The escape velocity from the Moon is much smaller than that from the Earth, only 2.38 km/s. At what temperature would hydrogen molecules (molar mass is equal to 2.016 g/mol) have a root-mean-square velocity vrms equal to the Moon’s escape velocity?

Answer:

(A.) Resonance Frequency

(B.) Totally resistance

Explanation:

A. The resonance frequency is the frequency at which the net reactance of the inductor and capacitor is zero, this is the frequency at which the capacitive reactance of the capacitor and inductive reactance of the inductor cancels each other out. Resistance is low here and the frequency is high

B. Since the net reactance of the inductor and capacitor is zero as explained in (A) above, the total reactance of the circuit will be entirely from the resistor (resistance)

Answer:

Goodness of fit

Explanation:

The word "goodness of fit" is characterized as the concept that growth depends on the level of correlation between the personality of children as well as the existence and needs of the community where they were born.

Goodness of fit, used in psychology as well as in parenting, defines a person's personality alignment with the characteristics of their specific social culture.

There are various features and requirements in all contexts, i.e. community, culture, workplace, etc. Goodness of Fit is very crucial component of any individual's psychological change.

Explanation:

Given that,

Mass of thee pool ball, m = 3 kg

Initial speed of the ball, u = -4.3 m/s

Impulse experienced by thee ball, J = -4 N-s

To find,

Speed of the object after impulse occurs and its direction.

Solution,

Let left side is negative and right side is positive. So, the change in momentum or the impulse is given by the following expression as :

[tex]J=m(v-u)[/tex]

[tex]v=\dfrac{J}{m}+u[/tex]

[tex]v=\dfrac{-4}{3}+(-4.33)[/tex]

v = -5.663 m/s

So, the speed of the object is 5.663 m/s and it is towards left.

The pool ball experiences an impulse in the opposite direction of its initial motion. After the impulse, the speed of the ball reduces to 2.97 m/s and it is still moving to the left.

The subject of this question is Physics, specifically impulse and momentum. Firstly, we need to calculate the initial momentum which is the mass of the object multiplied by its initial velocity, so that will be 3.00 kg * 4.30 m/s = 12.90 kg*m/s to the left since the direction is towards left.

An impulse is the change in momentum, and in this case, the impulse is -4.00 Ns which means it acts in the opposite direction of the initial motion, i.e to the right.

After the impulse, the final momentum will be the sum of initial momentum and impulse: 12.90 kg*m/s (to the left) + -4.00 kg*m/s (to the right) = 8.90 kg*m/s (to the left).

Finally, to calculate the final speed, divide the final momentum by the mass: 8.90 kg*m/s ÷ 3.00 kg = 2.97 m/s. So, after the impulse, the object's speed is 2.97 m/s and it is still moving to the left.

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Answer:

24.07415 rpm

Explanation:

[tex]\mu[/tex] = Coefficient of friction = 0.63

v = Velocity

d = Diameter = 4.9 m

r = Radius = [tex]\frac{d}{2}=\frac{4.9}{2}=2.45\ m[/tex]

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

Here the frictional force balances the rider's weight

[tex]f=\mu F_n[/tex]

The centripetal force balances the weight of the person

[tex]\mu m\frac{v^2}{r}=mg\\\Rightarrow \mu \frac{v^2}{r}=g\\\Rightarrow v=\sqrt{\frac{gr}{\mu}}\\\Rightarrow v=\sqrt{\frac{9.81\times 2.45}{0.63}}\\\Rightarrow v=6.17656\ m/s[/tex]

Velocity is given by

[tex]v=\omega r\\\Rightarrow \omega=\frac{v}{r}\\\Rightarrow \omega=\frac{6.17656}{2.45}\\\Rightarrow \omega=2.52104\ rad/s[/tex]

Converting to rpm

[tex]2.52104\times \frac{60}{2\pi}=24.07415\ rpm[/tex]

The minimum angular speed for which the ride is safe is 24.07415 rpm

The minimum angular speed for which the ride will be safe is ≈ 24.07 rpm

Given data :

