A block of mass M is connected by a string and pulley to a hanging mass m.The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. Find the acceleration of the system and tensions on the string.

Answers

Answer 1

The free body diagram for the block of mass M consists of four forces:

• the block's weight, Mg, pointing downward

• the normal force of the table pushing upward on the block, also with magnitude Mg

• kinetic friction with magnitude µMg = 0.2 Mg, pointing to the left

• tension of magnitude T pulling the block to the right

For the block of mass m, there are only two forces:

• its weight, mg, pulling downward

• tension T pulling upward

The m-block will pull the M-block toward the edge of the table, so we take the right direction to be positive for the M-block, and downward to be positive for the m-block.

Newton's second law gives us

T - 0.2Mg = Ma

mg - T = ma

where a is the acceleration of either block/the system. Adding these equations together eliminates T and we can solve for a :

mg - 0.2 Mg = (m + M) a

a = (m - 0.2M) / (m + M) g

a = 1.96 m/s²

Then the tension in the string is

T = m (g - a)

T = 78.4 N

A Block Of Mass M Is Connected By A String And Pulley To A Hanging Mass M.The Coefficient Of Kinetic

Related Questions

Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Determine how far vertically up the hill the ball reaches when it stops. Show your work.(2 points)

Answers

Answer:

The maximum height is 0.33 m.

Explanation:

initial velocity, u = 8 m/s

final velocity, v = 0 m/s

10% of  kinetic energy is lost in friction.

The kinetic energy used to move up the top,

KE = 10 % of 0.5 mv^2

KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m

Let the maximum height is h.

Use conservation of energy

KE at the bottom = PE at the top

3.2 m = m x 9.8 x h

h = 0.33 m  

The height traveled vertically up the hill by the ball when it stops is 0.327 meter.

Given the following data:

Velocity = 8.0 m/sKinetic energy = 10% lost to friction.

Scientific data:

Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

To determine how far (height) vertically up the hill the ball reaches when it stops:

By applying the law of conservation of energy, we have:

Kinetic energy lost at the bottom = Potential energy gained at the top.

Mathematically, the above expression is given by the formula:

[tex]0.1 \times \frac{1}{2} mv^2 = mgh\\\\0.1 \times \frac{1}{2} v^2 = gh\\\\h=\frac{0.1v^2}{2g}[/tex]

Substituting the given parameters into the formula, we have;

[tex]h=\frac{0.1 \times 8^2}{2\times 9.8} \\\\h=\frac{0.1 \times 64}{19.6} \\\\h=\frac{6.4}{19.6}[/tex]

Height, h = 0.327 meter.

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12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)

Answers

Answer:

See the answer below

Explanation:

After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem

1. Drop a few drops of immersion oil on the slide and view again under high the power objective.

2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.

When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?

Answers

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

A piston-cylinder device contains 5 kg of refrigerant-134a at 0.7 MPa and 60°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24°C. If the surroundings are at 100 kPa and-24°C, determine:
(a) the exergy of the refrigerant at the initial and the final states and
(b) the exergy destroyed during this process.

Answers

Answer:

Yes sure, keep it going, and never give up because your dreams are so important

A) The exergy of the refrigerant at the initial and final states are :

Initial state =  - 135.5285 kJ Final state =  -51.96 kJ

B) The exergy destroyed during this process is : - 1048.4397 kJ

Given data :

Mass ( M )  = 5 kg

P1 = 0.7 Mpa = P2

T1 = 60°C = 333 k

To = 24°C = 297 k

P2 = 100 kPa

A) Determine the exergy at initial and final states

At initial state :

U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k

exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )

                                           = 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)

                                           ≈ - 135.5285 kJ

At final state  :

U = 84.44 kJ / kg , V = 0.0008261 m³/kg,  S = 0.31958 kJ/kg.k

exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )

                                             = -51.96 kJ

B) Determine the exergy destroyed

  exergy destroyed = To * M ( S2 - S1 )

                                 = 297 * 5 ( 0.31958 - 1.0256 )

                                 = - 1048.4397 KJ

Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state =  - 135.5285 kJ, Final state =  -51.96 kJ  and The exergy destroyed during this process is : - 1048.4397 kJ

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A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is

Answers

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

Here,

v = 45m/s

t = 5s

d = v × t

Therefore,

d = 45 × 5

= 225m

Which parts of The Action Potential Are Represented On The ECG?

Answers

Which parts of The Action Potential Are Represented On The ECG?The QRS complex of the electrocardiogram corresponds to the action potential depolarization, while the T wave is associated with ventricular repolarization. Torsades de pointes is associated with the twisting of the QRS complex around the isoelectric line on the electrocardiogram.

