The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in magnitude and direction

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:

V = 1.7 x 10^{-3} m^{3}

Explanation:

apparent weight in ethyl alcohol = 16.5 N

apparent weight in water = 13.3 N

apparent weight = mg - ρgV

where

m = mass

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

ρ = density

V = volume

    we can solve the two equations above as simultaneous equations by subtracting equation 2 from equation 1

we have

16.5 - 13.3 = (mg - mg) - ρ₁gV - (-ρ₂gV)

3.2 = gV(ρ₂ - ρ₁)

V = \frac{3.2}{g(ρ₂-ρ₁)}

where

ρ₂ = density of water = 1000 kg/m^{3}

ρ₁ = density of ethyl alcohol = 806 kg/m^{3}

V = \frac{3.2}{9.8(1000 - 806)}

V = 1.7 x 10^{-3} m^{3}

The volume of the object can be found using Archimedes' principle and the given apparent weights in water and ethyl alcohol. The volume is computed from these values using the densities of the two fluids, yielding a volume of about 0.013 cubic meters.

To answer the question about the object's volume when it's submerged in ethyl alcohol and water, we'll need to use Archimedes' principle. This principle tells us that the apparent weight loss of the object in a liquid equals the weight of the fluid displaced by the object. In this case, the difference in apparent weights given for the object when in water and when in ethyl alcohol represents the weight of the fluids displaced.

We have to remember that weight can be calculated as the product of volume, density, and gravity. If you rearrange this equation, you get the volume as the quotient of weight and the product of density and gravity. Note that gravity cancels out in this problem since it remains constant for both fluids.

If we denote the volume of the object as V, the density of water as ρ_water (approximated to 1000 kg/m^3), the density of ethyl alcohol as ρ_alcohol (approximated to 789 kg/m^3), and the weight of the object in water and ethyl alcohol as W_water and W_alcohol respectively, we can write two equations:

We have the weights as the apparent weights from the problem, which are W_water = 13.3 N and W_alcohol = 16.5 N. If you solve these two equations, you should obtain the volume of the object, which should be around 0.013 m^3. This is the answer for the volume of the solid object.

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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:

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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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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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:

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:

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:

-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

Final answer:

Georges Lemaître, a Belgian cosmologist, proposed the initial concept of the Big Bang, predicting the universe's expansion and Hubble's Law before they were empirically observed. His ideas on the universe starting from a 'primeval atom' and his work on particle interactions in the early universe laid the foundations of modern cosmology.

Explanation:

Georges Lemaître was a Belgian priest and cosmologist who made significant contributions to the Big Bang theory. Born in 1894, Lemaître studied theology as well as mathematics and physics. He was instrumental in exploring the concept of the expanding universe and was the first to propose a concrete model of the Big Bang. This model suggested that the universe started as a single 'primeval atom', which eventually fragmented into smaller pieces, leading to the formation of the current atoms in the universe through a process akin to nuclear fission. Lemaître's work predated and anticipated the empirical findings that became known as Hubble's Law, which observed that galaxies are moving away from each other, implying that the universe is expanding. His insights laid the groundwork for the understanding of the universe's beginnings and its initial hot, dense state.

Additionally, Lemaître, alongside collaborators such as George Gamow, further developed the Big Bang theory. They predicted that as the universe expands and cools, the interactions among particles would lead to the formation of protons, neutrons, and eventually the nuclei of light elements such as deuterium, helium, and lithium. This theoretical framework is bolstered by measurements such as the cosmic microwave background radiation and the abundance of deuterium, supporting the Big Bang theory as a robust model of the universe's inception.

Answer:

v₃ = (10) i + (10) j

v₃ = 10√2

Explanation:

Given info

Before the explosion

ux = 0

uy = 0

After the explosion

v₁x = -30

v₁y = 0

v₂x = 0

v₂y = -30

We can use the Principle of Conservation of Momentum as follows

pi = pf

where

pix = M*ux = M*0 = 0

piy = M*uy = M*0 = 0

pfx = p₁x + p₂x + p₃x = (m*v₁x + m*v₂x + 3m*v₃x) = m*(-30) + m*(0) + 3m*v₃x

⇒   pfx = -30m + 3m*v₃x

if

pix = pfx     ⇒    0 = -30m + 3m*v₃x    ⇒  v₃x = 10

pfy = p₁y + p₂y + p₃y = (m*v₁y + m*v₂y + 3m*v₃y) = m*(0) + m*(-30) + 3m*v₃y

⇒   pfy = -30m + 3m*v₃y

if

piy = pfy     ⇒    0 = -30m + 3m*v₃y    ⇒  v₃y = 10

then we have

v₃ = (10) i + (10) j

and its module can be obtained as follows

v₃ = √(v₃x² + v₃y²) = √(10² + 10²) = 10√2

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 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:

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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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

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:

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.

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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Answer:

I and II

Explanation:

A management system that includes fire suppression will likely lead to

I. large quantities of biomass accumulating on the forest floor.

II. an increase in the likelihood of uncontrolled natural fires.

Biomass is the total quantity of weight of flora and fauna in a given area or volume. The recent fire in the Amazon rain forest is an example of uncontrolled natural forest fires.

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:

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:

2 [tex]\sigma[/tex] bonds and 2 [tex]\pi[/tex] bonds

Explanation:

If we consider the the bonding in the [tex]CO_{2}[/tex] molecule:

[tex]O = 1s^{2}\ 2s^{2}\ 2p^{4}[/tex]

[tex]C = 1s^{2}\ 2s^{2}\ 2p^{2}[/tex]

Thus carbon forms double bonds with oxygen:

O = C = O

Now,

We know that double bond comprises of a [tex]\sigma\ bond[/tex] and a [tex]\pi \ bond[/tex]

Since, in the [tex]CO_{2}[/tex], there are 2 double bonds thus there are 2 [tex]\sigma[/tex] bonds and 2 [tex]\pi[/tex] bonds in the molecules.