Diameter of hollow steel cylinder = 4.9 m.   Radius ( r ) = 4.9 / 2 = 2.45 m

coefficient of friction of clothing ( [tex]\alpha[/tex] ) = 0.63

g = 9.81 m/s²

First step : Determine the velocity using the centripetal forces relation

v = [tex]\sqrt{\frac{g*r}{\alpha } }[/tex] ----- ( 1 )

where ;  g = 9.81 m/s,  r = 2.45 m ,  [tex]\alpha = 0.63[/tex]

Insert values into equation 1

V = [√( 9.81 * 2.45 )/0.63 ]

   = 6.177 m/s

Next : convert velocity to rad/sec ( angular velocity )

V = ω*r

∴ ω = V / r

       = 6.177 / 2.45 = 2.52 rad/sec

Final step:  The minimum angular speed expressed in rpm

angular velocity ( ω ) * [tex]\frac{60}{2\pi }[/tex]

= 2.52 * [tex]\frac{60}{2\pi }[/tex]  ≈ 24.07 rpm

Hence the minimum angular speed in rpm = 24.07 rpm.

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The phase φ(x,t) of a wave represents its phase shift. It can be obtained from the wave function y (x, t) = A sin (kx - wt + p) used to model the wave's behavior. The expression for the phase is φ(x,t) = kx - wt + p.

The phase φ(x,t) of a wave can be expressed from the given sinusoidal wave function y (x, t) = A sin (kx - wt + φ). Here, φ represents the phase of the wave, which is a shift that allows for initial conditions other than x = +A and v = 0. The aforementioned variables stand for: A - amplitude of the wave, k - wave number, ω - angular frequency, x - position, and t - time.

Now, looking at the wave equation, the phase shift, φ, is the amount by which the wave function is shifted. This shift can be to the right or the left and is usually represented by the Greek letter phi (p). In the wave function, this phase shift may result because the motion being modeled by the function isn't starting from the rest position. For example, if considering the motion of a pendulum, if it was released from an elevated position, there would be a phase shift in the resulting wave function, unlike if it had been released from a resting position.

So for the waveform y (x, t) = A sin (kx - wt + p), the phase φ(x,t) = kx - wt + p.

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In physics, the phase ϕ of a wave accounts for the initial conditions and describes the state of oscillation relative to a specific reference point. It is part of the wave function formula: y (x, t) = A sin (kx - ωt + ϕ). Here, ϕ = kx - ωt + ϕ, with k and x related to position and ωt to elapsed time.

In physics, often when discussing waves, the term 'phase' is used. The phase of a wave, represented by ϕ in wave equations, is a measure of the relative position in a wave cycle of a particle, taking into account the initial point of the cycle (which can involve a phase shift) and the time elapsed since this initial point.

It allows us to describe the current state of oscillation of any point in the wave, relative to a specific reference point.

For a sinusoidal wave moving in the x direction, its wave function, which models the wave's displacement over time, can be represented as y (x, t) = A sin (kx - ωt + ϕ). Here, y (x, t) is the amplitude of the wave at any point x and at any time t. The variable A signifies the peak amplitude of the wave, k is the wave number, ω is the angular frequency, and ϕ is the phase.

The purpose of ϕ in this equation is to account for the initial state of the system. For example, if our wave is not starting at its equilibrium position, we can incorporate these initial conditions into the wave equation using the phase constant ϕ.

So, in conclusion, the phase ϕ of a wave in terms of the given variables would simply be ϕ = kx - ωt + ϕ, with k and x related to spatial position and ωt related to the elapsed time.

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Answer:

Elastic potential energy, E = 3.26 J

Explanation:

It is given that,

Force constant of the spring, k = 5.2 N/m

Relaxed length of the spring, X = 2.45 m

When the mass is attached to the end of the spring, the vertical length of the spring is, x' = 3.57 m

To find,

The elastic potential energy stored in the spring.

Solution,

The extension in the length of the spring is given by :

[tex]x=x'-X[/tex]

[tex]x=3.57\ m-2.45\ m[/tex]

x = 1.12 m

The elastic potential energy of the spring is given by :

[tex]E=\dfrac{1}{2}kx^2[/tex]

[tex]E=\dfrac{1}{2}\times 5.2\times (1.12)^2[/tex]

E = 3.26 J

So, the elastic potential energy stored in the spring is 3.26 joules.

