A plank 2.65 m long is supported by a cable, and a small ledge. The cable attaches 79.7 cm from the ledge, and makes an angle of 40.3 with the plank. It can hold 2648 N. How close (in meters) to the end of the plank can a 76.4 kg person walk before the cable breaks? Ignore the mass of the plank itself.

Answer:

10^-7 C

Explanation:

m = 1 g = 10^-3 kg, E = 200,000 N/C, a = 20 m/s^2, u = 0

Let q be the charge on bead

Force = m a = q E

a = q E / m

q = m a / E = (10^-3 x 20) / 200000 = 10^-7 C

Answer:

Part a)

F = 0.056 N

Part b)

a = 28.13 m/s/s

Explanation:

Part a)

Electrostatic force between two charged balls is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now we will have

[tex]q_1 = q_2 = 50 nC[/tex]

r = 2 cm

now we will have

[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.02^2}[/tex]

so here we have

[tex]F = 0.056 N[/tex]

Part b)

Now due to above force the acceleration of one ball which is released is given as

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

[tex]a = \frac{0.056}{2 \times 10^{-3}}[/tex]

[tex]a = 28.13 m/s^2[/tex]

Answer:

a) Left to right

b) 1.51 A

Explanation:

a)

The gravitational force on the rod due to its mass is in downward direction. hence to levitate the rod, the magnetic force on the rod must be in upward direction.

The magnetic field is inward to page and magnetic force must be upward. Using right hand rule, the current must be flowing from left to right.

Left to right

b)

L = length of the copper rod = 0.570 m

m = mass of the rod = 0.059 kg

B = magnitude of magnetic field in the region = 0.670 T

θ = Angle between the magnetic field and rod = 90

i = current flowing throw the rod = ?

The magnetic force on the rod balances the gravitational force on the rod. hence

Magnetic force = gravitational force

mg = i B L Sinθ

(0.059) (9.8) = i (0.670) (0.570) Sin90

i = 1.51 A

Answer:

Charge, q = 3.48 μC

Explanation:

It is given that,

Electric field, [tex]E=1.25\times 10^5\ N/C[/tex]

Distance form positive charge, r = 0.5 meters

We need to find the charge. It is given by the formula as follows :

[tex]E=\dfrac{kq}{r^2}[/tex]

[tex]q=\dfrac{Er^2}{k}[/tex]

k = electrostatic constant

[tex]q=\dfrac{1.25\times 10^5\ N/C\times (0.5\ m)^2}{9\times 10^9}[/tex]

q = 0.00000347 C

or

[tex]q=3.48\times 10^{-6}\ C[/tex]

[tex]q=3.48\ \mu C[/tex]

So, the charge is 3.48 μC. Hence, this is the required solution.

Answer:

Power, P = 112500 watts

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of the car, u = 0

Final velocity of the car, v = 30 m/s

Time taken, t = 6 s

The minimum average power delivered by the engine is given by :

P = W/t

Where

W = work done

t = time taken

Work done = change in kinetic energy

So, [tex]P=\dfrac{\Delta K}{t}=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (30\ m/s)^2}{6\ s}[/tex]

P = 112500 watts

So, the minimum average power delivered by the engine is 112500 watts. Hence, this is the required solution.

Answer:

v = 335.7 m/s

Explanation:

As we know that speed of sound in air is given by the formula

[tex]v = \sqrt{\frac{\gamma RT}{M}}[/tex]

now we have

[tex]\gamma = 1.4[/tex] For air

M = 29 g/mol = 0.029 kg/mol

T = 8 degree Celcius = 273 + 8 = 281 K

R = 8.31 J/mol K

now from above formula we have

[tex]v = \sqrt{\frac{(1.4)(8.31)(281)}{0.029}}[/tex]

[tex]v = 335.7 m/s[/tex]

Answer:

125.09 Hz

Explanation:

Given : A person stands 5.20 from one speaker and 4.10 m away from the other speaker.

Distance between the speakers is 5.20 - 4.10 =1.10 m

We know that

For destructive interferences

        n . λ / 2

where n =1,3,5,7....

