What is the escape speed on a spherical asteroid whose radius is 517 km and whose gravitational acceleration at the surface is 0.636 m/s2

Answers

Answer 1

Answer:

810.94 m/s

Explanation:

Applying,

v = √(2gR)............. Equation 1

Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth

From the question,

Given: g = 0.636 m/s², R = 517 km = 517000 m

Substitute these values into equation 1

v = √(2×0.636×517000)

v = √(657624)

v = 810.94 m/s

Hence, the escape velocity is 810.94 m/s


Related Questions

how many rings does saturn have

Answers

Answer:

From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Answers

Answer:

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Explanation:

Describe the change in motion and kinetic energy of the particles as thermal energy is added to a liquid. Which change of state might happen?
please ill put brainliest!!!

Answers

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away.

Answer:

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away

Select the correct answer.
What is abstraction?
OA. the concept that software architecture can be separated into modules and that each module can be examined independently
OB. the process of containing information within a module, preventing any crossover or access to Irrelevant information
OC. the process of splitting a program both horizontally and vertically
OD. the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains
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4:

Answers

Answer:

OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains

Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.

What is abstraction?

Abstraction is the practice of removing anything from a set of core features by eliminating or deleting attributes.

One of the three core ideas of object-oriented programming is abstraction order to decrease complexity and maximize efficiency, a programmer uses abstraction to conceal all but the important facts about an object.

Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.

Hence option D is correct.

To learn more about the abstraction refer to the link;

https://brainly.com/question/13072603

You drive 7.5 km in a straight line in a direction east of north.

a. Find the distances you would have to drive straight east and then straight north to arrive at the same point.
b. Show that you still arrive at the same point if the east and north legs are reversed in order.

Answers

Answer:

a)  a = 5.3 km, b) sum fulfills the commutative property

Explanation:

This is a vector exercise, If you drive east from north, we can find the vector using the Pythagorean theorem

              R² = a² + b²

where R is the resultant vector R = 7.5 km and the others are the legs

If we assume that the two legs are equal to = be

             R² = 2 a²

             r = √2 a

             a = r /√2

we calculate

             a = 7.5 /√2

             a = 5.3 km

therefore, you must drive 5.3 km east and then 5.3 km north and you will reach the same point

b) As the sum fulfills the commutative property, the order of the elements does not alter the result

         a + b = b + a

therefore, it does not matter in what order the path is carried out, it always reaches the same end point

write any two physical hazard occuring in the late choldhood​

Answers

Answer:

Hazards during late childhood

Health Problems: Chronic health ailments like T.B., Pneumonia etc will hinder the child's motor abilities.Accidents: School age children are more adventurous in nature, they run fast, play hard, ride bicycles and scooters and engage in a variety of sports.

The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.

Answers

Answer:

valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.

I hope this helps

what aspect of the US justice system has its roots in Jewish scripture?​

Answers

The aspect of the US justice system that has its roots in Jewish scripture is:

the idea that all people are subject to the same rules and laws.

It is the doctrine of "equality before the law."  Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc.  Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism.  It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.

Answer:

answer is C

the idea that all people are subject to the same rules and laws

Explanation:

hope this helps!

two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential​

Answers

Answer:

B has greater potential

Explanation:

We know;

Potential Energy (PE) = mgh

where, m=mass of body

g=acceleration due to gravity

h=height of body

From the formula,

PE is directly proportional to the mass of the body

so the body with greater mass has greater potential.

If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg? *

Answers

Assuming ideal conditions, Boyle's law says that

P₁ V₁ = P₂ V₂

where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂

==>   V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.​

Answers

Answer:

v² = u² + 2as

v² = 3600 + 6400

v² = 10000

v = 100

Explanation:

final velocity is 100 m/s

initial velocity=u=60m/sAcceleration=10m/s^2=aDistance=s=320mFinal velocity=v

According to third equation of kinematics

[tex]\boxed{\sf v^2-u^2=2as}[/tex]

[tex]\\ \sf\longmapsto v^2=u^2+2as[/tex]

[tex]\\ \sf\longmapsto v^2=(60)^2+2(10)(320)[/tex]

[tex]\\ \sf\longmapsto v^2=3600+3400[/tex]

[tex]\\ \sf\longmapsto v^2=10000[/tex]

[tex]\\ \sf\longmapsto v=\sqrt{10000}[/tex]

[tex]\\ \sf\longmapsto v=100m/s[/tex]

3 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Explanation:

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

Multiply both sides by 2 and then divide by m to get

[tex]\dfrac{2qV}{m} = v^2[/tex]

Take the square root of both sides to get

[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]

What is a measure between the difference in start and end positions?