The Elastic Potential Energy stored  in the  spring mass system

= 3.26 Joule

The Elastic Potential Energy Stored (E) in the  spring mass system  is given by equation (1)

E = (1/2) [tex]\times K \times x^2[/tex]........(1)

Where K is the Force Constant = 5.2 N/m

and  [tex]x[/tex] is the extension or compression of the spring.

Here as the spring length increases  so  [tex]x[/tex] = Final Length - Initial Length

= (3.57- 2.45) = 1.12 m

E    = (1/2) [tex]\times[/tex]5.2 [tex]\times[/tex][tex]1.12^2[/tex] = 3.26144 [tex]\approx[/tex] 3.26 Joule

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Answer:

3 x 10^18 kg

Explanation:

Time period, T = 3 days = 86400 x 3 = 259200 seconds

r = 7 x 10^5 m

Let M be the mass of planet

Use the formula of time period of satellite

[tex]T = 2\pi \sqrt{\frac{r}{GM}}[/tex]

Where, G be the universal gravitational constant.

[tex]M=\frac{4\pi ^{2}r^{3}}{GT^{2}}[/tex]

By substituting the values

[tex]M=\frac{4\times 3.14 \times 3.14\times \left ( 7\times 10^{5} \right )^{3}}{6.67\times 10^{-11}\times 259200\times 259200}[/tex]

M = 3 x 10^18 kg

Thus , the mass of planet is 3 x 10^18 kg.

Answer:

a)   U = - G m₁m₂ / r , b)  K = ½ m (v₀² + 2gy)²  or   K = 2 mg² y²   c)  Em = m g y (2 g y + 1)

Explanation:

Let's write the functions that are requested

a) Gravitational power energy

     U = - dF / dr

     F = G m₁ m₂ / r²

    U = - G m₁m₂ / r

    r is the height

b) The scientific enrgia

    K = ½ m v²

Cinematic

    v² = v₀² + 2 g y

    K = ½ m (v₀² + 2gy)²

If the initial velocity is zero, the ball is released

    K = ½ m 4 g² y²

    K = 2 mg² y²

c) Mechanical energy

    Em = K + U

    Em = 2 m g² y² + m g y

    Em = m g y (2 g y + 1)

The magnetic force on a wire in a bundled configuration can be calculated using the formula F = I × B. The force per unit length on a wire located 0.200 cm from the center of the bundle can be calculated using this formula. A wire on the outer edge of the bundle experiences a smaller force compared to a wire closer to the center.

a. The magnetic force per unit length on a wire located 0.200 cm from the center of the bundle can be calculated using the formula F = I × B. Since the current and magnetic field are perpendicular, we can simplify the formula to F = I × B × sin(90°) = I × B. Plug in the values to get F = (2.00 A) × (μ × I/(2πR)) = (2 × 10^(-7) T·m/A) × (2.00 A) / (2π × 0.005 m). Calculate this to get the magnitude of the force per unit length.

b. A wire on the outer edge of the bundle will experience a smaller force than the wire 0.200 cm from the center. This is because the magnetic field strength decreases as the distance from the wire increases. Therefore, the force on a wire decreases as its distance from the center of the bundle increases.

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The magnetic force per unit length on a wire [tex]0.200 cm[/tex] from the center is [tex]1.28 \times 10^{-2} \, \text{N/m}[/tex]. A wire at the outer edge experiences a greater force of [tex]1.6 \times 10^{-2} \, \text{N/m}\\ \\[/tex] due to the higher magnetic field there.

This problem involves calculating the magnetic force per unit length acting on a current-carrying wire located within a bundle of wires, as well as comparing it to the force on a wire at the bundle's outer edge.

