The magnitude of the tidal force between the International Space Station (ISS) and a nearby astronaut on a spacewalk is approximately

2GmMa/r3 . In this expression, M is the mass of the Earth, r=6.79×106m is the distance from the center of the Earth to the orbit of the ISS, m=135kg is the mass of the astronaut, and a=13m is the distance from the astronaut to the center of mass of the ISS. Calculate the magnitude of the tidal force for this astronaut. This force tends to separate the astronaut from the ISS if the astronaut is located along the line that connects the center of the Earth with the center of mass of the ISS.

Answer:

a) The displacement is -4.5 m.

b) The traveled distance is 11.7 m.

Explanation:

Hi there!

a)The velocity of the particle is the derivative of the displacement function, x(t):

v(t) = dx/dt = 5t - 9

Separating varibles:

dx = (5t - 9)dt

Integrating both sides from x = x0 to x and from t = 0 to t.

x - x0 = 1/2 · 5t² - 9t

x = 1/2 · 5t² - 9t + x0

If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):

x(3) = 1/2 · 5 · (3)² - 9(3) + 0

x(3) = -4.5

The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.

b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:

v = 5t - 9

0 > 5t - 9

9/5 > t

1.8 s > t

Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.

Let´s find the position of the particle at that time:

x = 1/2 · 5t² - 9t

x = 1/2 · 5(1.8 s)² - 9(1.8 s)

x = -8.1 m

From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.

Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.

The distance travelled by the particle during the given time interval is 45/2 meters.

The distance travelled by the particle can be found by integrating the absolute value of the velocity function over the given time interval:

The distance is, therefore, the integral of |5t - 9| from 0 to 3:

D = ∫(5t - 9)dt = [5(t^2/2) - 9t] from 0 to 3 = (45/2) meters

So, the distance travelled by the particle during the given time interval is 45/2 meters.

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Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

[tex]T=2\pi\sqrt{\dfrac{I}{mgh}}[/tex]

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

[tex]I=I_{cm}+md^2[/tex]

For a meter stick mass m , the rotational inertia about it's center of mass

[tex]I_{cm}-\dfrac{mL^2}{12}[/tex]

Where, L = 1 m

Put the value into the formula of time period

[tex]T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}[/tex]

[tex]T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}[/tex]

[tex]T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})[/tex]

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

[tex](\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0[/tex]

Put the value of T, L and g into the formula

[tex]4.028d^2-6.25d+0.336=0[/tex]

[tex]d = 0.056\ m, 1.496\ m[/tex]

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

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

Put the value into the formula

[tex]T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}[/tex]

[tex]T=1.35\ sec[/tex]

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Answer:

[tex]F = 1.489*10^{-7}  N[/tex]

Explanation: Weight of space probes on earth is given by:[tex]W= m*g[/tex]

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 [tex]\frac{m}{s^{2} }[/tex])

Therefore,

[tex]m_{1} = \frac{14500}{9.81}[/tex]

[tex]m_{1} = 1478.08  kg[/tex]

Similarly,

[tex]m_{2} = \frac{4800}{9.81}[/tex]

[tex]m_{2} = 489.29  kg[/tex]

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

[tex]F =  \frac{Gm_{1} m_{2}}{R^{2} }[/tex]

G= gravitational constant ([tex]6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}[/tex])

[tex]m_{1} , m_{2}[/tex]= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

[tex]F = 1.489*10^{-7}  N[/tex]

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

Answer:

Part a)

[tex]F_v = 4.28 N[/tex]

Part B)

[tex]L = 1.02 m[/tex]

Part C)

[tex]v = 1.25 m/s[/tex]

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

[tex]Tcos\theta = mg[/tex]

[tex]T sin\theta = F_v[/tex]

[tex]\frac{F_v}{mg} = tan\theta[/tex]

[tex]F_v = mg tan\theta[/tex]

[tex]F_v = 1.2\times 9.81 (tan20)[/tex]

[tex]F_v = 4.28 N[/tex]

Part B)

Here we can use energy theorem to find the distance that it will move

[tex]-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2[/tex]

[tex](-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2[/tex]

[tex](-3.5 + 2.54)L = - 0.98[/tex]

[tex]L = 1.02 m[/tex]

Part C)

At terminal speed condition we know that

[tex]F_v = mg[/tex]

[tex]bv^2 = mg[/tex]

[tex]2.5 v^2 = 3.9[/tex]

[tex]v = 1.25 m/s[/tex]

Answer:

a.Systems development life cycle

Explanation:

Of all the options the correct answer is a.Systems development life cycle.

