don't you ever feel like thursday is like the number 7?
and friday like 8? or 6?
and monday maybe 3?

(a) The amplitude of the wave is determined as 8 cm.

(b) The period of the wave motion is  20 s and the frequency of the wave is 0.05 Hz

(c) The wavelength of the wave is 500 cm.

(a) The amplitude of the wave is the maximum displacement of the wave.

from the graph, amplitude of the wave = 8 cm

(b) The period of the wave motion is calculated as;

T = 20 s

The frequency of the wave = 1/T = 1/20 s = 0.05 Hz

(c) The wavelength of the wave is calculated by applying the following wave formula.

λ = v / f

λ = 25 cm/s / 0.05 Hz

λ = 500 cm

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The circuit courts in Illinois have original jurisdiction as trial courts. There are 24 judicial circuits in the state, and each one includes one or more of the 102 counties that make up Illinois.

All cases filed in the State of Illinois other than those for which the Supreme Court has original jurisdiction are first heard by the Circuit Court, which is the trial court with wide jurisdiction.

A Supreme Court, an Appellate Court, and Circuit Courts are in charge of the legal system. (Source: Constitution of Illinois.) JUDICIAL DISTRICTS SECTION 2 For the purpose of choosing judges for the Supreme and Appellate Courts, the State is divided into five judicial districts.

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The direction of the third force, such that the object continues to

move with a constant velocity is  94.5N at 0.19° south of east.

The force that one thing exerts on another. A barrel being pushed by a human is an illustration of applied force. When the individual pushes the barrel, the barrel experiences an applied force. Newton's second law of motion defines the force formula as follows: A force is equal to an object's mass times its acceleration, or F = m a. You must use SI units when applying this formula: kilograms for mass, meters per second squared for acceleration, and newtons for force. a push or pull that an object has experienced. Here, an object is given a force by a person or any other thing. A chair being pushed to the opposite side of the room, for instance.

For the object to move with constant velocity the total acceleration must be zero.

So,

Total  F = \(F_{1} +F_{2} +F_{3}\)

  ma = 0

\(F_{1}\) = 84.4N

\(F_{2}\) = 42.7 N

\(F_{3} = -F_{1} - F_{2}\)

= 94.58 N

TanФ = 0.19°

Hence \(F_{3}\) = 94.5N at 0.19° south of east.

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Kenny’s ideas and their acceptance were due to open-mindedness because she led to the use of hot baths and exercise to treat polio victims.

Elizabeth Kenny was a world-renowned nurse from Australia who develop a treatment for poliomyelitis, a very serious viral disease caused by poliovirus.

Elizabeth Kenny's ideas were controversial because they may be considered to be against conventional medical treatments.

In conclusion, Kenny’s ideas and their acceptance were due to open-mindedness because she led to the use of hot baths and exercise to treat polio victims.

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Answer:   the acceleration of the masses is given by = 0, which means the angular acceleration of the pulley is zero. This implies that the masses m and 2m move with constant velocity,  they are in equilibrium.

a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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Answer:

leverage to open it

Explanation:

The mechanical advantage of the tab of a soda can is that, it acts as a second class lever.

Lever is defined as a simple machine that increases the force. Lever is used to lift loads or weight by reducing the work.

Here,

Caleb opened a soda can by puling the tab of the can. The  tab of the soda can here acts as a simple machine to open it.

The tab of the soda can is used to open the can such that it makes the task super easy. The tab of the can is an application of a second class lever.

Second class lever is the type of lever in which the fulcrum is located at one end and the load is in the middle such that the effort is applied in the other end. In the tab, the force is applied at one end upwards and the other end of the tab opens downwards to the can and thus it is opened. This is the working of the tab of a soda can.

Hence,

The mechanical advantage of the tab of a soda can is that, it acts as a second class lever.

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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Answer:

Electric field intensity is a Vector Field. Electric field intensity (E, N/C or V/m) is a vector field that quantifies the force experienced by a charged particle due to the influence of charge not associated with that particle

Explanation:

The direction and magnitude of the net vector is:

When vector A with magnitude of 5 meters pointing west and

vector B with magnitude of 5 meters pointing east are added,

they will result in a net vector with a magnitude of the difference between the magnitudes of the vectors and a direction equal to the direction of the vector with the largest magnitude.