First, we calculate the magnetic field at a distance of [tex]0.200 cm[/tex] from the center. The magnetic field inside the bundle can be found using Ampère's Law:

The current enclosed by a circular path of radius [tex]r[/tex] inside the bundle is:

[tex]I_{\text{enclosed}} = (I_{\text{total}}) \times \left( \frac{\pi r^2}{\pi R^2} \right) \\= 200 \, \text{A} \times \left( \frac{0.200^2}{0.500^2} \right) \\= 32 \, \text{A}[/tex]

The magnetic field B at this radius is given by:

[tex]B = \frac{\mu_0 I_{\text{enclosed}}}{2\pi r} \\= \frac{4\pi \times 10^{-7} \times 32}{2\pi \times 0.002} \\ = 6.4 \times 10^{-3} \, \text{T}[/tex]

The force per unit length acting on a wire carrying a current I₁ is:

[tex]\frac{F}{L} = I_1 \times B \\= 2 \, \text{A} \times 6.4 \times 10^{-3} \, \text{T} \\= 1.28 \times 10^{-2} \, \text{N/m}[/tex]

The direction of the force follows the right-hand rule and is perpendicular to both the wire's current and the magnetic field.

A wire on the outer edge will experience a magnetic field generated by all the currents inside the circle of radius R. Therefore:

[tex]I_{\text{total}} = 100 \, \text{wires} \times 2.00 \, \text{A} \\= 200 \, \text{A}[/tex]

The magnetic field at the edge [tex]B_{edge}[/tex] is:

[tex]B_{\text{edge}} = \frac{\mu_0 I_{\text{total}}}{2\pi R} \\= \frac{4\pi \times 10^{-7} \times 200}{2\pi \times 0.005} \\= 8 \times 10^{-3} \, \text{T}[/tex]

The force per unit length [tex]F_{new}[/tex] on a wire at the edge is:

[tex]\frac{F}{L}_{\text{new}} = I_1 \times B_{\text{edge}} \\= 2 \, \text{A} \times 8 \times 10^{-3} \, \text{T} \\= 1.6 \times 10^{-2} \,[/tex]

Thus, the force experienced by a wire at the edge is greater than the force calculated in part (a).

Answer:

option D

Explanation:

The correct answer is option D

The friction force is the force that holds back the surface from sliding. It acts in the opposite direction of the motion of the body.

It can be explained with an example,

when you throw a ball on the ground it stops automatically this is because of friction which is acting opposite to the ball.

Friction also has benefits like due to friction we can walk, run, etc.  

So, from the given option

The force that resists sliding is friction force.

Answer:

a

Explanation:

The most reasonable conclusion of the above phenomenon is that water is a poor conductor of heat. Basically water is an insulator. The heat from surface to the bottom of the beaker will take a lot of time. Moreover, no convection current is formed so, heat might not even reach the bottom surface. Hydrogen bonding also play a vital role in determining the thermal properties of water.

hence option A is correct

The most reasonable conclusion from the observation that only the surface of the water in a beaker boils is that water is a poor conductor of heat, which leads to inefficient heat transfer through its body. So the correct option is a.

If a beaker of water is placed under a broiler, with the observation that only the surface layer boils while the water at the bottom of the beaker remains close to the initial temperature, it can be concluded that water is a poor conductor of heat. This phenomenon happens because heat is not efficiently transferred through the body of the water from the hot surface at the top to the cooler sections below. In this scenario, the heat from the broiler predominantly warms the water by radiation directly at the surface, but the lack of conduction prevents the lower part of the water from reaching the boiling point.

Heat conduction in materials depends on the kinetic energy transfer between molecules. Good conductors, like metals, allow for a rapid heat flux due to the effective transfer of energy during molecular collisions. In contrast, poor conductors experience less efficient energy transfer, resulting in a slower rate of heat flow. Additionally, convection could play a role in heat distribution, but in this case, the observation that only the surface boils suggests conduction is the limiting factor for heat transfer to the bottom of the beaker.

Answer:

Explanation:

Given to find the pressure

v = 12.6 m/s , A = 3.0m * 1.8 m = 54 m^2

p air = 1.3 kg/m^3

F = 1/2 * p *v^2 *A

F = 1/2 *1.3 kg/m^3 * (12.6 m/s)^2 * 54m^2

F = 5572.476 N

The stress and the pressure can find across the area

α = F /A

α = 5572.476 N / 54 m^2

α = 103.194 Pa

Answer:

T = 23.92 m

Explanation:

given,

mass of the diver = 50.9 Kg

Resistant force from the water = f = 1492 N

diver come top rest under water at a distance = 6 m

acceleration due to gravity = 9.81 m/s²

final velocity = v  = 0 m/s

initial velocity = u = ?

total distance = ?