Answers

Answer:

Displacement

General Formulas and Concepts:

Kinematics

Displacement vs Total Distance

Explanation:

Displacement is the difference between the start position and end position.

Total Distance is the entire distance traveled between the start and end position.

Topic: AP Physics 1 Algebra-Based

Unit: Kinematics

Một vật được ném lên trên theo phương thẳng đứng. Người quan sát
thấy vật đó đi qua vị trí có độ cao h hai lần và khoảng thời gian giữa hai lần đó là
t. Tìm vận tốc ban đầu và thời gian chuyển động của vật từ lúc ném đến khi vật
rơi về vị trí ban đầu.

Answers

Answer:

Language -English plz I cant understand

Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?

Answers

Answer:

Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.

En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.

La persona A aplica una fuerza:

Fa = -3N

La persona B aplica una fuerza:

Fb = 5N

La persona C aplica una fuerza Fc, la cual aún no conocemos.

Pero sabemos que la caja está en equilibrio físico, por lo que:

Fa + Fb + Fc = 0N

reemplazando los valores que conocemos, obtenemos:

-3N + 5N + Fc = 0N

Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.

Fc = 0N + 3N - 5N

Fc = -2N

Podemos concluir que la persona C aplica una fuerza horizontal de -2N

Cho các máy cắt sử dụng trong công nghiệp có ký hiệu trên nhãn thiết bị: C350; B500. Hãy tính dòng điện bảo vệ ngắn mạch và dòng điện bảo vệ quá tải của từng thiết bị?

Answers

Answer:

ask in the English then I can help you

Explanation:

please mark me as brain list

A car whose tire have radii 50cm travels at 20km/h. what is the angular velocity of the tires?

Answers

Radius=r=50cm=0.5mVelocity=20km/h=v

Convert to m/s

[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]

We know

[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]

[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]

A nylon string on a tennis racket is under a tension of 285 N . If its diameter is 1.10 mm , by how much is it lengthened from its untensioned length of 29.0 cm ? Use ENylon=5.00×109N/m2.

Answers

Answer:

1.74×10⁻³ m

Explanation:

Applying,

ε = Stress/strain............. Equation 1

Where ε = Young's modulus

But,

Stress = F/A.............. Equation 2

Where F = Force, A = Area

Strain = e/L.............. Equation 3

e = extension, L = Length.

Substitute equation 2 and 3 into equation 1

ε = (F/A)/(e/L) = FL/eA............. Equation 4

From the question,

Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,

A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²

Substitute these values into equation 4

5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)

Solve for e

e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)

e = 82.65/4.74925×10⁴

e = 1.74×10⁻³ m

Để sử dụng nguồn điện xoay chiều 220V/50Hz thắp sáng bóng đèn 12V/3W, ta chọn điện trở giảm áp có giá trị:

Answers

Explanation:

Hi Linda,

How's it going?

Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?

How's the new job?

Looking forward to hearing from you!

Helga

A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnetic field travels in a circular path of radius 18.0 mm. What is the charge of the particle

Answers

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

[tex]a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C[/tex]

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

The force of gravity is an inverse square law. This means that, if you double the distance between two large masses, the gravitational force between them Group of answer choices weakens by a factor of 4. strengthens by a factor of 4. weakens by a factor of 2. also doubles. is unaffected.

Answers

Answer:

the force decreases by a factor of 4

Explanation:

The expression for the law of universal gravitation is

          F = [tex]G \frac{m_1m_2}{r^2}[/tex]

let's call the force Fo for the distance r

          F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]

They indicate that the distance doubles

          r ’= 2 r

we substitute

          F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]

          F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]

         

          F = ¼ F₀

consequently the correct answer is that the force decreases by a factor of 4

If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.

Let the initial force be F

Let the initial distance apart be r

Thus, we can obtain the final force as follow:

Initial force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = 2r

Final force (F₂) =?

F = GM₁M₂ / r²

Fr² = GM₁M₂ (constant)

Thus,

F₁r₁² = F₂r₂²

Fr² = F₂(2r)²

Fr² = F₂4r²

Divide both side by 4r²

F₂ = Fr² /4r²

F₂ = F / 4

From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4

Learn more: https://brainly.com/question/975812

How to calculate voltage U1 ?

Please help!