Systems development life cycle: The life cycle phases of systems development include preparation, system assessment, system design, advancement, application, inclusion and testing, as well as maintenance and support.

So, we can see that the Systems development life cycle enables staff to systematically go through every step in the development process and has a lower probability of missing important user requirements.

Final answer:

The Systems Development Life Cycle (SDLC) is a methodical approach that encompasses a thorough step-by-step process to ensure all user requirements are captured, reducing the risk of overlooking important needs.

Explanation:

The system acquisition method that requires staff to systematically go through every step in the development process and has a lower probability of missing important user requirements is the Systems Development Life Cycle (SDLC). SDLC is a structured process that involves detailed planning, building, testing, and deployment, ensuring that all user requirements are met comprehensively.

Unlike prototyping, which may be quicker but less thorough, or end-user development which might miss broader system requirements, SDLC's methodical approach reduces the chance of overlooking user needs. Furthermore, external acquisition and object-oriented development are not as specifically focused on capturing all user requirements through a step-by-step process.

Answer:

(A) time = 3.205 s

(B)time =4.04 s

Explanation:

mass (m) = 850 kg

power (P) = 40 hp = 40 x 746 = 29,840 W

final velocity (Vf) =  15 m/s

final height (Hf) = 3 m

since the car is starting from rest at the bottom of the hill, its initial velocity and initial height are both 0

(A) from the work energy theorem

        power x time = 0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}[/tex])

        time = [tex]\frac{0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}[/tex])}{power}[/tex]

time = [tex]\frac{0.5 x 850 x ([tex](15)^{2} - (0)^{2}[/tex])}{29,840}[/tex]

time = 3.205 s

(B) from the work energy theorem

      power x time = (mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}[/tex])

      time = [tex]\frac{(mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}[/tex])}[/tex])}{power}[/tex]

time = [tex]\frac{(850 x 9.8 x (3 - 0)) + (0.5 x 850 x [tex](15)^{2} - (0)^{2}[/tex])}[/tex])}{29,840}[/tex]

time =4.04 s

Answer:

a) [tex]\Delta t = 3.205\,s[/tex], b) [tex]\Delta t = 4.043\,s[/tex]

Explanation:

a) The time needed is determined by the Work-Energy Theorem and the Principle of Energy Conservation:

[tex]K_{1} + \Delta E = K_{2}[/tex]

[tex]\Delta E = K_{2} - K_{1}[/tex]

[tex]\dot W \cdot \Delta t = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]\Delta t = \frac{m\cdot v^{2}}{2\cdot \dot W}[/tex]

[tex]\Delta t = \frac{(850\,kg)\cdot \left(15\,\frac{m}{s} \right)^{2}}{2\cdot (40\,hp)\cdot \left(\frac{746\,W}{1\,hp} \right)}[/tex]

[tex]\Delta t = 3.205\,s[/tex]

b) The time is found by using the same approach of the previous point:

[tex]U_{1} + K_{1} + \Delta E = U_{2} + K_{2}[/tex]

[tex]\Delta E = (U_{2}-U_{1})+(K_{2} - K_{1})[/tex]

[tex]\dot W \cdot \Delta t = m\cdot \left(g\cdot \Delta h + \frac{1}{2}\cdot v^{2} \right)[/tex]

[tex]\Delta t = \frac{m\cdot\left(g\cdot \Delta h + \frac{1}{2}\cdot v^{2}\right)}{\dot W}[/tex]

[tex]\Delta t = \frac{(850\,kg)\cdot \left[\left(9.807\,\frac{m}{s^{2}} \right)\cdot (3\,m) + \frac{1}{2}\cdot \left(15\,\frac{m}{s} \right)^{2}\right]}{(40\,hp)\cdot \left(\frac{746\,W}{1\,hp} \right)}[/tex]

[tex]\Delta t = 4.043\,s[/tex]

Answer:

The angular acceleration is  

=

15.71

r a d s −  2  and the number of revolutions is  = 419.9

Explanation:

a)  The angular acceleration of the automobile  is 9000 rev/min².

b)   The engine makes 420 revolution during this 12 s interval.

The temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second.

Rotational acceleration is another name for angular acceleration. It is a numerical representation of the variation in angular velocity over time.

Initial angular speed = 1200 rev/min.

Final  angular speed = 3000 rev/min.