Since both vectors have the same magnitude (5 meters), the net vector will have a magnitude of 0 meters.

This means that the vectors completely cancel each other out and there is no net vector remaining.

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The density of water at STP, which is \(997 kg/m^3\), we can see that Saturn is less dense than water.

To determine whether Saturn is less dense than water, we must compute its density and compare it to the density of water at standard temperature and pressure (STP), which is \(997 kg/m^3\).

Saturn's density can be computed using the following formula:

density equals mass divided by volume

Saturn's mass and volume may be computed given its mass and radius.

The volume of Saturn can be determined using the sphere volume formula:

volume =\((4/3) \pi (r^3)\)

where r is Saturn's radius.

Filling in the blanks:

volume = \((4/3) \pi (5.6 \times 107) m^3\)

8.27 x 1023 \(m^3\)volume

Saturn's mass is given as \(5.68 \times 10^{26} kg.\)

We can now compute Saturn's density:

density equals mass divided by volume

density= \((5.68 x 10^{26 }kg\)) /\((8.27 \times 10^{23 }\)m³) a density of\(687 kg/m^3\)

This is due to the fact that Saturn is mostly made up of hydrogen and helium, which are far less dense than water. In reality, Saturn is the least dense planet in the Solar System, and it would float in a large enough body of water.

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When a body of mass 12kg travelling at 4.2m/s² collides with a second body of mass 18kg at rest, their common velocity after the collision is 1.68 m/s.

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Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = \(0.12 \times 10^{-2} \ m\)

\(x = 1.17 \ cm = 1.17 \times 10^{-2}\ m\)

m = 149 kg

\(\delta = 7.87 \ g/cm^3\)

\(da = 2.28 \times 10^{-10}\ m\)

\(F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg \\ \\\)

∴

\(k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m\)

\(N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}\)

\(N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2\)

\(N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2\)

\(N_{chain} = 2.77 \times 10^{13}\)

\(N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}\)

\(\text{Finally; the stiffness of a single interatomic spring is:}\)

\(k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s\)

\(k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)\)

\(\mathbf{k_{si} =45.46 \ N/m}\)

the pressure of the water after the contraction is -50000 Pa (or 50 kPa below atmospheric pressure), and the velocity of the water after the contraction is 15.0 m/s.

The continuity equation states that the product of the cross-sectional area and the velocity of an incompressible fluid is constant along a pipe, so we can use it to relate the pressure and velocity before and after the contraction:

A₁v₁ = A₂v₂

where A₁ and v₁ are the area and velocity of the pipe before the contraction, and A₂ and v₂ are the area and velocity of the pipe after the contraction.

We can also use the Bernoulli equation, which relates the pressure and velocity of a fluid along a streamline:

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

where P₁ and v₁ are the pressure and velocity of the fluid before the contraction, and P₂ and v₂ are the pressure and velocity of the fluid after the contraction, and ρ is the density of the fluid, which we assume to be constant.

Solving for the pressure and velocity after the contraction, we can use the continuity equation to express v₁ in terms of v₂ and substitute it into the Bernoulli equation:

A₁v₁ = A₂v₂

v₁ = (A₂/A₁) v₂

P₁ + 1/2 ρ((A₂/A₁) v₂)² = P₂ + 1/2 ρv₂²

Simplifying and solving for P₂, we get:

P₂ = P₁ + 1/2 ρ(v₁² - v₂²)

Substituting the given values, we get:

A₂ = (1/3) A₁

v₁ = 5.0 m/s

P₁ = 450000 Pa

ρ = 1000 kg/m³

Using the continuity equation, we can find the value of v₂:

A₁v₁ = A₂v₂

v₂ = (A₁/A₂) v₁

v₂ = 3 × 5.0 m/s

v₂ = 15.0 m/s

Substituting this value into the Bernoulli equation, we can find the pressure P₂:

P₂ = P₁ + 1/2 ρ(v₁² - v₂²)

P₂ = 450000 Pa + 1/2 × 1000 kg/m³ × (5.0 m/s)² - (15.0 m/s)²

P₂ = 450000 Pa - 500000 Pa

P₂ = -50000 Pa

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One important takeaway from Dan Ariely's talk is that our decision-making processes are frequently irrational and influenced by variables outside of our conscious consciousness.