Answers

Answer:

he is a baby art and design

In first case a mass M is split into two parts with one part being 1/6.334 th of the original mass. In second case M is split into two equal parts. In both the cases the two parts are separated by same distance. What ratio of the magnitude of the gravitational force in first case to the magnitude of the gravitational force in the second case

Answers

Answer:

[tex]F_r=0.132:0.25[/tex]

Explanation:

From the question we are told that:

[tex]M_1=M*\frac{1}{6.334}[/tex]

Therefore

[tex]M_2=M-M*\frac_{1}{6.334}[/tex]

[tex]M_2=M*\frac{5.334}{6.334}[/tex]

Generally the equation for Gravitational force of attraction is mathematically given by

For Unequal split

[tex]F=\frac{GM_1M_2}{d^2}[/tex]

[tex]F=\frac{G(M*\frac_{1}{6.334})(M*\frac{5.334}{6.334})}{d^2}[/tex]

[tex]F=\frac{GM^2}{d^2}*(0.132)[/tex]

For equal split

[tex]F=\frac{GM_1M_2}{d^2}[/tex]

[tex]F=\frac{G(\frac{M}{2})((\frac{M}{2}}{d^2}[/tex]

[tex]F=0.25 \frac{GM^2}{d^2}[/tex]

Therefore the ratio of the gravitational force is

[tex]F_r=0.132:0.25[/tex]

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________

Answers

Answer:  

Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].

Explanation:  

E inside [tex]= KQr_{1} /R^{3}[/tex]  

so if r1 will be the same then  

E  [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

Answer:

The electric field is E/8.

Explanation:

The electric field due to a solid sphere of uniform charge density inside it is given by

[tex]E =\frac{\rho r}{3}[/tex]

where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.

For case I:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]

So, electric field at a distance r is

[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]

Case II:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]

So, the electric field at a distance r is

[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]

A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

B = 0.013(-j) T

Explanation:

Given that,

The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]

Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]

We need to find the magnitude and the direction of the magnetic field. At equilibrium,

[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]

The velocity is in +z direction, force in +x direction, then the field must be in -y direction.

A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.

Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.

Answers

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

          ∑ τ = 0

          60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

           90 - 117 + 30 x = 0

           x = 27/30

           x = 0.9 m       

In summary the center of mass is on the side of the lightest weight x = 0.9 m

A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?

Answers

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds


(I was only able to do A and B)

Answer:

Explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

Answers

Answer:

Explanation:

The formula for determining the Emf induced in a loop is:

[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]

[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]

[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]

[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]

where;

square area A = ( l²)

l² = 6.0 cm = 6.0 × 10⁻²

[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]

[tex]\varepsilon =18 \times 10^{6} \ V[/tex]

Recall that:

The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m

We can as well say that the length of the copper wire = perimeter of the square loop;

The perimeter of the square loop = 4L

Thus, the length of the copper wire  = 4 (6.0 × 10⁻² )m

= 24× 10⁻² m

Finally, the current in the loop is determined from the formula:

V = IR

where,

V = voltage

I = current and R = resistance of the wire

Making "I" the subject:

I = V/R

where;

[tex]R = \dfrac{\rho \times l}{A}[/tex]

[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]

[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]

[tex]R = 0.001283 \ ohms[/tex]

[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]

I = 14.029 mA

You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.

Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?

Answers

Answer:

0.37 m

Explanation:

Given :

Window height, [tex]h_1[/tex] = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex]   seconds.

Assuming that the speed of the pot just above the window is v then,

[tex]h_2=ut+\frac{1}{2}gt^2[/tex]

[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]

[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]

[tex]$v= 2.69$[/tex] m/s

Initially the pot was dropped from rest. So,  u = 0.

If it has fallen from a height of h above the window then,

[tex]$h = \frac{v^2}{2g}$[/tex]

[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]

h = 0.37 m

The exponent of the exponential function contains RC for the given circuit, which is called the time constant. Use the units of R and C to find units of RC. Write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify until you get something simple. Show your work below.

Answers

Answer:

The unit of the time constant RC is the second

Explanation:

The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.

Since R = V/I

Unit of R = unit of V/unit of I

Unit of R = V/A

Ω = V/A

Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V

Since C = Q/V

Unit of C = unit of Q/unit of V

Unit of C = C/V

F = C/V

Now the time constant equals RC.

So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A

Also. we know that the  1 Ampere = 1 Coulomb per second

1 A = 1 C/s

So, substituting 1 A in the denominator, we have

unit of RC =  C/A = C ÷ C/s = s

So, the unit of RC = s = second

So, the unit of the time constant RC is the second

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