Time taken = 12 second = 0.2 minute.

a)  its angular acceleration is = (final angular speed - Initial angular speed )/ Time taken

= ( 3000 rev/min - 1200 rev/min)/0.2 minute

= 9000 rev/min²

b)  The engine makes during this 12 s interval = (Initial angular speed + Final  angular speed) × time interval/2

= (1200 + 3000)× 0.2/2 revolution

= 420 revolution.

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To balance a seesaw with a 90 kg person on one end and a 60 kg person on the other end, the pivot should be placed 1.2 m from the 90 kg person. This is calculated using concepts of physics, specifically torque and equilibrium, assuming the force is applied at the person's center of mass.

In physics, this problem can be solved using the concept of torque and the conditions for equilibrium. For the seesaw to be in balance or equilibrium, the total torque about the pivot point must be zero. Torque (τ) is defined as the product of the force (F) applied and the distance (d) from the pivot point where the force is applied, i.e., τ = Fd.

In this case, let's assume the pivot is placed x meters from your end of the seesaw. The weights of you and your friend can be represented as forces through multiplication by gravity (approx. 9.81 m/s^2). So, for you, the torque is (90 kg x 9.81 m/s^2)  x and for your friend, it is (60 kg x 9.81 m/s^2) (3 m - x).

In equilibrium, these two torques should be equal, so we get the equation: (90 kg x 9.81 m/s^2)  x = (60 kg x 9.81 m/s^2)  (3 m - x). Solving this equation gives x = 1.2 m. So, the pivot should be placed 1.2 m from your end (90 kg person) for the seesaw to balance.

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Answer: Gradient Wind

Explanation:

Gradient wind, is the wind that accounts for air flow along a curved trajectory. It is an extension of the concept of geostrophic wind; for example the wind assumed to move along straight and parallel isobars (lines of equal pressure). The gradient wind represents the actual wind better than the geostrophic wind, especially when both wind speed and trajectory curvature are large, because they are in hurricanes and jet streams.

An Olympic diver is on a diving platform 8.60 m above the water. The diver's total speed just before entering the water is [tex]14.45\ m/s[/tex].

Horizontal motion:

The horizontal component of her velocity remains constant throughout the motion.

Horizontal velocity [tex](v_{horizontal}) = 1.23\ m/s[/tex]

Vertical motion:

The diver is subject to free fall in the vertical direction, starting from rest. The equations of motion for vertical free fall are:

[tex]h = (1/2) \times g \times t^2\\v_{vertical} = g \times t[/tex]

Where:

h is the vertical displacement [tex](8.60\ m)[/tex],

g is the acceleration due to gravity,

t is the time of flight.

The first equation for t:

[tex]t^2 = (2 \times h) / g\\t = \sqrt{((2 \times 8.60 ) / 9.8 )}\\t = 1.47 s[/tex]

Then, use the second equation to find the vertical velocity:

[tex]v_{vertical} = g \times t\\v_{vertical} = 9.8 \times 1.47 \\v_{vertical} = 14.406\ m/s[/tex]

Now, the Pythagorean theorem to find the total speed just before entering the water:

[tex]Total speed = \sqrt{((v_{horizontal})^2 + (v_{vertical})^2)}\\Total speed = \sqrt{((1.23 )^2 + (14.406 )^2)}\\Total speed = 14.45 m/s[/tex]

So, the diver's total speed just before entering the water is [tex]14.45\ m/s[/tex].

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Final answer:

To find the diver's speed just before she enters the water, we need to determine her horizontal and vertical velocities. By solving equations related to vertical motion and using the values provided, we can calculate the diver's vertical velocity. Her speed is the magnitude of the sum of her horizontal and vertical velocities.

Explanation:

To find the diver's speed just before she enters the water, we need to determine her horizontal and vertical velocities. Since she runs off the platform horizontally, her horizontal velocity remains constant. The vertical velocity can be found using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. In this case, the acceleration is due to gravity, which is approximately 9.8 m/s². Initially, the diver has no vertical velocity, so vi = 0 m/s. The time it takes for the diver to reach the water can be found using the equation d = vi × t + 0.5 × a × t², where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. In this case, the distance is equal to the height of the platform, which is 8.6 m. By solving these equations, we can find the diver's vertical velocity just before she enters the water. The diver's speed is the magnitude of the sum of her horizontal and vertical velocities.

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

The total distance a toy robot travels in a straight line and back again is double the distance from start to the turning point, regardless of speed changes. Displacement, however, is different as it considers only the final and initial points, thus it would be zero in this case.