Answer:

The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.

Answers

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]

For equipotential surface, dV = 0 so

[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

When an object is in free fall, ____________________.

Answers

Answer:

Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.

Explanation:

Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.

A pump lifts 400 kg of water per hour a height of 4.5 m .
Part A
What is the minimum necessary power output rating of the water pump in watts?
Express your answer using two significant figures.
Part B
What is the minimum necessary power output rating of the water pump in horsepower?
Express your answer using two significant figures.

Answers

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

Question 2 of 10
Which of the following statements is true of an isolated system?
A.The system has energy but no matter.
B.Energy separates the matter in the system from outside matter.
C.The matter within the system does not interact with matter outside the system
D.The matter within the system does not interact with other matter in the system

Answers

Answer:

D

Explanation:

because it is system removed from system and it dosent interact with them

what is the force of a body which have mass of 7 kg​

Answers

Answer:

Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N

Answer:

10 m/s2 or 70 newtons.

Explanation:

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

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

A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.

Answers

Answer:

The acceleration of the payload is 26 m/s2.

Explanation:

length, L = 65 cm =  0.65 m

angular acceleration = 40 rad/s^2

The acceleration is given by

a = angular acceleration x length

a = 40 x 0.65

a = 26 m/s^2

A boy is playing with a water hose, which has an exit area of
10 cm2 and has water flowing at a rate of 2 m/s. If he covers
the opening of the hose with his thumb so that it now has an
open area of 2 cm2, what will be the new exit velocity of the
water?

Answers

Answer:

The exit velocity of water is  B. 15 m/s.

Explanation:

According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.

If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity  is v₂, then,

Rewrite the expression for v₂.

Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.

The exit speed of water from the hose is 15 m/s.

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration

Answers

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]

Therefore, the frequency of this mode of vibration is 138.87 Hz


If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?

Answers

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]

Therefore, the current that flows through the component is 2.42 A

b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velocity of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.

Answers

Answer:

The correct answer is "6.96 rad/s".

Explanation:

The given values are:

Length,

L = 0.6 m

Mass,

m₁ = 0.5 kg

m₂ = 0.2 kg

Initial velocity,

V = 8 m/s

Now,

The final angular velocity will be:

⇒ [tex]\omega =\frac{6m_1V}{(4m_1+3m_2)L}[/tex]

By substituting the values, we get

⇒     [tex]=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}[/tex]

⇒     [tex]=\frac{9.6}{1.38}[/tex]

⇒     [tex]=6.96 \ rad/s[/tex]

An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine

Answers

Answer:

If efficiency is .22 then  W = .22 * Q   where Q is the heat input

Heat Input    Q = 2510 / .22 = 11,400 J

Heat rejected = 11.400 - 2510 = 8900  J of heat wasted

Also, 8900 J / (4.19 J / cal) = 2120 cal

An efficiency is the measure of productivity of an engine. The heat rejected by the engine is 8900 Joules.

What is efficiency?

An efficiency of a heat engine is the ratio of the work done and heat supplied.

Given is the automobile engine has the efficiency 22% and Work done is 2510 Joules.

The efficiency is written as,

η= W / Qs.

The work done is W= Qs - Qr, where Qr is the rejected heat.

The heat rejected can be represented as

Qr = W ( 1/η -1)

Substituting the value into the equation, we get the rejected heat.

Qr = 2510 (1/0.22 -1)

Qr = 8900 Joules.

Thus, the heat rejected by the engine is 8900 Joules.

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Differences between angle of twist and angle of shear

Answers

Answer:

idek

Explanation:

A block of mass 10kg is suspendet at a diameter of 20cm from the centre of a uniform bar im long, what force is required to balance it at its centre of gravity by applying the fore at the other end of the bar?​

Answers

Answer:

4 kg of force

Explanation:

Force = (mass x distance to fulcrum) / length of fulcrum to end

Subsitute values

F = (10 x 20)/50

F =4

A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.

Answers

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i  the question;

mass of sun = M

minimum and maximum distance = r1 and r2 respectively

Now, using Kepler's third law,

" the square of period T of any planet is proportional to the cube of average distance "

T² ∝ R³

average distance a = ( r1 + r2 ) / 2

we know that

T² = 4π²a³ / GM

T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]

T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]

T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

T = 2π √[( r1 + r2 )³ / 8GM ]

Therefore, the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

An elevator with its occupants weighs 2400 N and is supported by a vertical cable. What is the tension in the cable if the elevator is moving up with its speed decreasing at a rate of 1.7

Answers

Answer:

Hope you find it useful. please correct me if I am wrong

The tension in the cable if the elevator is moving upward with its speed decreasing at a rate of 1.7 m/s² is equal to 1983.67 N.

