A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both
move forward so that they are one-half their original distance from the pivot point,what will happen to the seesaw? Assume that both people are small enough compared to the length of the seesaw to be thought of as point masses.
A. It is impossible to say without knowing the masses.
B. It is impossible to say without knowing the distances.
C. The side the boy is sitting on will tilt downward.
D. Nothing will happen; the seesaw will still be balanced.
E. The side the girl is sitting on will tilt downward.

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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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:

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

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.

               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

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.

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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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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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:

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:

[tex]I=5369.375[/tex]

Explanation:

Given:

Considering merry-go-round as a disk, its moment of inertia is given as:

[tex]I_d=\frac{1}{2} M.r^2[/tex]

[tex]I_d=0.5\times 155\times 5.5^2[/tex]

[tex]I_d=2344.375\ kg.m^2[/tex]

Considering children as point masses, their moment of inertia is given as:

[tex]I_C=5(m.r^2)[/tex]

since there are 5 children

[tex]I_C=5\times20\times 5.5^2[/tex]

[tex]I_C=3025\ kg.m^2[/tex]

Now, total moment of inertia:

[tex]I=I_C+I_d[/tex]

[tex]I=3025+2344.375[/tex]

[tex]I=5369.375[/tex]

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:

[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:

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:

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

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

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:

[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.

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:

3.53 second

Explanation:

The formula for the height is

[tex]h=-16t^{2}+55t+5[/tex]

When it hits the ground, the height is zero.

So, put h = 0 in the above equation

[tex]0=-16t^{2}+55t+5[/tex]

[tex]16t^{2}-55t-5=0[/tex]

[tex]t=\frac{+55\pm \sqrt{55^{2}+4\times 5\times 16}}{2\times 16}[/tex]

[tex]t=\frac{+55\pm 57.84}{2\times 16}[/tex]

Take positive sign

t = 3.53 second.

Thus, the time taken to hit the ground is 3.53 second.

Answer:

do your best

Explanation:

babyboiiii∛√

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:

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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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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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:

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:

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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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:

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.

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.

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:

The rubber ball

Explanation:

In order to understand this, let's begin with the fact that momentum it's a vector quantity that has mass, sense, direction and a numerical value.

Now, we have the ball and the putty, in both hands. Both of them, has the same mass, let's say they have a mass of 20 g each.

In this point, you throw both balls against the wall, and they have a speed of 10 m/s (I'm assuming these values); from the moment that you let go the balls, they are both have a momentum, and as they have the same speed and mass, the momentum it's the same for both of them.

Now, they hit the wall. The putty sticked to the wall, so it's movement finished. At this point it's momentum becomes zero. Even though it still has mass, but it's not moving, so momentum equals zero here. However, inthe ball bounces back to you, at this point, the ball even with a reduced speed, it still has a momentum, so, it's greater than the one that the putty has because it becomes zero. Therefore, the ball has a greater change in momentum.

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

The gravitational force that Mars would exert on a man with a mass of 68.0 kg standing on its surface is approximately 252.28 N (Newtons).

To calculate the gravitational force (Weight) that Mars would exert on a man standing on its surface, we can use the formula for the weight which is 'W = mg', where 'm' is the mass of the man and 'g' is the acceleration due to gravity. However, on Mars, the value of 'g' (acceleration due to gravity) is different than on Earth. On Mars, 'g' is approximately 3.71 m/s².

Therefore, by substituting the given and calculated values into the formula we get:
W = mg = 68.0 kg x 3.71 m/s² = 252.28 N.
So, the gravitational force that Mars would exert on the person would be approximately 252.28 N (Newtons).

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