Now acceleration of body under water

f = m a

[tex]a = \dfrac{1492}{50.9}[/tex]

[tex]a = 29.31\ m/s^2[/tex]

using equation of motion

v² = u² + 2 a s

0 = u² - 2 x 29.31 x 6

[tex]u = \sqrt{2\times 29.31\times 6}[/tex]

u = 18.75 m/s

now.

calculating distance of the diver in air

v² = u² + 2 a s

0 = 18.75² - 2 x 9.81 x s

s = 17.92 m

total distance

T = 17.92 + 6

T = 23.92 m

the total distance between the diving board and the diver’s stopping point underwater T = 23.92 m

Answer:

s = 3.68 m

Explanation:

given,

mass of the car = 6 x 10³ Kg

angle of slope with horizontal = 19.3°

Average frictional force = 4.4×10³ N

Speed of the car at the bottom = 4.3 m/s

acceleration due to gravity = 9.81 m/s²

length of driveway = ?

first calculating the net force acting on the car

F = mg sin θ - f

(mg sin θ) is the weight component of the body

(f) is the frictional force which is acting opposite to the car

F = 6000 x 9.81 x  sin 19.3° - 4400

F = 19454 - 4400

F = 15054 N

we know,

Force = mass  x  acceleration

15054 =  6000 x a

a = 2.51 m/s²

using formula

v² = u² + 2 as

4.3² = 0² + 2 x 2.51 x s

s = 3.68 m

the length of the driveway = s = 3.68 m

Final answer:

To find the length of the driveway, we can use the equations of motion. Calculate the acceleration using trigonometry, find the net force, and use the equation v^2 = u^2 + 2as to determine the distance.

Explanation:

To find the length of the driveway, we can use the equations of motion. The acceleration of the car can be found by considering the forces acting on it. The force of gravity can be resolved into two components: one along the slope and one perpendicular to the slope. The force along the slope, which is responsible for the acceleration of the car, can be found using trigonometry. The net force acting on the car is the force along the slope minus the frictional force. We can then use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, to find the length of the driveway.

Let's break the problem down step by step:

Substituting the given values into the equations will give you the length of the driveway.

Answer:

option (d)

Explanation:

The extra solar planet is also called as exo planet.

the planet which orbits around the star other than sun is called exo planet.

So, a planet that orbits a star that is not our own sun.

Thus, option (d) is correct.

Answer:

x = 2000 Km

Explanation:

Given

y = 10 km

Slope: 1 : 200

x = ?

We can apply the formula

y / x = 1 / 200   ⇒     x = 200*y = 200*10 Km

⇒     x = 2000 Km

Given the slope ratio of 1:200 for a warm front and an observed cloud height of 10 km, the warm front is around 2000 km away.

In meteorological terms, a warm front slope can be visualized as a ramp, approximating a 1:200 ratio in this case, where the height (vertical change) is comparable to the 'rise' and the horizontal distance (or reach) is comparable to the 'run'. Here, you mentioned cirrus clouds at an approximate altitude of 10 kilometers - this is the vertical elevation or 'rise'. The slope ratio 1:200 is a simplification indicating that for every 1 km rise, the front extends 200 km in the horizontal direction or the 'run'.

So, if the 'rise' is 10 km (the altitude at which you observe the cirrus clouds), we can calculate the 'run' by multiplying the 'rise' by the slope ratio (200). Thus, the approaching warm front is approximately 10 km * 200 = 2000 km away.

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Final answer:

The fifth line of the Balmer series, with a wavelength of 397 nm, occurs in the ultraviolet region of the electromagnetic spectrum. Option D is correct.

Explanation:

The fifth line of the Balmer series with a wavelength of 397 nm falls into the ultraviolet (UV) region of the electromagnetic spectrum. The visible part of the spectrum ranges approximately from 380 nm to 740 nm. Since the wavelength of the fifth line is just below 400 nm, it is just outside the range for visible light and therefore must be in the ultraviolet range.