Social norms, emotional states, and cognitive biases, for example, can all influence our choices. As a result, one method to put this knowledge to use is to become more aware of these influences and actively consider them when making decisions.

This may entail devoting more time to decision-making, seeking out diverse views, and being aware of our own biases and tendencies.

Another key takeaway from the talk is the significance of considering the long-term consequences of our choices rather than focusing solely on short-term gains or losses.

This may entail weighing the risks and benefits of various options, as well as considering the effect of decisions on others as well as ourselves.

Overall, the information presented by Dan Ariely can be used to make more thoughtful and informed decisions by becoming more aware of our own biases and tendencies, considering diverse viewpoints, and thinking critically about the potential repercussions of our decisions.

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Answer:

hi

Explanation:

hey

Answer:

(A) potential energy

I think it helps you

The kinetic energy of the brick equal the elastic potential energy at the top of the hill. Using this, the speed of the brick is 14 m/s.

The elastic potential of a spring is directly proportional to the squire of the displacement.

Then,

p = 1/2 k x²

Given that, spring constant k = 500 N/m

height of the hill x = 2 m

weight of the load = 100 N

then mass = 100N/9.8 m/s² =10.20 Kg.

At the top of the hill, kinetic energy of the hill is equal to the elastic potential.

then, 1/2 mv² =  1/2 k x²

speed v of the brick  = √kx²/m

v = √(500 N/m × 2 m²/10.20 kg)

  = 14 m/s.

Therefore, the speed of the brick at the top of the hill is 14 m/s.

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The surface charge density of the ring is given by, n = Q / πR^2. and also it is given by n = 2λ / R.

In this context, "w << R" means that the width (w) of the circular ring is much smaller than the radius (R) of the ring. Mathematically, this can be expressed as w/R << 1.

To find the surface charge density (n) of the ring, we can use the formula:

n = Q / A

where Q is the total charge on the ring and A is the area of the ring. For a circular ring, the area is given by:

A = π(R^2 - r^2)

where R is the outer radius of the ring and r is the inner radius of the ring. Since the ring is very thin, we can assume that the inner radius is negligible compared to the outer radius (i.e., r ≈ 0). Thus, we can simplify the formula for the area as:

A ≈ πR^2

Substituting this expression for A into the formula for surface charge density gives:

n = Q / πR^2

Therefore, the surface charge density of the ring is n = Q / πR^2.

Note that since the charge is uniformly distributed over the ring, we can also express the total charge Q in terms of the linear charge density λ (charge per unit length) and the circumference of the ring C:

Q = λC

Substituting this expression for Q into the formula for surface charge density gives:

n = λC / πR^2

However, since the ring is a complete circle, we know that the circumference is equal to 2πR, so we can simplify this expression further:

n = λ(2πR) / πR^2

n = 2λ / R

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The region a represents the distance of the electron from the nucleus.

According to the wave mechanical model of the atom, the probability of finding an electron within a given volume element (representing the atom) is the square of the wave function psi.

Since a is the region in space where there is the greatest probability of finding the electron in the atom, it follows that distance of the electron form the atom is always a.

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The quantity of heat energy that must be transferred to 2.25 kg of brass to raise its temperature from 20 °C to 240 °C is 195030 J

First, we shall list out the given parameters from the question. This is shown below:

The quantity of heat energy that must be transferred can be obtained as follow:

Q = MCΔT

= 2.25 × 394 × 220

= 195030 J

Thus, we can conclude quantity of heat energy that must be transferred is 195030 J

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Answer:

the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg

the initial acceleration is a = 753.47 m/s²

the  an instrument package could they carry a required mass of 4.64 kg

Explanation:

a)  What should be the radius of these balloons so they just hover above the surface of Mars?

Given that :

The density of the Martian atmosphere, ρ = 0.0154 kg/m³

The volume of the sphere, V = (4/3)πr³

The area of the sphere, A = 4πr²

The mass of the balloon is m = (4.60 g/m²)A

m = (4.60×10⁻³ kg/m²)(4πr²)

The formula for the buoyant force is expressed as :

F = ρVg

m×g = ρ×V×g

m = ρ×V

Now;

(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³

r = 3(4.60×10⁻³ kg/m²)/ ρ

r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³

r = 0.896 m

Thus; the radius of the balloon r = 0.896 m

The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)

m = (4.60×10⁻³ kg/m²)(4π×0.896²)

m = 4.64 × 10⁻² kg

b) what would be its initial acceleration assuming it was the same size as on Mars?