The total distance that a toy robot travels is calculated by adding the overall length of the path it followed, regardless of its direction. If the toy robot moved in a straight line from a starting point, traveled at different speeds, and then turned around and returned to the starting point, its total distance is double the distance from its starting point to its farthest point.

For example, if the toy robot traveled 2km in a straight line from its starting point, turned around, and returned to its starting point, the total distance traveled is 2km + 2km = 4km, regardless of changes in its speed during the journeys.

Note that this concept differs from displacement, which would be zero in this case as the robot ended up at its initial point.

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To compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.

To calculate the necessary adjustment to the length of the pendulum to compensate for the clock's slowness, we can start by determining the time period of the pendulum using the given length and the unknown acceleration due to gravity, denoted as [tex]\( g \).[/tex] The time period [tex]\( T \)[/tex] of a simple pendulum is given by the formula:

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

where [tex]\( L \)[/tex] is the length of the pendulum.

Given that the clock loses 19 seconds per day, we convert this to seconds per period, as [tex]\( 1 \) day has \( 24 \) hours, \( 60 \)[/tex] minutes per hour, and [tex]\( 60 \)[/tex] seconds per minute:

[tex]\[ \text{Seconds lost per period} = \frac{19 \, \text{s}}{24 \times 60 \times 60 \, \text{s/day}} \][/tex]

Let's calculate the time period [tex]\( T \)[/tex]  and the adjustment needed.

[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \][/tex]

Now, we'll solve for [tex]\( g \):[/tex]

[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \]\[ T^2 = 4\pi^2 \frac{0.9930 \, \text{m}}{g} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{T^2} \][/tex]

Substitute [tex]\( T \) with \( 24 \, \text{hours} \)[/tex] converted to seconds:

[tex]\[ T = 24 \times 60 \times 60 \, \text{s} = 86,400 \, \text{s} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{(86,400 \, \text{s})^2} \][/tex]

Calculate [tex]\( g \):[/tex]

[tex]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{7.48224 \times 10^9 \, \text{s}^2} \]\[ g = 0.0000016512 \, \text{m/s}^2 \][/tex]

Now that we have [tex]\( g \)[/tex], we can find the adjusted length of the pendulum. The new time period [tex]\( T' \)[/tex] can be calculated as follows:

[tex]\[ T' = 2\pi \sqrt{\frac{L'}{g}} \][/tex]

Where [tex]\( L' \)[/tex] is the adjusted length. Rearrange to solve for [tex]\( L' \):[/tex]

[tex]\[ L' = \frac{T'^2 \times g}{4\pi^2} \][/tex]

Given that the clock loses [tex]\( 19 \)[/tex] seconds per day and we want it to lose [tex]\( 0 \)[/tex]  seconds, the new time period [tex]\( T' \) is \( 86,400 \)[/tex] seconds. Substitute into the formula:

[tex]\[ L' = \frac{(86,400 \, \text{s})^2 \times 0.0000016512 \, \text{m/s}^2}{4\pi^2} \][/tex]

[tex]\[ L' = 0.9939 \, \text{m} \][/tex]

So, to compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.

Complete Question:
Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).

Answer:

[tex]\frac{v_1}{v_2} = 0.698[/tex]

Explanation:

As we know that the two students are standing on skates

So there is no external force on the system of two students

So we can say that momentum is conserved

So here initially both students are at rest and hence initial momentum is zero

So we have

[tex]P_i = P_f[/tex]

[tex]m_1v_1 + m_2v_2 = 0[/tex]

[tex]\frac{v_1}{v_2} = \frac{m_2}{m_1}[/tex]

[tex]\frac{v_1}{v_2} = \frac{65}{93}[/tex]

[tex]\frac{v_1}{v_2} = 0.698[/tex]

The correct ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity is 65:93.

To find the ratio of the magnitudes of the velocities of the two students after they push each other away, we can use the principle of conservation of momentum. According to this principle, in the absence of external forces, the total momentum of a system remains constant.

 Let's denote the velocities of the first and second students as[tex]\( v_1 \)[/tex]and [tex]\( v_2 \)[/tex] respectively. Since the students push each other in opposite directions, their momenta will be equal in magnitude but opposite in direction. We can write the conservation of momentum as:

[tex]\[ m_1 \cdot v_1 = m_2 \cdot v_2 \][/tex]

 where [tex]\( m_1 = 93 \)[/tex] kg is the mass of the first student and[tex]\( m_2 = 65 \) kg[/tex] is the mass of the second student.