Therefore difference between the speakers is

5.10 - 4.20 = 1 X 0.5 λ

λ = 2.2

Given velocity of sound is V = 344 m/s

Therefore frequency, f = [tex]\frac{v}{\lambda }[/tex]

                                      = [tex]\frac{344}{2.2 }[/tex]

                                      = 156.36 Hz

Now, the third lowest frequency is given by

              λ = (5.20-4.10) X 5 X 0.5

                 = 2.75 m

Therefore frequency, f = [tex]\frac{v}{\lambda}[/tex]

                                      = [tex]\frac{344}{2.75}[/tex]

                                      = 125.09 Hz

Therefore third lowest frequency is 125.09 Hz

The current induced in the loop can be found using Faraday's law of electromagnetic induction. By calculating the rate of change of magnetic flux through the loop and using Ohm's law, we can determine the induced current in the loop. The given values for the solenoid's current, number of turns, radius, and the loop's resistance are used in the calculations.

The current induced in the loop can be found using Faraday's law of electromagnetic induction.

Based on the given information, the rate of change in current (di/dt) in the solenoid is 5 A/s. The number of turns per unit length of the solenoid is 17 turns/cm, so the total number of turns in the solenoid is 17 * 2π * 6 cm. The radius of the loop is 8 cm and its resistance is 4Ω.

Using Faraday's law, the induced emf in the loop is given by the equation ε = -N * (dΦ/dt), where N is the total number of turns in the solenoid and (dΦ/dt) is the rate of change of magnetic flux through the loop. The magnetic flux through the loop depends on the magnetic field produced by the solenoid and the area of the loop.

By substituting the given values and using the formula for magnetic flux, we can find the induced emf. Finally, Ohm's law can be used to calculate the induced current in the loop by dividing the induced emf by the loop's resistance.

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Answer:

6538.8 Angstrom

Explanation:

work function, w = 1.9 eV = 1.9 x 1.6 x 10^-19 J = 3.04 x 10^-19 J

Let the longest wavelength is λ.

W = h c / λ

λ = h c / W

λ = (6.626 x 10^-34 x 3 x 10^8) / (3.04 x 10^-19)

λ = 6.5388 x 10^-7 m = 6538.8 Angstrom

Thus, the longest wavelength is 6538.8 Angstrom.

Answer:

Voltage, V = 0.0741 volts

Explanation:

It is given that,

Width of membrane, [tex]d=8.47\ nm=8.47\times 10^{-9}\ m[/tex]

Electric field strength, [tex]E=8.76\ MV/m=8.76\times 10^6\ V/m[/tex]

We need to find the voltage across membrane. Electric field is given by :

[tex]E=\dfrac{V}{d}[/tex]

[tex]V=E\times d[/tex]

[tex]V=8.76\times 10^6\ V/m\times 8.47\times 10^{-9}\ m[/tex]

V = 0.0741 V

So, the voltage across the membrane is 0.0741 volts. Hence, this is the required solution.

Answer:

[tex]f = 8.89 cm[/tex]

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]

[tex]-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}[/tex]

[tex]f = 8.89 cm[/tex]

Answer:

Ratio of charge to mass is 8 x 10⁶

Explanation:

q = charge on the ion

V = potential difference = 10000 Volts

B = magnitude of magnetic field = 1 T

m = mass of the ion

[tex]v[/tex] = speed gained by the ion due to potential difference

using conservation of energy

kinetic energy gained by ion = electric potential energy lost

[tex](0.5)mv^{2}=qV[/tex]

[tex]mv^{2} = 2qV[/tex]                       eq-1

r = radius of the circular path described = 5 cm = 0.05 m

radius of circular path is given as

[tex]r = \frac{mv}{qB}[/tex]

taking square both side

[tex]r^{2} = \frac{m^{2}v^{2}}{q^{2}B^{2}}[/tex]

using eq-1

[tex]r^{2} = \frac{2mqV}{q^{2}B^{2}}[/tex]

[tex]r^{2} = \frac{2mV}{qB^{2}}[/tex]

[tex]\frac{q}{m} = \frac{2V}{r^{2}B^{2}}[/tex]

inserting the values

[tex]\frac{q}{m} = \frac{2(10000)}{(0.05)^{2}(1)^{2}}[/tex]

[tex]\frac{q}{m}[/tex] = 8 x 10⁶

Ratio of charge to mass is 8 x 10⁶

Answer:

a) Average velocity of the car for the time interval t=0 to t = 1.95 s is 2.62 m/s.

b) Average velocity of the car for the time interval t=0 to t = 3.96 s is 4.92 m/s.

c) Average velocity of the car for the time interval t=1.95 to t = 3.96 s is 7.15 m/s.

