What is the free-fall acceleration at the surface of the jupiter?

Answers

Answer 1

The free-fall acceleration at the surface of Jupiter is approximately 24.79 m/s², which is more than two and two thirds times the gravitational pull experienced on Earth. A person weighing 150 pounds on Earth would weigh around 400 pounds on Jupiter.

The free-fall acceleration at the surface of Jupiter is approximately 24.79 m/s². This value is derived from using Newton's Law of Universal Gravitation and considering Jupiter's mass and radius. Jupiter has a mass about 300 times that of Earth and a radius approximately 11 times larger. The gravitational acceleration (g) at a planet's surface is given by the formula:

g = G x (mass of the planet) / (radius of the planet)²,

where G is the gravitational constant. Since the mass of Jupiter is much greater than that of Earth, and despite its larger radius, the acceleration due to gravity would be significantly higher on Jupiter. Therefore, an astronaut entering Jupiter's atmosphere would fall faster compared to falling through the Earth's atmosphere. If an astronaut who weighs 150 pounds on Earth were to stand on a scale on Jupiter, she would weigh approximately 400 pounds, which is more than two and two thirds times her weight on Earth. However, this value may vary slightly depending on whether she is near Jupiter's pole or equator, due to its oblateness.


Related Questions

The purpose of a master production schedule (MPS) is to break down the aggregate planning decisions into such details as order sizes and schedules for individual subassemblies and resources by week and day. True or False?

Answers

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.

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).1. coasting universe2. recollapsing universe3. critical universe

Answers

Answer:

do your best

Explanation:

babyboiiii∛√

A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam. Then it drives the beam 12.0 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest. a) m=2100 kg b) Xi=5.00m c) Xf=12.0 cm =.12m

Answers

Answer:

   f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

from work done = force x distance work done = change in potential energy of the pile driverequating the two equations above we have

               force x distance = m x g x (s - d)

              f x 0.12 = 2100 x 9.8 x (5- (-0.12))

              d = - 0.12 because the steel beam went down at we are taking its  

              initial position to be an origin point which is 0

              f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12

                   f = 878,080 N

Final answer:

The average force that the pile driver exerted on the steel beam can be calculated using energy considerations, specifically by using the principle of conservation of energy. The potential energy of the pile driver is converted into the work done to drive the beam into the ground. This results in an average force of approximately 857,500 Newtons.

Explanation:

To solve this problem, we can first consider the principle of conservation of energy. The energy of the pile driver, when it starts falling, is purely potential energy, and when it has driven the steel beam into the ground, it's all been converted to work done against the resistance of the ground.

Firstly, calculate the potential energy of the pile driver as it begins to fall. The formula for potential energy (P.E.) is mass (m) times the acceleration due to gravity (g), which is about 9.8 m/s², times the height (h, the distance fallen): P.E. = m * g * h = 2100 kg * 9.8 m/s² * 5m = 102,900 Joules.

Secondly, the work done (W) in driving the steel beam into the ground can be calculated using this energy. Since this work was done to overcome the force of the beam as it went into the ground, we can also write W = F * d, where F is the average force and d is the distance it drove the beam down (0.12m).

Lastly, solve for the average force (F) by rearranging the equation to F = W / d = 102,900 Joules / 0.12 m = approx. 857,500 Newtons. Assuming all the energy was used in driving the steel beam into the ground, the pile driver would have had to exert an average force of around 857,500 N.

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Two students on roller skates stand face-toface, then push each other away. One student has a mass of 93 kg and the second student 65 kg. Find the ratio of the magnitude of the first student’s velocity to the magnitude of the second student’s velocity.

Answers

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]

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a 13.0 kg block and release the block from rest. As the block falls, the cable unwinds without stretching or slipping. How far will the mass have to descend to give the cylinder 510 J of kinetic energy?

Answers

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

[tex]Kc=510=1/2*Ic*\omega c^2[/tex]

[tex]\omega c=\sqrt{510*2/Ic}[/tex]

Where the inertia is given by:

[tex]Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2[/tex]

Replacing this value:

[tex]\omega c=106.46rad/s[/tex]

Speed of the block will therefore be:

[tex]V_b=\omega_c*R_c=106.46*0.15=15.969m/s[/tex]

By conservation of energy:

Eo = Ef

Eo = 0

[tex]Ef = 510+1/2*m_b*V_b^2-m_b*g*h[/tex]

So,

[tex]0 = 510+1/2*m_b*V_b^2-m_b*g*h[/tex]

Solving for h we get:

h=16.67m

The mass would have to descend from a height of 13.01 meters.