To find the ratio of the velocities, we divide both sides of the equation by[tex]\( m_2 \cdot v_2 \)[/tex]:

[tex]\[ \frac{m_1 \cdot v_1}{m_2 \cdot v_2} = 1 \][/tex]

[tex]\[ \frac{v_1}{v_2} = \frac{m_2}{m_1} \][/tex]

Substituting the given masses:

[tex]\[ \frac{v_1}{v_2} = \frac{65 \text{ kg}}{93 \text{ kg}} \][/tex]

Simplifying the ratio, we get:

[tex]\[ \frac{v_1}{v_2} = \frac{65}{93} \][/tex]

Answer:

False.

Explanation:

Master production schedule (MPS) is nothing but plan for the individual commodities to be produced in a factory, during to a time period. MPS includes Planning, production, staffing , inventory, etc. It preferably used in  places where it is know that when and how each product is demanded. It has nothing to deal with decision and breaking down of aggregate planning.

The definition of MPS given in question is wrong. There the given statement is false.

Answer:

a) r eq = -a/(2b)

b) k = a/r eq = -2b

Explanation:

since

U(r) = ar + br²

a) the equilibrium position dU/dr = 0

U(r) = a + 2br = 0 →  r eq= -a/2b

b) the Taylor expansion around the equilibrium position is

U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!  

,where Un(a) is the nth derivative of U respect with r , evaluated in a

Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative

U(r) = U(r eq) + dU/dr(r eq) (r- r eq) +  d²U/dr²(r eq) (r- r eq)² /2

since dU/dr(r eq)=0

U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2

comparing with an energy balance of a spring around its equilibrium position

U(r) - U(r eq)  = 1/2 k (r-r eq)² → U(r) = U(r eq)  + 1/2 k (r-r eq)²

therefore we can conclude

k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq

thus

k= a/r eq

Answer:

It moves faster at the lowest point in its swing than the other one.

Final answer:

The pendulum with the larger amplitude has slightly more energy than the one with the smaller amplitude. However, the mass, length, and speed of the pendulum at the lowest point do not necessarily differ between the two.

Explanation:

The pendulum with the larger amplitude must have more energy than the one with the smaller amplitude. The amplitude of the pendulum is directly related to the maximum displacement from the equilibrium position. The greater the amplitude, the greater the potential energy stored in the pendulum. Therefore, option d) It has slightly more energy than the other one is true for the pendulum with the larger amplitude.

However, the mass and length of the pendulum do not affect the amplitude of the pendulum. Therefore, options a) It has more mass than the other one and b) It is longer than the other one are not necessarily true.

Regarding the motion of the pendulum at the lowest point in its swing, both pendulums have the same period or time taken to complete one oscillation. This means that both pendulums have the same time to travel from the highest point to the lowest point. Therefore, option c) It moves faster at the lowest point in its swing than the other one is not true as both pendulums have the same speed at the lowest point in their swing.

Answer:

v = 37.4 m/s , h = 71.39m

Explanation:

To find the velocity given:

m = 0.4 kg

F =190.4 N

d = 1.47 m

g = 9.8 m/s^2

So use the equation of work to solve the kinetic energy

W = F *d = 190.4 N * 1.47m

W = 279.88 J

Ke = 1 / 2 * m* v^2

v = √2*Ke / m =√ 2 *279.88 / 0.4 kg

v = 37.4 m/s

Now to find the high to rise can use the conserved law so:

Ke = Pe

279.88 = m*g*h

Solve to h'

h = 279.88 / 0.4kg * 9.8m/s^2

h =71.39 m

Final answer:

The arrow leaves the bow with a speed of approximately 37.49 m/s and, when shot straight up, rises to a maximum height of about 71.4 m.

Explanation:

To determine the speed with which the arrow leaves the bow, we apply the work-energy principle, which states that work done on the arrow is converted into its kinetic energy. The work done W by the bow can be calculated by multiplying the force F exerted by the distance d over which the force is applied: W = F × d. It is given that F is 190.4 N and d is 1.47 m; thus, W = 190.4 N × 1.47 m = 279.888 J.

The kinetic energy KE of the arrow can be given by KE = ½ mv², where m is the mass of the arrow and v is its velocity. Since work done equals the kinetic energy, we get 279.888 J = ½ × 0.4 kg × v². Solving for v gives us a velocity of approximately 37.49 m/s.

