The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the station so she accelerates at 1.50 m/s2. What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall? Give your answer relative to an observer who is space walking and therefore does not accelerate with the space station due to the push.

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:

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:

(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)

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:

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:

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

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:

Part a)

W = 16.7 N

Part b)

r = 2.45 R

Explanation:

Part a)

As we know that acceleration due to gravity on the surface of moon is 1/6 times the gravity on the surface of earth

So the force due to gravity will decrease by the factor of 6

so we will have

[tex]W_{moon} = \frac{1}{6}W_{earth}[/tex]

[tex]W_{moon} = \frac{1}{6}(100)[/tex]

[tex]W_{moon} = 16.7 N[/tex]

Part b)

For the same value of the weight as the surface of moon the acceleration due to gravity of earth must be 1/6 times

so we have

[tex]\frac{GM}{r^2} = \frac{GM}{6R^2}[/tex]

[tex]r^2 = 6R^2[/tex]

[tex]r = 2.45 R[/tex]

The object must be [tex]\( \sqrt{6} \)[/tex]  Earth radii from the center of Earth to weigh the same as it does on the Moon.

The weight of an object on the Moon's surface is approximately 1/6 of its weight on Earth's surface due to the Moon's lower gravitational acceleration. Therefore, the object that weighs 100 N on Earth will weigh:

[tex]\[ W_{\text{Moon}} = \frac{1}{6} \times W_{\text{Earth}} = \frac{1}{6} \times 100 \text{ N} = 16.67 \text{ N} \][/tex]

For the second part of the question, we need to find the Earth radius multiple where the gravitational acceleration is the same as that of the Moon. The gravitational acceleration g on Earth is approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex], and on the Moon, it is [tex]\( \frac{9.81}{6} \, \text{m/s}^2 \)[/tex]. The gravitational acceleration ( g' ) at a distance r from the center of Earth is given by:

[tex]\[ g' = g \left( \frac{R_{\text{Earth}}}{r} \right)^2 \][/tex]

where [tex]\( R_{\text{Earth}} \)[/tex] is the radius of Earth. We want to find r such that ( g' ) is equal to the gravitational acceleration on the Moon:

[tex]\[ g \left( \frac{R_{\text{Earth}}}{r} \right)^2 = \frac{g}{6} \][/tex]

Solving for r:

[tex]\[ \left( \frac{R_{\text{Earth}}}{r} \right)^2 = \frac{1}{6} \] \[ \frac{R_{\text{Earth}}}{r} = \sqrt{\frac{1}{6}} \] \[ r = R_{\text{Earth}} \sqrt{6} \][/tex]

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.

a) For an EM wave traveling in a vacuum, this equation holds true:

c = fλ

c is the speed of light in a vacuum, f is the frequency, and λ is the wavelength.

Given values:

c = 3×10⁸m/s

λ = 6m

Plug in the values and solve for f:

3×10⁸ = f(6)

f = 50MHz

b) The direction of an EM wave's Poynting vector determines the direction of the wave's propagation.

S = 1/μ₀(E×B)

S is the Poynting vector, μ₀ is the magnetic constant, E is the electric field vector, and B is the magnetic field vector. Note that we are taking the cross product between E and B, not taking the product of two scalar quantities.

Since S depends on the cross product of E and B, you may use the right hand rule in the following way to determine the direction of B:

You can conclude that B must point along the z axis, so you can represent B with the k unit vector.

a) The frequency f of the wave is 50 MHz.

b) The direction of the magnetic field associated with the wave is along Z-axis with the k unit vector.

The Poynting vector is a measurement that expresses the strength and direction of the energy flow in electromagnetic waves.

Mathematically;

S = 1/μ₀(E×B)

Where: S =  Poynting vector,  μ₀ = magnetic constant, E = electric field vector, and B = the magnetic field vector

a) For an EM wave traveling in a vacuum, it can be written that:

c = fλ

Where c is the speed of light in a vacuum, f is the frequency, and λ is the wavelength.

