-6}N/m^2[/tex]

b)  [tex]a=5.4*10^{-4}m/s[/tex]

c)  [tex]v=46.65m/s[/tex]

Explanation:

From the question we are told that:

Intensity I= 1.36 kW/m2=>1360W/m

b)Average mass per square meter m = 0.170 kg

c) [tex]T=24hrs[/tex]

a)

Generally the equation for force per square meter  is mathematically given by

[tex]F=\frac{2E}{C}[/tex]

[tex]F=\frac{2*1360}{3*10^8}[/tex]

[tex]F=9.2*10^{-6}N/m^2[/tex]

b)

Generally the equation for force  is mathematically given by

F=ma

Therefore

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{9.2*10^{-6}N/m^2}{0.0170}[/tex]

[tex]a=5.4*10^{-4}m/s[/tex]

c)

Generally the Newton's equation for Motion is mathematically given by

[tex]v=u+at[/tex]

[tex]v=0+5.4*10^{-4}m/s*(24*3600)[/tex]

[tex]v=46.65m/s[/tex]

Imagine that you and your lab partner are standing on smooth ice holding onto opposite ends of a long rope. There is no friction between your feet and the ice. You tug gently on the rope. (a)Describe the motion of you and your lab partner before, during, and after the tug. Be as specific as you can. Expla

Answers

Answer:

Following are the complete solution to the given question:

Explanation:

In this question, before tug they are at rest, thus their initial momentum is 0 [tex]\pi=0[/tex], according to the law of conservation [tex]p_f=0[/tex] thus both should move in the opposite direction whose sum of the momentums is equaled to zero as there is no friction they will continue to move with their velocities until their collision. Its collision doesn't happen it will continue up to an infinite distance.

The motion of both you and your partner depends on the tension in the rope.

We must note that frictional force is the force that acts between two surfaces in contact with each other. The frictional force depends on the nature of the surfaces in contact.

Now, since there is no friction between your leg and the ice. The motion of both you and your partner depends on the tension in the rope.

Learn more about tension force: https://brainly.com/question/1195122

If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?

Answers

Answer:

opposite direction

Explanation:

An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.

Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.

Answer:

Explanation:

negatively charged particle is placed inside uniform electric field

The force on the charge due to the electric field is

F = q E

when the charge is negative so the force on the charge is opposite to the direction of electric field.

The electric field is opposite to the force.

Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.

a. This low value is possible because the magnetic field of Saturn is so strong.
b. This low value is possible because the magnetic field of Saturn is so weak.
c. This low value is possible because the density of Saturn is so high.
d. This low value is possible because the density of Saturn is so low.

Answers

Answer:

Explanation:

That is an amazing fact.

The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.

The answer is D

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B

Answers

Answer:

The speed at B is 16.18 ft/s .

Explanation:

Speed at A, u = 11 ft/s

Speed at C, v' = 18 ft/s

Time from A to C = 5 s

Time from B to C = 1.3 s

Let the speed of car at B is v.

Let the acceleration is a.

From A to B

Use first equation of motion

v = u + a t

18 = 11 + a x 5

a = 1.4 ft/s^2

Let the time from A to B is t' .

t' = 5 - 1.3 = 3.7 s

Use first equation of motion from A to B

v = 11 + 1.4 x 3.7 = 16.18 ft/s  

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