Given the following data:

Kinetic energy = 510 Joules.Mass of cylinder = 8.00 kg.Diameter = 30.0 cm.Mass of block = 13.00 kg.

How to calculate the height.

First of all, we would determine the moment of inertia for the solid cylinder by using this formula:

[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 8 \times 0.15^2\\\\I=4 \times 0.0225[/tex]

I = 0.09 [tex]Kgm^2[/tex]

Next, we would determine its angular velocity by using this formula:

[tex]K.E =\frac{1}{2} I\omega^2\\\\\omega=\sqrt{\frac{2K.E}{I} } \\\\\omega=\sqrt{\frac{2 \times 510}{0.09} }\\\\\omega=\sqrt{11,333.33} \\\\\omega=106.46\;rad/s.[/tex]

For the speed:

[tex]V=r \omega\\\\V= 0.15 \times 106.46[/tex]

V = 15.97 m/s.

Now, we would calculate the height by applying the law of conservation of energy:

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\2gh=v^2\\\\h=\frac{v^2}{2g} \\\\h=\frac{15.97^2}{2\times 9.8} \\\\h=\frac{255}{19.6}[/tex]

h = 13.01 meters.

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A toy robot moved from a starting point, traveling in a straight line at different speeds and then turning around and returning to the starting point at different speeds. What is the total distance that the toy robot traveled?

Answers

Final answer:

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.

Explanation:Calculating Total Distance Traveled

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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The Pony Express was a mail delivery system in the Old West that used a series of men on horseback to deliver mail from St. Joseph, MO to Sacramento, CA along a trail that was 2000 miles long. True or false: If each rider traveled 100 miles, got a fresh horse every 10 miles, and maintained an average speed of 10 mi/hr, it took 150 horses for each delivery.A. TrueB. False

Answers

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

(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m high hill in the process?

Answers

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

work = 0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}[/tex])  (change in kinetic energy)work = power x timetherefore

        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

work = change in potential energy + change in kinetic energywork = (mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}[/tex])work = power x timetherefore

      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]

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind is blowing. If the wind's force on the rope is negligible, what drag force in Newtons does the wind exert on the ball? B)A box is sliding down an incline tilted at a 15° angle above horizontal. The box is initially sliding down the incline at a speed of 1.4 m/s. The coefficient of kinetic friction between the box and the incline is 0.37. How far does the box slide down the incline before coming to rest?C)An object weighing 3.9 N falls from rest subject to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b = 2.5 N ∙ s2/m2. What terminal speed will this object approach?

Answers

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]

Let's begin by determining the equilibrium position of a seesaw pivot. You and a friend play on a seesaw.Your mass is____
90 kg, and your friend’s mass is 60 kg. The seesaw board is 3.0 m long and has negligible mass. Where should the pivot be placed so that the seesaw will balance when you sit on the left end and your friend sits on the right end?

Answers

Final answer:

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.

Explanation:

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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"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in groups dozens of millions of lightyears away. What method do astronomers use with the Hubble to find such distances

Answers

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.

A balloon filled with helium gas has an average density of rhob = 0.27 kg/m3. The density of the air is about rhoa = 1.23 kg/m3. The volume of the balloon is Vb = 0.084 m3. The balloon is floating upward with acceleration a.

Answers

Final answer:

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.

Explanation:

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

A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.718 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.34 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. (a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round? stay the same increase decrease (b) Calculate the initial and final kinetic energies for this system.Ki = kJKf = kJ

Answers

Answer:

The kinetic energy of the system decrease

Ki = 5.78 KJ

Kf = 4.55 KJ

Explanation:

For answer this question we will use the law of the conservation of the angular momentum so,

Li = Lf

Where Li is the inicial momentum of all the system, and Lf is the final momentum of the system.

also, the angular momentum L can be calculated in two ways

L = IW

where I is the momentum of inertia and the W is the Angular velocity.

or,

L = MVD

where M is the mass, V is the lineal velocity and the D is the lever arm.