"The CO₂ molecule has 2 σƒ (sigma) bonds and 2 π (pi) bonds.

To determine the number of sigma and pi bonds in CO₂, we need to consider its Lewis structure. Carbon dioxide has a linear molecular geometry with carbon at the center and two oxygen atoms double-bonded to it.

In CO₂:

- The carbon atom forms two double bonds with the two oxygen atoms.

- Each double bond consists of one sigma bond and one pi bond.

Therefore, for each carbon-oxygen double bond, there is:

- One sigma bond (σƒ), which is the head-on overlap of atomic orbitals.

- One pi bond (π), which is the side-to-side overlap of p-orbitals.

Since there are two carbon-oxygen double bonds in CO₂, we have:

- A total of 2 sigma bonds from the double bonds.

- Additionally, each double bond includes a sigma bond from the overlap of the sp hybrid orbital of carbon with the sp2 hybrid orbital of oxygen, contributing another 2 sigma bonds.

In summary, CO₂ has 4 sigma bonds in total:

- 2 sigma bonds from the sp-sp² hybrid orbital overlaps.

- 2 sigma bonds from the head-on overlap of p-orbitals that form part of the double bonds.

And CO₂ has 2 pi bonds in total:

- 2 pi bonds from the side-to-side overlap of p-orbitals in the double bonds.

Thus, the final count is 4 σƒ bonds and 2π bonds in the CO₂ molecule. However, it is important to note that the question specifically asks for the number of sigma and pi bonds, and the correct answer should reflect the number of each type of bond individually, not the total number of bonds. Therefore, the correct answer is 2σ bonds and 2π bonds.

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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Answer:

Explanation:

Answer:

3300J

Explanation:

Work done is the energy that is lost by the skater

Formula for workdone = 1/2*mV^2

m = 66kg

V = 10m/s

Work done = 1/2 * 66 * 10^2

= 3300J

The work done by the wall to stop a 66.0-kg ice skater moving at 10.0 m/s is calculated using the work-energy theorem and is found to be 3300 joules.

The student is asking how much work is done by the wall to stop a 66.0-kg ice skater who is moving at a speed of 10.0 m/s. To solve this, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Since the skater is coming to rest, the final kinetic energy is 0. The initial kinetic energy can be calculated using the equation KE = 0.5 × m × v^2, where m is the mass and v is the velocity. After plugging in the values, we get KE = 0.5 × 66.0 kg × (10.0 m/s)^2 = 3300 J. Therefore, the padded wall does 3300 joules of work to bring the skater to rest.

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

Coal-fired power plants produce electricity by burning coal to boil water into steam, which drives a turbine connected to a generator. The efficiency of energy conversion is low, with significant heat loss to the environment and a large CO2 emission as one of the main environmental impacts.

A coal-fired power plant converts the energy stored in coal into electricity through a multi-step process. First, coal is mined and processed to be suitable for burning. When coal is combusted in the plant, it heats water to turn it into steam. The steam at high pressure then drives a turbine, which is connected to a generator. As the turbine blades turn, they rotate the generator, which converts the kinetic energy into electricity. This process involves significant heat transfer to the surroundings, which is an inherent part of energy production from combustion.

During the energy conversion process, the efficiency of coal power stations is quite low, with only about 42% of the energy being used for electricity generation and the rest being lost as heat transfer to the environment. The chemical reaction during the combustion of coal is C + O2 → CO2, and a significant amount of CO2 is emitted into the atmosphere. This contributes to the warming of our planet, and coal power plants are known for being the least efficient and most CO2-emitting fossil fuel energy sources.

Answer:

1.162 x 10^-7 Nm

Explanation:

length of wire, l = 25 cm

l = 2 π r

where, r is the radius of circular loop

25 = 2 x 3.14 x r

r = 3.98 cm

Magnetic field, B = 5.55 mT = 5.55 x 10^-3 T

Current, i = 4.121 mA = 4.21 x 10^-3 A

Torque, τ = i x A x B

τ = 4.21 x 10^-3 x 3.14 x 0.0398 x 0.0398 x 5.55 x 10^-3

τ = 1.162 x 10^-7  Nm

Thus, the maximum torque in the coil is 1.162 x 10^-7 Nm.

Answer:

Time taken to accelerate to 48 m/s =  8 seconds

Explanation:

Initially the object is moving south at = [tex]16 m/s[/tex]

So, initial velocity of the object = [tex]16 m/s[/tex]

Acceleration caused by the force = 4 [tex]m/s^2[/tex]

Final velocity towards south = [tex]48 m/s[/tex]

Using equation of motions:

[tex]v_f=v_i+at[/tex]

where

[tex]v_f\rightarrow[/tex] final velocity

[tex]v_i\rightarrow[/tex] initial velocity

[tex]a\rightarrow[/tex] acceleration

[tex]t\rightarrow[/tex] time

Plugging in values.

[tex]48=16+(4)t[/tex]

[tex]48=16+4t[/tex]

Subtracting both sides by 16

[tex]48-16=16+4t-16[/tex]

[tex]32=4t[/tex]

Dividing both side by 4.

[tex]\frac{32}{4}=\frac{4t}{4}[/tex]

[tex]8=t[/tex]

∴ [tex]t=8[/tex]

Time taken to accelerate to 48 m/s =  8 seconds

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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