In the context of the Balmer series, there are four lines within the visible spectrum, specifically at wavelengths of 656 nm, 486 nm, 434 nm, and 410 nm. The line at 397 nm would be the first one in the UV part of the spectrum, making the answer to the student's question d) ultraviolet.

The Balmer series is a series of spectral lines in the hydrogen emission spectrum. Four of the wavelengths of the Balmer series occur in the visible spectrum: 656 nm, 486 nm, 434 nm, and 410 nm. Since the fifth line has a wavelength of 397 nm, it falls in the ultraviolet region of the electromagnetic spectrum.

Explanation:

a) Period of simple pendulum

                 [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

   l₁ = length of pendulum in Earth = ?

   l₂ = length of pendulum in Mars = ?

   T = Period of pendulum = 1.2 s

    g₁ = Acceleration due to gravity in Earth = 9.80 m/s²

    g₂ = Acceleration due to gravity in Mars = 3.70 m/s²

For Earth :-

     [tex]T=2\pi \sqrt{\frac{l_1}{g_1}}\\\\1.2=2\pi \sqrt{\frac{l_1}{9.8}}\\\\l_1=0.36m[/tex]

     Length of pendulum in Earth = 0.36 m      

For Mars :-

     [tex]T=2\pi \sqrt{\frac{l_2}{g_2}}\\\\1.2=2\pi \sqrt{\frac{l_2}{3.7}}\\\\l_2=0.13m[/tex]

     Length of pendulum in Mars = 0.13 m

b) Period of spring

                 [tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

 Here period is independent of acceleration due to gravity, so mass value is same in Earth and Mars.

                m = Mass = ?

                 T = Period = 1.2 s

                  k = Spring constant = 20 N/m

              [tex]T=2\pi \sqrt{\frac{m}{k}}\\\\1.2=2\pi \sqrt{\frac{m}{20}}\\\\m=0.73kg[/tex]

             Mass in Earth = Mass in Mars = 0.73 kg

The length of a simple pendulum and the mass to suspend the spring are mathematically given as

a L1 = 0.36m for the earth and L2 = 0.13m for mars

b m=0.73kg

Question Parameter(s):

period of 1.2 s on Earth

where the acceleration due to gravity is 9.80 m/s2

and on Mars, where the acceleration due to gravity is 3.70 m/s2?

a force constant of 20 N/m

Generally, the equation for the Period of a simple pendulum  is mathematically given as

[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

therefore the length of Earth is

[tex]\\\\1.2=2\pi \sqrt{\frac{l_1}{9.8}}\\\\l_1=0.36m[/tex]

and for mars

[tex]\\\\1.2=2\pi \sqrt{\frac{l_2}{3.7}}\\\\l_2=0.13m[/tex]

In conclusion length of a simple pendulum is

L1 = 0.36m for earth and L2 = 0.13m for mars

Period of spring

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]}\\\\1.2=2\pi \sqrt{\frac{m}{20}}\\\\m=0.73kg[/tex]

In conclusion, the mass to suspend the spring

m = 0.73kg

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Answer:

[tex]p2=0.667\\p1=0.167\\p3=0.167\\p=1[/tex]

Explanation:

Let's start writing the sample space for this exercise :

Let be ''M'' an abbreviation for Macrostate

Ω = { M1 , M2 , M3 }

Let be P(M1) the probability of Macrostate 1.

Reading the exercise, we know that ⇒

[tex]P(M1)=P(M3)[/tex]

Let's note this probability as ''p''.

[tex]P(M1)=P(M3)=p[/tex]

Macrostate 2 is four times more likely to occur than either of the other two macrostates ⇒

[tex]P(M2)=4p[/tex]

The sum of all probabilities must be equal to 1 for this sample space.Therefore,

[tex]p+p+4p=1[/tex]

[tex]6p=1[/tex]

[tex]p=\frac{1}{6}[/tex]

Finally :

[tex]P(M1)=\frac{1}{6}[/tex]

[tex]P(M2)=\frac{4}{6}[/tex]

[tex]P(M3)=\frac{1}{6}[/tex]

For Part A :

[tex]p2=\frac{4}{6}=0.667[/tex]

For Part B and C :

[tex]p1=p3=0.167[/tex]

For Part D :

The sum of the probabilities for all macrostates is equal to 1 :

[tex]p=p1+p2+p3=\frac{1}{6}+\frac{4}{6}+\frac{1}{6}=\frac{6}{6}=1[/tex]

Final answer:

The probability that equilibrium state 2 occurs is approximately 0.667. The probability that macro state 1 or macro state 3 occurs is each approximately 0.167. The sum of the probabilities for all macro states is exactly 1, as expected for a complete set of possible outcomes.