Therefore,

Li = Ld ( merry-go-round) + Lp ( person )

Lf = Ls

Where Ld is the angular momentum of the merry go round, Lp is the angular momentum of the person and Ls is the angular momentum of the sistem (merry-go-round +  person)

so,

[tex]L_d=I_dW_d[/tex]

Ld =  [tex]\frac{1}{2}M_dR^{2}W_d[/tex]

Ld = [tex]\frac{1}{2}(155) (2.63)^{2}(0.718*2\pi)[/tex]

Ld = 2418.43

and,

[tex]L_p=M_pV_pD[/tex]

Lp = (59.4)(3.34)(2.63)

Lp = 521.78

then,

Lf = Ls

L_d=I_sW_s

Lf = [tex](\frac{1}{2}(155)(2.63)^{2}+(59.4)(2.63^2))(W_s)[/tex]

[tex]Lf = 946.92W_s[/tex]

so, solving for Ws

Lf = Li

[tex]946.92W_s = 521,78 + 2418.43[/tex]

Ws =  3.1 rad/s

Finally, the inicial and the final Kinetic energy

Ki = [tex]\frac{1}{2}I_d(W_d)^2 + \frac{1}{2}M_p(V_p)^2[/tex]

Ki = 5786.284 J = 5.78 KJ

Kf = [tex]\frac{1}{2}I_s(W_s)^2[/tex]

Kf =  4549.97 J = 4.55 KJ

Then, The kinetic energy of the system decrease because Kf < Ki

A 49 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 0.7 m/s. The acceleration of gravity is 9.8 m/s2. Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole.

Answers

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]

Final answer:

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


Explanation:

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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The acceleration due to gravity on the surface of a planet is five times as large as it is on the surface of Earth. The mass density of the planet is known to be four times that of Earth. What is the radius of this planet in terms of Earth's radius?

Answers

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.

Flow around curved height contours requires the incorporation of the centrifugal force. What is the general term to describe the winds that flow along a curved trajectory above the level where friction plays a role?

Answers

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.

Which of the following systems acquisition methods requires staff to systematically go through every step in the development process and has a lower probability of missing important user requirements?
a.Systems development life cycle
b. Prototyping
c. End-user development
d. External acquisition
e. Object-oriented development

Answers

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.

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center?

a) 7/5 I
b) 3/5 I
c) 2/5 I
d) 1/7 I
e.2/7 I

Answers

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

During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by 2.00ºC . What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is 1050 kg/m

Answers

Answer:

-293 W

Explanation:

mass = density * volume = 1050 kg/m^3   *  0.002 m^3/min = 2.1 kg/min

Heat transferred = mass * specific heat capacity * change in temperature

                            = mcΔT

                            = 2.1 kg/min * 4186 J/kg-°C * -2 °C

                            = -17 581.2 kJ/min

                            = -17 581.2 kJ/60s

                            = -293 J/s

                            =  -293 W

The negative sign shows us that the heat is being given off the blood

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m and may be treated as uniform spherical objects.
Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

Answers

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.

A person in a car during a sudden stop can experience potentially serious chest injuries if the combined force exerted by the seat belt and shoulder strap exceeds 16,000 N. Assume the mass of the passenger is 80 kg and the initial speed of the car is 16 m/s. Describe what it would take to avoid injury.

Answers

Answer:

minimum time interval to stop = 0.08 seconds

minimum stopping distance  = 0.64 m

Explanation:

maximum force (F) = 16,000 N

mass (m) = 80 kg

initial velocity (U) =  16 m/s

what it would take for the passenger to avoid in this case refers to how long it would take the vehicle to come to a full stop and the stopping distance it would also take to come to a full stop. Therefore we are to find the time (t) and the distance (s)

from the impulse momentum equation,

impulse = change in momentum

Ft = m(V-U)   (V-U is the change in velocity Δv)

where V is the final velocity = 0

and t = time

16000 x t = 80 (0 - 16)

16000t = -1,280 (he negative sign tell us there is a decrease in momentum, so we would not be using it further)

t = 0.08 seconds   ( this is also the difference between the initial time when the vehicle started to come to a stop and the final time when it came to a full stop)

assuming the acceleration is constant, the stopping distance (s) would be given by the kinetic relation

change in distance (Δs) = \frac{(ΔV) x (Δt)}{2}

(Δ refers to change, that is final value - initial value)

Δs =  \frac{16 x 0.08}{2}

Δs = 0.64 m

** You pull a rope oriented at a 37° angle above the horizontal. The other end of the rope is attached to the front of the first of two wagons that have the same 30-kg mass. The rope exerts a force of magnitude T1 on the first wagon. The wagons are connected by a second horizontal rope that exerts a force of magnitude T2 on the second wagon. Determine the magnitudes of T1 and T2 if the acceleration of the wagons is 2.0 m⁄s2.