Speed of light in vacuum: c = 3×10⁸m/s

Wavelength of the given EM Wave: λ = 6m

Hence, the frequency of the EM wave: f =  3×10⁸m/s/  6m = 50MHz.

b) The EM wave travels in the x direction, therefore S points in the x direction.

Electric field acts along the y axis.

Hence, The direction of the magnetic field associated with the wave is along Z-axis with the k unit vector.

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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.

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.

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:

The ball's maxium height after 1.49 seconds will be 103.52 feet.

Explanation:

Vi= 48 ft/sec

hi= 32 ft

hf= ?

time of maxium height :  

t= Vi/g    t= 48 ft/sec  /  32.17 ft/sec²

t= 1.49 sec

maxium height:

hf= hi + Vi * t

hf= 32ft + 48 ft/sec * 1.49 sec

hf= 103.52 ft

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:

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:

W = 907963.50 J = 907.96 J

Explanation:

Note: Refer to the figure attached

Now, from the figure we have similar triangles ΔAOB and ΔCOD

we have

[tex]\frac{5}{4}=\frac{x}{r}[/tex]

or

[tex]r=\frac{4x}{5}[/tex]

Now, the work done to empty the tank can be given as:

[tex]W = \int\limits^4_0 {(5-x)\rho\times g A} \, dx[/tex]

or

[tex]W = \int\limits^4_0 {(5-x)1080\times 9.8 (\pi r^2)} \, dx[/tex]

or

[tex]W = \int\limits^4_0 {(5-x)\times10584\times (\pi (\frac{4x}{5})^2)} \, dx[/tex]

or

[tex]W = 6773.76\pi\int\limits^4_0 {(5-x)x^2)} \, dx[/tex]

or

[tex]W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0 [/tex]

or

[tex]W = 6773.76\pi[\frac{128}{3}] [/tex]

or

W = 907963.50 J = 907.96 J

The work required to empty the tank by pumping the hot chocolate over the top of the tank is 907.96 J.

Work done is the force applied on a body to move it over a distance. The work required to lift a body through a height h is given as,

[tex]W=Fh[/tex]

Here, (F) is the magnitude of force and (f) is the height.

It can be rewritten as,

[tex]W=mgh[/tex]

Here, (m) is the mass of the body.

The tank is in the shape of an inverted right circular cone with height 5 meters and radius 4 meters. As, It is filled with 4 meters of hot chocolate.

Here the ratio of height to radius should be equal to the height at another point (say x), to the radius of that point. Thus,

[tex]\dfrac{5}{4}=\dfrac{x}{r}\\r=\dfrac{4x}{5}[/tex]

For the distance x the work done to empty the cone can be given as,

[tex]W=\int^4_0 (5-x)\rho g \pi r^2 dx\\W=\int^4_0 (5-x)1080\times9.81 \pi (\dfrac{4x}{5})^2 dx\\W=22280.4\int^4_0(5-x)x^2 dx\\W=22280.4[\dfrac{128}{3}]\\W=907.96\rm J[/tex]

Thus, the work required to empty the tank by pumping the hot chocolate over the top of the tank is 907.96 J.

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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

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:

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:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

[tex]r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}[/tex]

here we have

[tex]m_1 = 10.2 kg[/tex]

[tex]m_2 = 4.6 kg[/tex]

[tex]r_1 = (0, 0)[/tex]

[tex]r_2 = (8.1cm, 0)[/tex]

now we have

[tex]r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}[/tex]

[tex]r_{cm} = {(37.26, 0)}{14.8}[/tex]

[tex]r_{cm} = (2.52 cm, 0)[/tex]

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

[tex]r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}[/tex]

here we have

[tex]m_1 = 10.2 kg[/tex]

[tex]m_2 = 4.6 kg[/tex]

[tex]r_1 = (0, 0)[/tex]

[tex]r_2 = (8.1cm, 0)[/tex]

now we have

[tex]r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}[/tex]

[tex]r_g_{cm} = {(37.26, 0)}{14.8}[/tex]

[tex]r_g_{cm} = (2.52 cm, 0)[/tex]

So center of mass is same as center of gravity because value of gravity is constant here

The center of mass and center of gravity for the two given masses are both located at x = 2.49 cm from the origin. In typical situations where gravitational forces are uniform, the center of gravity will coincide with the center of mass.