To find how high the arrow goes if it's shot straight up, we use the conservation of energy, where the initial kinetic energy is converted to gravitational potential energy at the highest point. The potential energy PE at maximum height can be given by PE = mgh, where g is the acceleration due to gravity (9.8 m/s²) and h is the height. Setting KE equal to PE, we have 279.888 J = 0.4 kg × 9.8 m/s² × h. Solving for h gives us a maximum height of approximately 71.4 m.

Answer:

The answer is C.

Explanation:

For acceleration to be achieved so to speak there must be a force acting on it. The only force on the ball before the air drag increases is gravity. As the air resistance increases the force resisting gravity increases. This means that the forces start to cancel out. Therefore the acceleration must get smaller.

These forces will continue to cancel until it reaches terminal velocity.

What happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.

The downward motion of the ball is reduced by frictional force opposing the motion. The frictional force opposing the motion is the drag force of the air or air resistance.

The net downward force on the ball is given as;

[tex]W -F_D = ma\\\\[/tex]

when the drag force on the ball equals force of gravity, the acceleration of the ball will be zero.

[tex]W - W = ma\\\\(W = F_D)\\\\0 = ma\\\\a = 0[/tex]

Thus, we can conclude that what happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.

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The gauge pressure at the water/mercury interface is simply the "head" due to the water:

p = ρgh = 1000kg/m³ * 9.8m/s² * 0.19m = 1862 Pa

If we take the specific gravity of mercury to be 13.6, then the difference in height between the water and mercury columns is

h' = h(1 - 1/s.g.) = 19cm * (1 - 1/13.6) = 17.6 cm

Answer:

1.25 R

Explanation:

Acceleration due to gravity on earth, ge = g

Acceleration due to gravity on planet, gP = 5 times the acceleration due to gravity on earth

gP = 5 g

Density of planet = 5 x density of earth

Let the radius of earth is R

Let the radius of planet is Rp.

Use the for acceleration due to gravity

[tex]g = \frac{4}{3}G\pi R\rho[/tex]

where, G s the universal gravitational constant and ρ be the density of planet.

For earth

[tex]g = \frac{4}{3}G\pi R\rho[/tex] .... (1)

For planet

[tex]g_{P} = \frac{4}{3}G\pi R_{P}\rho_{P}[/tex]

According to the question

gp = 5 g, ρP = 4 ρ

Substitute the values

[tex]5g = \frac{4}{3}G\pi R_{P}\4rho[/tex]   .... (2)

Divide equation (2) by equation (1), we get

[tex]5=\frac{R_{p\times 4\rho }}{R\rho }[/tex]

Rp = 1.25 R

Thus, the radius of planet 1.25 R.

Answer:

C

Explanation:

The correct answer C part.

The phenemonon mention the question above happens only because Fusion of iron nuclei into heavier nuclei requires energy rather than producing excess energy and therefore will not produce the additional gas pressure to halt the collapse, hence the process does not work beyond the silicon- fusion cycle that produces iron.

Answer:

50 J

Explanation:

Work = force × distance

W = Fd

W = (5 N) (10 m)

W = 50 J

The amount of work done is 50 J (Joule)

To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.

By applying the formula, we get:

Work = force × distance

Force = 5 newtons                       ...(given)

distance = 10 meters                    ...(given)

⇒ W = Fd

⇒ W = (5 N) (10 m)

⇒ W = 50 J

Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

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Answer:

Height will be 5.127 m

Explanation:

We have given mass m = 49 kg

Speed over the bar v = 10 m /sec

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

Kinetic energy on the ground [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 10^2=2450J[/tex]

Potential energy on the ground [tex]=mgh=0[/tex] (as height will be zero )

Speed above the bar = 0.7 m /sec

So kinetic energy above the bar [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 0.7^2=12J[/tex]

Potential energy above the bar = mgh

From energy conservation

Total kinetic energy = total potential energy

So [tex]2450+12=0+49\times 9.8\times h[/tex]

[tex]h=5.127m[/tex]

Using the principle of conservation of energy, we can calculate the pole vaulter's height above the bar as being approximately 5.02 meters.

In this question, we're dealing with conservation of energy. The energy of the pole vaulter is kinetic energy when she is running (1/2*mass*speed^2), and as she goes over the bar, her energy becomes potential energy (mass*gravity*height) and a little kinetic energy (1/2*mass*speed^2).

Let's use the following formula: Kinetic Energy initial + Potential Energy initial = Kinetic Energy final + Potential Energy final.