The density of the air on earth, ρ = 1.20 kg/m³  

The volume of the balloon is V = (4/3)(π)(0.896 m)³

V = 3.01156 m³

Considering the net force acting on the balloon ; we have

ΣF = ρVg - mg = ma

However; making the initial acceleration  a of the balloon the subject ; we have:

a = (ρVg - mg)/m

a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)

a = 753.47 m/s²

c) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?

The volume of the total system is V' = (4/3)π(5r)³

V' =  (4/3)π(5)³(0.896 m)³

V' =376.446 m³

The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)

m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²

m = 1.159587 kg

We can then say that  the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:

F = (m + m')g

ρV'g = (m + m')g

ρV' = (m + m')

Thus; the required mass m' is =  ρV' - m

m'  =  ρV' - m

m' = (0.0154 kg/m³)(376.446 m³) - (1.159587 kg)

m' = 4.64 kg

Thus; the  an instrument package could they carry a required mass of 4.64 kg

The specific latent heat of fusion of ice obtained from this experiment is approximately 583.33 J/g.

To determine the specific latent heat of fusion of ice using the given experiment, we need to consider the energy transferred during the process.First, we need to calculate the energy lost by the water to cool down from 25 °C to 0 °C. The energy lost is given by:

Q1 = m1 * c * ΔT1

Where:

m1 = mass of water = 100 g

c = specific heat capacity of water = 4.2 J/g °C

ΔT1 = change in temperature = (0 °C - 25 °C) = -25 °C

Q1 = 100 g * 4.2 J/g °C * (-25 °C) = -10,500 J

Next, we calculate the energy released by the water to freeze and cool the remaining ice. The energy released is given by:

Q2 = m2 * Lf

Where:

m2 = mass of ice = 18 g

Lf = specific latent heat of fusion of ice (to be determined)

Q2 = 18 g * Lf

Since energy is conserved in the system, the energy lost by the water (Q1) is equal to the energy released by the water (Q2):

-10,500 J = 18 g * Lf

Solving for Lf:

Lf = -10,500 J / 18 g = -583.33 J/g

The negative sign indicates that energy is being released during the process of freezing.

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Answer:    Vf = Vi + A(t)

Vf= -2+ (-9.8)(5)

Vf = -51 m/s

Explanation:

The final velocity is equal to the initial velocity plus the acceleration multiplied by the time

(-9.8) is used for the acceleration for this question because that is the speed at which things in free fall accelerate when they are on earth

-2 is used as the initial velocity because the ball was thrown in the negative direction which is down.

The time of 5 was given in the question so you can plug it in for time

Answer:

Explanation:

Given:

V₀ = 2 m/s

g = 9.8 m/s²

t = 5 s

____________

V - ?

Axis OX direct vertically down. Then:

V = Vₓ = V₀ₓ + g·t

V = 2 + 9.8*5 = 51 m

The acceleration due to gravity with a mass of the ball of 0.055 kg and a length of wire is 0.95m with a linear density of 1.2×10⁻⁴kg/m is 7.69m/s².

Acceleration due to gravity is defined as the object getting accelerated due to gravitational force acting on the freely falling body. The unit of acceleration is m/s².

From the given,

mass of ball (m) = 0.055kg

length of the wire (l) = 0.95 m

linear density = 1.2×10⁻⁴kg/m

time (t) = 0.016 s

acceleration due to gravity (g) =?

The tension of the wire, v = √(F)/m/L

The force F= v²×L/m

To find speed (v) the ratio between Lenght and time gives speed.

v = L/t

  = 0.95 / 0.016

 = 59.37 m/s.

The speed v is 59.37 m/s.

F = v²×m×L

 = 59.37×59.37×1.2×10⁻⁴

 = 0.423 N

The force is F = 0.423 N.

Force , F = W

                = m×g

    F/m = g

  0.423 / 0.055 = g

g = 7.69 m/s².