Explanation:

We have  x(t)= α t²− β t³

That is x(t)= 1.44 t²− 5 x 10⁻² t³

Average velocity is ratio of distance traveled to time.

a)Average velocity of the car for the time interval t=0 to t = 1.95 s

  x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

  x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

  Time difference = 1.95 - 0 = 1.95 s

  Average velocity [tex]=\frac{5.10}{1.95}=2.62m/s[/tex]

b)Average velocity of the car for the time interval t=0 to t = 3.96 s

  x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

  x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

  Time difference = 3.96 - 0 = 3.96 s

  Average velocity [tex]=\frac{19.48}{3.96}=4.92m/s[/tex]

c)Average velocity of the car for the time interval t=0 to t = 3.96 s

  x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

  x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

  Time difference = 3.96 - 1.95 = 2.01 s

  Average velocity [tex]=\frac{19.48-5.10}{2.01}=7.15m/s[/tex]

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

[tex]\mu mg = \frac{mv^2}{R}[/tex]

here we have

[tex]\mu g = \frac{v^2}{R}[/tex]

now we know that

[tex]v = \sqrt{\mu Rg}[/tex]

here we have

[tex]\mu = 0.600[/tex]

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

[tex]v = \sqrt{(0.600)(35)(9.81)}[/tex]

[tex]v = 14.35 m/s[/tex]

Explanation:

It is given that,

Mass of the artillery shell, m = 1100 kg

Acceleration of he shell, [tex]a=2.4\times 10^4\ m/s^2[/tex]

(a) The magnitude of force exerted on the shell can be calculated using Newton's second law of motion as :

F = ma

[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]

F = 26400000 N

or

[tex]F=2.6\times 10^7\ N[/tex]

(b) The magnitude of the force exerted on the ship by the artillery shell will be same but the direction is opposite as per Newton's third law of motion. i.e [tex]2.6\times 10^7\ N[/tex]

The magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.

(a) Net External Force on the Artillery Shell:

We can use Newton's second law of motion (ΣF = ma) to solve this problem. Here, ΣF represents the net external force acting on the object (artillery shell), m is the mass of the shell (1100.0 kg), and a is the acceleration of the shell (2.40 × 10^4 m/s²).

Force (F) = mass (m) × acceleration (a)

F = 1100.0 kg × 2.40 × 10^4 m/s²

F = 2.64 × 10^7 N

Therefore, the net external force exerted on the artillery shell is 2.64 × 10^7 N.

(b) Force on the Ship and Explanation:

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the force exerted on the artillery shell to accelerate it (calculated in part (a)) is equal in magnitude but opposite in direction to the force exerted by the shell on the ship.

Therefore, the magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.

Explanation:

There are two types of collision. First one is elastic and other one is inelastic collision.

The linear momentum of two objects before and after the collision remains the same in case of elastic collision. Also, the kinetic energy in elastic collision is conserved. But in case of inelastic collision, the momentum of two objects before and after the collision remains constant but the kinetic energy is in conversed. It changes form one form of energy to another.

Hence, the correct option is (E) "None of the above".

C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)

In an ideal elastic collision, the following happens:

Total Kinetic Energy is Conserved: The total kinetic energy of the system (colliding objects) before the collision is equal to the total kinetic energy after the collision. No energy is lost or gained in terms of kinetic energy.

Here's why the other options are incorrect:

A) Heat and Deformation: While real collisions might involve some energy loss due to heat and deformation, an ideal elastic collision is considered perfectly elastic, meaning no energy is lost in these ways.

B) Potential Energy: The collision doesn't necessarily involve a change in potential energy of the objects.

C) Momentum: Momentum is always conserved in any collision, elastic or inelastic.

D) All of the Above: Only the conservation of total kinetic energy applies to an ideal elastic collision.

E) None of the Above: In an ideal elastic collision, kinetic energy is conserved.