Explanation:

To find the probability of each macro state, we first recognize that the probabilities must sum to 1 for all macro states. Given that the equilibrium state, Macro state 2, is four times more likely than either of the other two Macro states 1 and 3, we can assign a probability 'p' to Macro states 1 and 3 and '4p' to Macro state 2.

Let's assume:

Probability of Macro state 1, p1 = p

Probability of Macro state 2, p2 = 4p

Probability of Macro state 3, p3 = p

Since the total probability must equal 1, we write the equation:

p1 + p2 + p3 = 1

Substituting the assumed probabilities gives:

p + 4p + p = 1

6p = 1

Therefore, p = 1/6

Now, we can calculate each probability:

p1 = 1/6 ≈ 0.167

p2 = 4/6 ≈ 0.667

p3 = 1/6 ≈ 0.167

The sum of all probabilities, p, is as expected:

p = p1 + p2 + p3 = 0.167 + 0.667 + 0.167 = 1

Answer:

259.62521 seconds

Explanation:

[tex]m_1[/tex] = Mass of astronaut = 87 kg

[tex]m_2[/tex] = Mass of wrench = 0.57 kg

[tex]v_1[/tex] = Velocity of astronaut

[tex]v_2[/tex] = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

[tex]m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s[/tex]

Time = Distance / Speed

[tex]Time=\frac{38.1}{0.14675}=259.62521\ s[/tex]

The time taken to reach the ship is 259.62521 seconds

a) The translational kinetic energy of the disk across the level surface is 17.58J.

b) The rotational kinetic energy of the disk is 8.79J.

Given the data in the question;

a)

Translational kinetic energy.

Translational kinetic energy of an object is the work needed to accelerate the object from rest to a given velocity. It is expressed as:

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

We substitute our given values into the equation

[tex]Translational\ K_E = \frac{1}{2}*2.50kg\ *\ (3.75m/s)^2\\\\Translational\ K_E = \frac{1}{2}*2.50kg\ *\ 14.0625m^2/s^2\\\\Translational\ K_E = 17.58kg.m^2/s^2\\\\Translational\ K_E = 17.58J[/tex]

Therefore, the translational kinetic energy of the disk across the level surface is 17.58J

b)

Rotational kinetic energy

Rotational kinetic energy is the energy of rotation of a rotating rigid object or system of particles. its is expressed:

[tex]Rotational\ K_E = \frac{1}{2} Iw^2[/tex]

Where is moment of inertia around the axis of rotation and ω is the angular velocity.

Also, [tex]Moment\ of\ Inertia\ I = \frac{1}{2}mr^2[/tex]

Angular velocity ω is analogous to linear velocity v

So, [tex]v = wr \ and\ w = \frac{v}{r}[/tex]

Hence;

[tex]Rotational\ K_E = \frac{1}{2} * \frac{1}{2}mr^2* (\frac{v}{r})^2\\\\Rotational\ K_E = \frac{1}{2} * \frac{1}{2}mr^2* \frac{v^2}{r^2}\\\\Rotational\ K_E = \frac{1}{2} * \frac{1}{2}mv^2[/tex]

We substitute in our values

[tex]Rotational\ K_E = \frac{1}{2} * \frac{1}{2}*2.50kg * (3.75m/s)^2\\\\Rotational\ K_E = \frac{1}{2} * \frac{1}{2}*2.50kg * 14.0625m^2/s^2\\\\Rotational\ K_E = 8.79kg.m^2/s^2\\\\Rotational\ K_E = 8.79J[/tex]

Therefore, the rotational kinetic energy of the disk is 8.79J

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Answer:

Carbon electrons are arranged in a way that makes it possible for them to form long chains. That also helps carbon maintain its properties.

Explanation:

Carbon electrons are arranged in a way that makes it possible for them to form long chains. That also helps carbon maintain its properties.