Thus, the acceleration due to gravity is g = 7.69 m/s².

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The speed of the water jet emerging from the nozzle with a water density of 1000 kg/m³ is 30.107 m/s.

Speed can be defined as the rate of change in the distance of an object. It is a scalar quantity due to only magnitude and no direction.

Height = 3.3 m

Diameter = 2.03 inch

Gauge pressure = 450.228 kPa

We need to calculate the speed of the water jet emerging from the nozzle

Using Bernoulli's equation,

1/2ρ(Vn² - Vp²) = P gauge - ρgh

(Vn² - Vp²) = (2/P) Pgauge - 2gh

Vn² - (An/ Ap)²Vn² = (2/P) Pgauge - 2gh

Vn² - (rn/ rp)⁴Vn² = (2/P) Pgauge - 2gh

Vn² = \(\sqrt{(2/P) Pgauge - 2gh/ (1- (rn/ rp)^4)}\)

Vn = \(\sqrt{[((2/1000)(450.228 X 10^3)) - 2 X 9.8 X 3.3]/ 1 - (1.01/ 1.91)^4}\)

Vn = \(\sqrt{(2 X 450.228 - 64.68)/ 1- (1.04/ 13.30)}\)

Vn = \(\sqrt{(900.45 - 64.68)/ (13.30- 1.04)/13.30}\)

Vn = \(\sqrt{(835.776)/ (12.26/ 13.30)}\)

Vn = \(\sqrt{(835.776)/ 0.922}\)

Vn = \(\sqrt{906.48}\)

Vn = 30.107 m/s

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According to seismic waves, the Earth's interior is made up of a number of concentric shells, including a solid inner core, a liquid outer core, a mantle, and a thin outer crust.

The wave's course will be bent or refracted as it enters the new layer if it moves at a different speed there. The wave will be somewhat curved in the direction of the contact between the two layers if it can move through it more quickly in the new layer.

Therefore, We can learn about just the layers that make up the Earth by understanding how waves behave as they pass through various materials.

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a. The two cars are moving at the same speed at t = 3.88 seconds.

b. The two autos are moving at a speed of 4.18 cm/s at that moment.

c. The cars pass each other at time t = 1.05 s and t = 2.54 s, respectively.

d. At t = 1.05 seconds, the first car is located at 12.15 cm, and the second car is located at 16.55 cm. At t = 2.54 seconds, the first car is located at 10.69 cm, and the second car is located at 17.39 cm.

What is the speed of the two cars?

The speed of the cars is derived using the formula for linear velocity.

The velocity of the first car is calculated as follows:

v = v₀ + at

where;

The velocity of the second car can be calculated as:

v = v₀

where;

v₀ is the initial velocity since there is no acceleration.

We equate the two velocities of the two cars and solve for time:

-4.40 + 2.70t = 6.10

2.70t = 10.50

t = 3.88 seconds

b. To find the speeds of the cars at time t, we can substitute t = 3.88 seconds in the velocity equation for the first car:

For the first car:

v = -4.40 + 2.70 * 3.88

v = 4.18 cm/s

c. Solving for the time when the positions of the cars are equal allows us to determine when they pass one another.

The position of the first car will be:

x = x₀ + v₀t + 1/2 at²

where x₀ is the initial position.

The position of the second car will be:

x = x₀ + v₀t

equating the two positions equal and solving for time:

13.5 - 4.40t + 1/2 * 2.70t² = 10.5 + 6.10t

1.35t² + 1.70t + 3.00 = 0

Solving the quadratic equation, we get two times:

t = 1.05 s and t = 2.54 s

d. We can re-insert the time values into the position equation for each car to determine where the automobiles were at the times indicated in (c).

For t = 1.05 seconds:

First car:

x = 13.5 - 4.40 * 1.05 + 1/2 * 2.70 * 1.05²

x = 12.15 cm

Second car:

x = 10.5 + 6.10 * 1.05

x = 16.55 cm

For t = 2.54 seconds:

First car:

x = 13.5 - 4.40 * 2.54 + 1/2 * 2.70 * 2.54²

x = 10.69 cm

Second car:

x = 10.5 + 6.10 * 2.54

x = 17.39 cm

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Answer:

b

Explanation:

IF- force you use

OF- force the machine give you