Therefore, the correct answer is:

C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)

Answer:

83.3 rad/s^2

Explanation:

r = 3.6 cm = 0.036 m

a = 3 m/s^2

the relation  between the linear acceleration and angular acceleration is given by

a = r α

where, α is angular acceleration, a be the linear acceleration.

α = a / r

α = 3 / 0.036 = 83.3 rad/s^2

Answer:

In metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.

Explanation:

Here we asked to divide 1 meter in to 100 equal parts.

Let us find out what is the length of piece when 1 m is divided in to 100 equal parts.

Length

[tex]l = \frac{1}{100} = 0.01m[/tex]

That is length of 1 m divided into 100 equal parts is 0.01m.

We know that 0.01 m is 1 centimeter.

So, in metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.

Final answer:

A meter divided into 100 equal-sized subunits is called a centimeter, with 'centi-' indicating one-hundredth of a meter in the metric system.

Explanation:

If a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter. The prefix centi- in the metric system indicates a division by 100.

Therefore, one centimeter is one-hundredth of a meter. This is consistent with metric prefixes being based on powers of ten, facilitating easy conversion and understanding of different scales of measurement.

The International System of Units, or SI, uses these prefixes to create a unified system for measuring length, and indeed all types of measurements.

Answer:

86.5m

Explanation:

first convert km/h

then

81.4*1000/60*60=22.6

17.6*1000/60*60=4.89

then, x1/t1=x2/t2

we get

x2=400*4.89/22.6=86.5//

Answer:

The answer is E) 20kPa  

Explanation:

Here we use the definition of Pressure: Pressure is a physical quantity that measures the projection of force in the perpendicular direction per unit area. In this case, wehave a 200N Force over a 100cm^2 Area

In order to have the pressure in international units we need to convert those cm^2 to m^2 this way:

We know 100cm is equal to 1m

[tex](100cm^2)*((1m^2)/(100cm)^2)[/tex]

In this case, the area will be equal to 0,01m^2

With this information we use the Pressure equation, which based on the definition will be:

[tex]P=F/A[/tex]

Solving:

[tex]P=(200N)/(0,01m^2)[/tex]

[tex]P=20000Pa[/tex]

Then P is the same as 20kPa  

Answer:

(a): The power required to accelerate the car is F= 2756.12 N.

(b): The power required to accelerate the car is F= 3603.67 N.

Explanation:

V= 27.2 m/s

Vi= 0 m/s

t= 8.3 s

g= 9.8 m/s²

W1= 8.26 * 10³ N

W2= 1.08 * 10⁴ N

a= (V-Vi)/t

a= 3.27 m/s²

m1= W1/g

m1= 842.85 kg

m2= W2/g

m2= 1102.04 kg

F= mi*a

F1= 2756.12 N (a)

F2= 3603.67 N (b)

Final answer:

The total magnetic flux through the coil before rotation is calculated using the formula Φ = B × A × cos(θ), with B being the Earth's magnetic field strength, A the area of the coil, and θ the angle between the field and the normal to the coil, resulting in a flux of 7.2 × 10⁻⁸ Tm².

Explanation:

The total magnetic flux Φ through the coil before it is rotated is given by the formula Φ = B × A × cos(θ), where B is the magnetic field strength, A is the area enclosed by the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil. Before the rotation, the plane of the coil is perpendicular to Earth's magnetic field, making θ = 0° (or cos(θ) = 1), so the cos(θ) term does not affect the calculation.

To calculate the magnetic flux, convert the area from cm² to m² by multiplying by 10⁻⁴, and use the given values:

Thus, the total magnetic flux through the coil before it is rotated is 7.2 × 10⁻⁸ Tm².