What is tension?

Tension can be described as a force acting along the length of a medium such as a rope, mainly a force carried by a flexible medium.

Tension can be defined as an action-reaction pair of forces acting at each end of the elements. The tension force is in every section of the rope in both directions, apart from the endpoints. Each endpoint of the rope experience tension and force from the weight attached.

Given the force due to the weight of the elevator = mg = 2400N

m = 2400/9.8 Kg

The elevator deaccelerating while moving upward, a = -1.7 m/s²

According to Newton's 3rd law: T - mg = ma

T - 2400 = (2400/9.8) × (-1.7)

T = 2400 - 416.32

T = 1983.67 N

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Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are
separated by a distance of d=0.05 m. At location 'P' between the
two charges, the net electric field is found to be zero.

b. [10 points] The distance between charge qı and location 'P' is
considered to be 'x'. Find the value of 'x' in [cm]

Answers

Answer:

wareffctgggyyggghhhh

A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?

Answers

Answer:

[tex]X=0.0389m[/tex]

Explanation:

From the question we are told that:

Period of spring [tex]T_s=2.25s[/tex]

Initial Position of Mass [tex]x=0.0480m[/tex]

Final Mass period [tex]T_f=5.85s[/tex]

Generally the equation for the Mass location is mathematically given by

[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]

[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]

[tex]X=0.0389m[/tex]

in what part of the plant is glucose suger made?​

Answers

[tex]\large \mid \underline {\bf {{{\color{navy}{Leaf \: \: \: Chloroplast \: ...}}}}} \mid[/tex]

More Information :

Green plants manufacture glucose through a process that requires light, known as photosynthesis.

Glucose is stored in the form of starch in plants.

What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?​

Answers

Answer:

Heat is flowing into the metal.

Explanation:

From the question given above, the following data were obtained:

Mass (M) of iron = 150 g

Initial temperature (T₁) = 25.0°C

Final temperature (T₂) = 73.3°C

Direction of heat flow =?

Next, we shall determine the change in the temperature of iron. This can be obtained as follow:

Initial temperature (T₁) = 25.0 °C

Final temperature (T₂) = 73.3 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 73.3 – 25

ΔT = 48.3 °C

Next, we shall determine the heat transfered. This can be obtained as follow:

Mass (M) of iron = 150 g

Change in temperature (ΔT) = 48.3 °C

Specific heat capacity (C) of iron = 0.450 J/gºC

Heat (Q) transfered =?

Q = MCΔT

Q = 150 × 0.450 × 48.3

Q = 3260.25 J

Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.

A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction

Answers

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)

[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]f_{s}[/tex] - Static friction force, in newtons.

[tex]f_{k}[/tex] - Kinetic friction force, in newtons.

[tex]m[/tex] - Mass, in kilograms.

[tex]g[/tex] - Gravitational constant, in meters per square second.

If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:

[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{s} = 0.273[/tex]

[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.181[/tex]

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 and 1510 N, respectively. The acceleration of the cable is 0.620 m s 2 , upward. What is the tension in the cable (a) below the worker and ( b) above the worker

Answers

Answer:

Explanation:

a)

Below  the worker , the tension in cable  is pulling  the crate . Let the tension be T₁ .

weight of  crate is acting downwards .

Total weight  1510  N.

Net force acting on both = T₁ - 1510

Applying second law of Newton ,

T₁ - 1510 = 1510 / 9.8 x 0.62                [ 1510 / 9.8 = mass of  crate ]

T₁ - 1510 = 95.5

T₁  = 1605.5  N.

b )

Above the worker , the tension in cable  is pulling both the worker and the crate . Let the tension be T₂ .

weight of both worker and crate is acting downwards .

Total weight = 965 + 1510 = 2475 N.

Net force acting on both = T₂ - 2475

Applying second law of Newton ,

T₂ - 2475 = 2475 / 9.8 x 0.62  [ 2475 / 9.8 = mass of both worker and crate ]

T₂ - 2475 = 156.6

T₂  = 2631.6 N.

Which number has four significant figures?

A. 4000
B. 3.008
C. 86.012
D. 0.0001

Answers

a. 4000

This has 4-digits.

Answer:

in my opinion letter d.

Explanation:

Sana pi tama

What are the differences among elements, compounds, and mixtures?

Answers

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm​

Answers

Answer:

The expected radius of the Earth is 3.883 meters.