Initially, her kinetic energy is 1/2 * 49 kg * (10 m/s)^2 = 2450 J. She has no potential energy, because she's on the ground (height=0). When she's above the bar, her kinetic energy is 1/2 * 49 kg * (0.7 m/s)^2 = 12.075 J. We don't yet know her final potential energy, because we're trying to find her height. So, we'll call her final potential energy m*g*h, or 49 kg * 9.8 m/s^2 * h.

We then plug these values into our energy equation: 2450 J + 0 J = 12.075 J + 49 kg * 9.8 m/s^2 * h. Solving for h gives us h = (2450 J - 12.075 J) / (49 kg * 9.8 m/s^2) = 5.02 meters, so this would be her altitude as she crosses the bar.

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Answer:

the given statement is False

Explanation:

given,                                                          

distance of the trail = 2000 miles long          

each rider traveled = 100 miles                        

every fresh horse travel = 10 miles                    

to maintain speed of = 10 mile/hr                      

the given statement is                                            

150 horses is used for each delivery.                    

if each horse is allowed to travel 10 miles to travel

distance traveled using 150 horses = 150 x 10

                                                            = 1500 miles

to travel 2000 miles horse required is equal to 200.

so, the given statement is False

Answer:

from los angeles distance plan a = 1111.08 mi

from los angeles distance plan b =  888.92 mi

Explanation:

given data

new york to los angeles distance = 2000 miles

Plane b speed = 100 mph faster than plane a

Plane b takes off  time = 1 hour after plane a

to find out

How far are they from los angeles when they pass

solution

we consider speed of plan a is = x mph

so speed of plan b will be = x + 100 mph

and we know plan b take here 1 hour less time than plan a so it mean time is distance divide speed i.e

[tex]\frac{2000}{x} - 1 =\frac{2000}{x+100}[/tex]

solve it we get x = 400 mph

it mean here

plan a speed is 400 mph

and plan b speed is 500 mph

and

now we consider they meet at time = t hour  after a take off

then plan a travel  = 400 t

and plan b travel = 500 ( t - 1 )

add both distance that is equal to 2000 mi

so 400 t + 500 ( t -1 ) = 2000

400 t + 500 ( t -1 ) = 2000

400 t + 500 ( t-1) = 2000

solve we get

t = 2.777

so total distance travel plan a = 400 × 2.777 = 1111.08 mi

total distance travel plan b = 2000 - 1111.08

total distance travel plan b = 888.92 mi

The student is asking about a helium-filled balloon's behavior in relation to its density and the density of air. To determine whether the balloon rises or falls, we compare the buoyant force with the gravitational force. The buoyant force is calculated using the formula (rhoa - rhob) * Vb * g.

The subject of this question is Physics.

The student is inquiring about a balloon filled with helium gas that has an average density of 0.27 kg/m3. The density of air is approximately 1.23 kg/m3. The volume of the balloon is 0.084 m3. The balloon is floating upwards with an acceleration.

To determine if the balloon is floating upwards or downwards, we need to compare the buoyant force with the gravitational force. If the buoyant force is greater, the balloon will rise; if it is less, the balloon will descend.

The buoyant force can be calculated using the formula:

Buoyant force = (rhoa - rhob) * Vb * g

where rhoa is the density of air, rhob is the density of the balloon, Vb is the volume of the balloon, and g is the acceleration due to gravity.

If the buoyant force is greater than the gravitational force (given by the formula mg, where m is the mass of the balloon and g is the acceleration due to gravity), then the balloon will float upwards.

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The balloon's acceleration is about 34.84 m/s².

First, let's calculate the buoyant force (F[tex]_b[/tex]) acting on the balloon using Archimedes' principle:

F[tex]_b[/tex] = ρ[tex]_a[/tex] × g × V[tex]_b[/tex],

where:

ρ[tex]_a[/tex] is the density of air = 1.23 kg/m³,

g is the acceleration due to gravity = 9.8 m/s²,

V[tex]_b[/tex] is the volume of the balloon = 0.084 m³.

Therefore, F[tex]_b[/tex] = 1.23 kg/m³ × 9.8 m/s² × 0.084 m³ = 1.012488 N.

Next, calculate the gravitational force (weight) of the balloon (W[tex]_b_a_l_o_o_n[/tex]):

W[tex]_b_a_l_o_o_n[/tex] = m[tex]_b_a_l_o_o_n[/tex]  × g,

where, m[tex]_b_a_l_o_o_n[/tex] is the mass of the balloon given by the product of its volume and density:

m[tex]_b_a_l_o_o_n[/tex] = ρ[tex]_b[/tex] × V[tex]_b[/tex] = 0.27 kg/m³ × 0.084 m³ = 0.02268 kg.