The location of the center of mass (xcom) and the center of gravity (xcog) for two bodies or particles can be calculated using the following formula: xcom = m1x1 + m2x2 / (m1 + m2) where m1, m2 are the masses and x1, x2 are their respective positions. In this case, m1 = 10.2 kg, x1 = 0 cm (since it's at the origin), m2 = 4.6 kg, and x2 = 8.1 cm.

Using these values, the formula becomes xcom = (10.2 kg * 0 cm + 4.6 kg * 8.1 cm) / (10.2 kg + 4.6 kg) = 36.86 cm / 14.8 kg = 2.49 cm. So, the center of mass is at x = 2.49 cm from the origin.

The concept of center of gravity is a specific application of the center of mass in the presence of a gravitational field. In everyday circumstances, where gravitational forces are uniform, the center of gravity coincides with the center of mass. Therefore, for this case, the center of gravity (xcog) will also be at x = 2.49 cm from the origin.

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Answer:

390 N

Explanation:

The net acceleration is:

a² = aₓ² + aᵧ²

a² = (900 m/s²)² + (870 m/s²)²

a = 1250 m/s²

So the net force is:

F = ma

F = (0.31 kg) (1250 m/s²)

F = 390 N

Answer:

QC = 122 KJ

QH = 2.64 x 122 = 322 KJ

Explanation:

TH = 500 Degree C = 500 + 273 = 773 K

TC = 20 degree C = 20 + 273 = 293 K

W cycle = 200 KJ

Use the formula for the work done in a cycle

Wcycle = QH - QC

200 = QH - QC    ..... (1)

Usse

TH / TC = QH / QC

773 / 293 = QH / QC

QH / QC = 2.64

QH = 2.64 QC     Put it in equation (1)

200 = 2.64 QC - QC

QC = 122 KJ

So, QH = 2.64 x 122 = 322 KJ

Answer:

Speed of the car after the collision is 1.2 m/s

Explanation:

It is given that,

Mass of first car, m₁ = 2 kg

Velocity of first car, v₁ = 6 m/s

Mass of second car, m₂ = 3 kg

Velocity of second car, v₂ = -2 m/s (it is travelling in opposite direction)

The two cars lock together and move along the road. This shows an inelastic collision. Let their common velocity is V. On applying the conservation of momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

[tex]V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]V=\dfrac{2\ kg\times 6\ m/s+3\ kg\times (-2\ m/s)}{5\ kg}[/tex]

V = 1.2 m/s

So, after collision the speed of the cars is 1.2 m/s. Hence, this is the required solution.

Answer:

t = 206 sec

Explanation:

m = mass of the ocean liner = 39000 Mg = 39000 x 10⁶ g = 3.9 x 10⁷ kg

F = constant force applied by the tugboat = 210 kN = 210000 N

v₀ = initial velocity of the liner = 4 km/h = 1.11 m/s

v = final velocity of the liner = 0 m/s

a = acceleration of the liner

Acceleration of the liner is given as

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

[tex]a = - \frac{210000}{3.9\times 10^{7}}[/tex]

a = - 0.0054 m/s²

Using the equation

v = v₀ + at

0 = 1.11 + (- 0.0054) t

t = 206 sec

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: [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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The force of the ramp on the block is 2.45 N.

To find the force of the ramp on the block, we need to consider the forces acting on the block. The only force acting on the block along the ramp is the component of the weight of the block that is parallel to the ramp. This force can be calculated using the equation:

F = mg sin(theta)

where F is the force, m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the ramp. Plugging in the values, we get:

F = (2.0 kg)(9.8 m/s^2) sin(30 degrees) = 4.9 N sin(30 degrees) = 2.45 N

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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]