Explanation:

The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:

[tex]U = K[/tex] (1)

Where:

[tex]U[/tex] - Gravitational potential energy, in joules.

[tex]K[/tex] - Translational kinetic energy, in joules.

Then, we expand the formula by definitions of potential and kinetic energy:

[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

Where:

[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.

[tex]M[/tex] - Mass of the Earth. in kilograms.

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

[tex]r[/tex] - Radius of the Earth, in meters.

[tex]v[/tex] - Escape velocity, in meters per second.

Then, we derive an expression for the escape velocity by clearing it within (2):

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)

If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]

[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]

[tex]r = 3.883\,m[/tex]

The expected radius of the Earth is 3.883 meters.

Other Questions
3. Compare Plato's and Aristotle's views on whether art has a positive or a negative influence, anddiscuss the topic of violence in films and on television with reference to Plato's and Aristotle'stheories. Incorporate into your answer how each might response to a specific film or storyyou've seen or read. Also, consider Nussbaum's comment "We have never lived enough. Ourexperience is, without fiction, too confined and too parochial. Literature extends it, making usreflect and feel about what might otherwise be too distant for feeling." when formulating yourresponse. AYUDAAAA, LES DOY CINCO ESTRELLAS SI LO RESUELVEN BIEN a watch was bought for RS 4444. at what price should it be sould to gain rs444? Please help me this is due in like five minutes Solve the equation and enter the value of d below. 6d + 7 d= 39 in what way do you think overseas commitment " influence the rise and fall of nations in Europe how to Write a simple Java socket programming that a client sends a text and server receives and print. Stuck on this question Floataway Tours has $420,000 that can be use to purchase new rental boats for hire during the summer. The boats can be purchased from two different manufacturers. Floataway Tours would like to purchase at least 50 boats and would like to purchase the same number from Sleekboat as from Racer to maintain goodwill. At the same time,Floataway Tours wishes to have a total seating capacity of at least 200. Required:Formulate this problem as a linear program. Choose the evidence from the text that best supports the ballerinas are graceful and beautiful. "Yup," said George. He tried to think a little about the ballerinas. They weren't really very good-no better than anybody else would have been, anyway. They were burdened with sashweights and bags of birdshot, and their faces were masked, so that no one, seeing a free and graceful gesture or a pretty face, would feel like something the cat drug in. George was toying with the vague notion that maybe dancers shouldn't be handicapped. But he didn't get very far with it before another noise in his ear radio scattered his thoughts.A. George was toying with the vague notion that maybe dancers shouldn't be handicapped.B. They were burdened with sashweights and bags of birdshot, and their faces were masked, so that no one, seeing a free and graceful gesture or a pretty face, would feel like something the cat drug in.C. But he didn't get very far with it before another noise in his ear radio scattered his thoughts.D. "Yup," said George. He tried to think a little about the ballerinas. They weren't really very good-no better than anybody else would have been, anyway. Describe the difference between a benign and malignant tumour Schneider Inc. had salaries payable of $61,400 and $90,700 at the end of Year 1 and Year 2, respectively. During Year 2, Schneider recorded $620,000 in salaries expense in its income statement. Cash outflows for salaries in Year 2 were: -24w^2+(-4w^2)Helppp what did the compromise of 1877 mean for african americans? If 6 playes cost 54$ how much do 30 plates cost In its first year of operations, Roger Company purchased trading securities at a total cost of $53,000. On December 31, the end of Roger's fiscal year, the fair market value of those investments totaled $57,000. As a result of these investments, Roger Company will report Jansen Company reports the following for its ski department for the year 2019. All of its costs are direct, except as noted. Sales $ 605,000 Cost of goods sold 425,000 Salaries 115,000 ($25,200 is indirect) Utilities 14,500 ($5,800 is indirect) Depreciation 48,600 ($17,500 is indirect) Office expenses 28,200 (all indirect) 1. Prepare a departmental income statement for 2019. 2. You want to buy a car and a local bank will lend you $18,500. The loan will be fully amortized over 5 years, and the nominal interest rate would be 4.65%, with interest paid monthly. What is the monthly loan payment Samantha acored 15 points in her laatbasketball game. She made 3 free throwathat are worth 1 point each. The rest ofher pointa came on 2 point field goala,Write an equation that can be used to findthe number of 2 point field goals thatSamantha made(uae p as your variable)Help fasttt Jeff and Cameron are arguing about which one of them is faster. Jeff says "I can run 777 kilometers per hour!" and Cameron says "I can run 100100100 meters per minute!