Answer:

Yes circuit is in resonance

Explanation:

As we know that the circuit consist of a resistance of 10 ohm and an inductive reactance of 5 ohm

This circuit is connected to a load with having value of resistance of 10 ohm and capacitive reactance of 5 ohm

now they are all in series so as we know that capacitor and inductor is in opposite phase so here reactance of inductor is cancelled by the reactance of capacitor due to the opposite phase of them also they are same in magnitude

so we have

[tex]x_L = x_C[/tex]

so the circuit is purely resistive and hence we can say that it must be in resonance

Answer:

Charge on each plate = 2.31 x 10⁻⁸ C

Explanation:

We have the equations

           [tex]E=\frac{V}{d}\texttt{ and }V=\frac{Q}{C}[/tex]

Combining both equations

           [tex]E=\frac{\left (\frac{Q}{C}\right )}{d}=\frac{Q}{Cd}[/tex]

We also have the equation for capacitance

           [tex]C=\frac{\epsilon A}{d}[/tex]

That is

          [tex]E=\frac{Q}{\frac{\epsilon A}{d}\times d}=\frac{Q}{\epsilon A}\\\\Q=\epsilon AE[/tex]

Substituting

           [tex]Q=8.85\times 10^{-12}\times 31.5\times 10^{-4}\times 8.30\times 10^5=2.31\times 10^{-8}C[/tex]

Charge on each plate = 2.31 x 10⁻⁸ C

Answer:

Increase in temperature =  269.54 °C

Explanation:

We have equation for thermal expansion

          ΔL = LαΔT

Change in length, ΔL = 0.08 m

Length, L = 56 m

Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1

Change in temperature, ΔT = T - 253

Substituting

          0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)

         (T - 253) = 269.54

           T = 522.54 °C

Increase in temperature =  269.54 °C      

Answer:

(a)  1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm  

Answer:

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Explanation:

We have expression for sound intensity level (SIL),

              [tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]

Here we need to find the intensity of sound (I).

               [tex]L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}[/tex]

Substituting

          L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

          [tex]I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2[/tex]

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Answer:

1 Hz

Explanation:

For rope fixed on both ends the length corresponds to λ/2  (λ is wavelength)\

 Thus L =  λ/2

=> λ = 16 m

We know that frequency and wavelength are related as

 f x λ = v   where f is frequency and v is speed of the wave

thus f = v/λ

        f = 16/16 =1 Hz

The fundamental frequency of an 8 m rope tied at both ends with waves traveling at a speed of 16 m/s is 1 Hz, calculated using the formula for the fundamental frequency of a standing wave.

The fundamental frequency of a string fixed at both ends (like an 8 m rope with waves traveling at 16 m/s) can be found using the formula for the fundamental frequency of a standing wave, which is given by f = v / (2L), where f is the fundamental frequency, v is the speed of the wave, and L is the length of the string. In this case, L = 8 m and v = 16 m/s. Plugging these values into the formula gives us:

f = 16 m/s / (2 × 8 m) = 1 Hz.

Therefore, the fundamental frequency of the 8 m rope tied at both ends with a wave speed of 16 m/s is 1 Hz.

Answer: [tex]2(10)^{42}kg [/tex]

Explanation:

Approaching the orbit of the Large Magellanic Cloud around the Milky Way to a circular orbit, we can use the equation of velocity in the case of uniform circular motion:

[tex]V=\sqrt{G\frac{M}{r}}[/tex] (1)  

Where:

[tex]V=300km/s=3(10)^{5}m/s[/tex] is the velocity of the Large Magellanic Cloud's orbit, which is assumed as constant.

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]  

[tex]M[/tex] is the mass of the Milky Way

[tex]r=160000ly=1.51376(10)^{21}m[/tex] is the radius of the orbit, which is the distance from the center of the Milky Way to the Large Magellanic Cloud.  

Now, if we want to know the estimated mass of the Milky Way, we have to find [tex]M[/tex] from (1):

[tex]M=\frac{V^{2} r}{G}[/tex] (2)  

Substituting the known values:

[tex]M=\frac{(3(10)^{5}m/s)^{2}(1.51376(10)^{21}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (3)  

[tex]M=\frac{1.362384(10)^{32}\frac{m^{3}}{s^{2}}}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex]  

Finally:

[tex]M=2.0416(10)^{42}kg\approx 2(10)^{42}kg[/tex] >>>This is the estimated mass of the Milky Way

To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.

To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The formula for this law is v = sqrt(GM/r), where v is the velocity, G is the gravitational constant, M is the Milky Way's mass, and r is the distance from the center. Rearranging the equation to solve for M, we get M = v^2 * r / G.

Using the given values of v = 300 km/s and r = 160,000 light-years, we need to convert them to appropriate units. 1 light-year is approximately 9.461 × 10^15 meters. After performing the necessary conversions and calculations, the estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.

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