Thus, W[tex]_b_a_l_o_o_n[/tex] = 0.02268 kg × 9.8 m/s² = 0.222264 N.

The net force (F[tex]_n_e_t[/tex]) acting on the balloon is the difference between the buoyant force and the weight of the balloon:

F[tex]_n_e_t[/tex] = F[tex]_b[/tex] - W[tex]_b_a_l_o_o_n[/tex] = 1.012488 N - 0.222264 N = 0.790224 N.

Finally, use Newton's second law to find the acceleration (a) of the balloon:

F[tex]_n_e_t[/tex] = m[tex]_b_a_l_o_o_n[/tex] × a

⇒ a = F[tex]_n_e_t[/tex] / m[tex]_b_a_l_o_o_n[/tex],

a = 0.790224 N / 0.02268 kg ≈ 34.84 m/s².

Therefore, the acceleration of the balloon is approximately 34.84 m/s²

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

[tex]I_{CM} = \dfrac{2}{5}mR^2[/tex].....(1)

now, moment of inertia about tangent

[tex]I= \dfrac{2}{5}mR^2 + mR^2[/tex]

[tex]I= \dfrac{7}{5}mR^2[/tex]...........(2)

dividing equation (1)/(2)

[tex]\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}[/tex]

[tex]\dfrac{I_{CM}}{I}=\dfrac{2}{7}[/tex]

[tex]I_{CM}=\dfrac{2}{7}I[/tex]

the correct answer is option E

Answer:

do your best

Explanation:

babyboiiii∛√

Answer:

a) h = 0.0088 m

b) Kb = 794.2J

c) Kt = 0.88J

Explanation:

By conservation of the linear momentum:

[tex]m_b*V_b = (m_b+m_p)*Vt[/tex]

[tex]Vt = \frac{m_b*V_b}{m_b+m_p}[/tex]

[tex]Vt=0.42m/s[/tex]

By conservation of energy from the instant after the bullet is embedded until their maximum height:

[tex]1/2*(m_b+m_p)*Vt^2-(m_b+m_p)*g*h=0[/tex]

[tex]h =\frac{Vt^2}{2*g}[/tex]

h=0.0088m

The kinetic energy of the bullet is:

[tex]K_b=1/2*m_b*V_b^2[/tex]

[tex]K_b=794.2J[/tex]

The kinetic energy of the pendulum+bullet:

[tex]K_t=1/2*(m_b+m_p)*Vt^2[/tex]

[tex]K_t=0.88J[/tex]

a. The vertical height through which the pendulum rises is equal to 0.9 cm.

b. The initial kinetic energy of the bullet is equal to 794.2 Joules.

c. The kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum is equal to 0.883 Joules.

Given the following data:

a. To determine the vertical height through which the pendulum rises:

First of all, we would find the final velocity by applying the law of conservation of momentum:

Momentum of bullet is equal to the sum of the momentum of bullet and pendulum.

[tex]M_bV_b = (M_b + M_p)V[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]0.011\times 380 = (0.011+10)V\\\\4.18 = 10.011V\\\\V = \frac{4.18}{10.011}[/tex]

Final speed, V = 0.42 m/s

Now, we would find the height by using this formula:

[tex]Height = \frac{v^2}{2g} \\\\Height = \frac{0.42^2}{2\times 9.8} \\\\Height = \frac{0.1764}{19.6}[/tex]

Height = 0.009 meters.

In centimeters:

Height = [tex]0.009 \times 100 = 0.9 \;cm[/tex]

b. To compute the initial kinetic energy of the bullet:

[tex]K.E_i = \frac{1}{2} M_bV_b^2\\\\K.E_i = \frac{1}{2} \times 0.011 \times 380^2\\\\K.E_i = 0.0055\times 144400\\\\K.E_i = 794.2 \; J[/tex]

Initial kinetic energy = 794.2 Joules

c. To compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum:

[tex]K.E = \frac{1}{2} (M_b + M_p)V^2\\\\K.E = \frac{1}{2} \times(0.011 + 10) \times 0.42^2\\\\K.E = \frac{1}{2} \times 10.011 \times 0.1764\\\\K.E = 5.0055 \times 0.1764[/tex]

Kinetic energy = 0.883 Joules.

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