Answers

Answer:

T2= 60 N and T1= 150,25 N

Explanation:

a free body diagram has to be done

The magnitude of the forces T1 and T2 if the acceleration is 2m/s² is 60N and 150.26N respectively.

Find the free body diagram attached. According to newton's second law;

[tex]\sum F_x = ma_x[/tex]

∑Fx is the sum of applied force in the horizontal direction

m is the mass of the object

ax is the acceleration of the object

For the body of mass 30kg

∑T = ma

T2 = ma

T2 = 30 * 2

T2 = 60N

For the sum of force acting on the second body;

T1 cos θ - T2 = ma

T1 cos 37 - 60 = 30(2)

T1 cos 37 = 120

T1 = 120/cos37

T1 = 120/0.7986

T1 = 150.26N

This shows that the magnitude of the forces T1 and T2 if the acceleration is 2m/s² is 60N and 150.26N respectively.

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"A high-mass star near the end of its life undergoes successive cycles of energy generation within its core in which gravitational collapse increases the temperature to the point where a new nuclear fusion cycle generates sufficient energy to stop the collapse. This process does not work beyond the silicon-fusion cycle that produces iron. Why is this?A. Electrostatic forces between the highly charged iron nuclei are sufficient to overcome the collapse and stabilize the stellar core.B. Iron nuclei are so large that they occupy all remaining space and so the collapse cannot continue.C. 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.D. The pressure from high-energy photons and neutrinos at the very high core temperatures reached at this stage of development is finally sufficient to halt the collapse.

Answers

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.

A 11.0 g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 10.0 kg, suspended from a cord 70.0 cm long.
a) Compute the vertical height through which the pendulum rises.(cm)
b) Compute the initial kinetic energy of the bullet;(j)
c) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.(j)

Answers

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:

Mass of bullet = 11.0 gSpeed = 380 m/sMass of pendulum = 10.0 kgLength of cord = 70.0 cm

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:

[tex]M_b[/tex] is the mass of bullet.[tex]M_p[/tex] is the mass of pendulum.[tex]V_b[/tex] is the velocity of bullet.V is the final velocity.

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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A passenger on a balloon drops a baseball over the side of the gondola. As the baseball falls faster, the drag force from air resistance increases.
Which of these describes what happens to the motion of the ball from the time the ball is dropped to the time when the drag force becomes equal to the force of gravity?

A The acceleration of the ball remains constant.
B The speed of the ball decreases.
C The acceleration of the ball decreases.
D The speed of the ball remains constant.

Answers

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 velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the displacement. -9/2 Correct: Your answer is correct. m (b) Find the distance traveled by the particle during the given time interval.

Answers

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.

Final answer:

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

Explanation:

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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An Olympic diver is on a diving platform 8.60 m above the water. To start her dive, she runs off of the platform with a speed of 1.23 m/s in the horizontal direction. What is the diver's speed (the sum of her horizontal and vertical velocities), in m/s, just before she enters the water?

Answers

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.

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is hW = 19.0 cm .

Answers

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

The angular speed of an automobile engine is increased at a constant rate from 1200 rev/min to 3000 rev/min in12 s.(a) What is its angular acceleration in revolutions per minute­squared?(b) How many revolutions does theengine make during this 12 s interval?

Answers

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.

What is angular acceleration?

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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Two pendulums have identical periods. One has a slightly larger amplitude than the other, but both swing through small angles compared to vertical. Which of the following must be true of the pendulum that has the larger amplitude?
Check all that apply.
a) It has more mass than the other one.
b) It is longer than the other one.
c) It moves faster at the lowest point in its swing than the other one.
d) It has slightly more energy than the other one.

Answers

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.

Other Questions
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