This Physics problem involves two parts using Newton's law of gravitation. For part (a) calculate the gravitational force between the 32 kg object and each of the other objects, and then find the net force. For part (b), set the forces as equal and solve for the appropriate distance, considering the force direction.

To solve these particular physics problems, which are based on Newton's law of gravitation, the following formula can be used: F = G* (m1*m2)/ r^2, where F stands for force, G is the gravitational constant (approximately 6.67 x 10^-11 N(m/kg)^2), m1 and m2 are masses of the objects and r is the distance between them.

For part (a) of the question, we need to calculate the gravitational force between the 32 kg object and each of the other objects, and then find the net force. And for part (b), we set the forces as equal and solve for the appropriate distance.

Remember, it is important to take into account the direction of the force. The two larger weights will each exert a force on the 32 kg weight that tries to pull it towards them. That means that if you measure distance from one of the larger weights, the forces exerted by it will be positive, and the forces from the other weight will be negative.

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Answer:

B. Path B

Explanation:

Since the velocity v is tangent to the circular path, the rock will follow the path B.

You can see the paths in the pic.

To find the speed of the package of mass m right before the collision, we can use the principle of conservation of energy. The common speed of the packages after the collision, if they stick together, can be found using the principle of conservation of momentum. If the collision is perfectly elastic, the rebounding height of the package of mass m can be determined using the principle of conservation of mechanical energy.

To answer the given question, we can use the principles of conservation of energy, momentum, and Newton's laws of motion.

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The package will hit the ground with a speed of 7.67 m/s. If the packages stick together, they will move with a speed of 2.56 m/s. In the case of a perfectly elastic collision, the package of mass m will rebound to the original height.

The subject of this question is Physics, specifically it applies the topics of kinetic energy, potential energy, momentum conservation and collisions in two scenarios: inelastic and elastic scenarios.

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The acceleration is 2 mph/min

Explanation:

The acceleration of an object is the rate of change of velocity. It is defined as follows:

[tex]a=\frac{v-u}{t}[/tex]

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

For the car in this problem, we have (taking North as positive direction):

u = 25 mph

v = 35 mph

t = 5 min

Substituting into the equation, we find the acceleration:

[tex]a=\frac{35 mph-25mph}{5min}=2 mph/min[/tex]

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Answer:

[tex]\omega_s=12.8886\ rad.s^{-1}[/tex]

Explanation:

Given that:

Now using the law of conservation of angular momentum:

angular momentum of tucked body=angular momentum of straight body

[tex]I_t.\omega_t=I_s.\omega_s[/tex]

[tex]16\times 15.708=19.5\times \omega_s[/tex]

[tex]\omega_s=12.8886\ rad.s^{-1}[/tex]

Answer:

weight

*the objects weight*

Answer:2155 J

Explanation:

Given

Change in Internal energy [tex]\Delta U=-1477 J[/tex] i.e. decrease in Internal Energy

Heat added to system [tex]Q=678 J[/tex]

According First law for a system

[tex]dQ=dU+dW[/tex]

[tex]678=-1477+dW[/tex]

[tex]dW=2155 J[/tex]

Thus 2155 J of work is done by system                      

Answer:

-24.76 kJ/mol

Explanation:

given,

mass of solid magnesium burned = 0.1375 g

the temperature increases by(ΔT) 1.126°C

heat capacity of of bomb calorimeter (C_{cal})= 3024 J/°C

heat absorbed by the calorimeter

    [tex]q_{cal} = C_{cal}\DeltaT[/tex]

    [tex]q_{cal} = 3024 \times 1.126[/tex]

    [tex]q_{cal} =3405.24\ J[/tex]

    [tex]q_{cal} =3.405\ kJ[/tex]

heat released by the reaction

    [tex]q_{rxn} = -q_{cal}[/tex]

    [tex]q_{rxn} = -3.405\ kJ[/tex]

energy density will be equal to heat released by the reaction divided by the mass of magnesium

Energy density = [tex]\dfrac{-3.405}{0.1375}[/tex]

Energy density = -24.76 kJ/mol

heat given off by burning magnesium is equal to -24.76 kJ/mol