the pressure in a horizontal pipe is 2.02x105 pa, and the speed of the water flowing in it is 0.40 m/s, and the diameter of the pipe is 7 centimeters. if the diameter of the pipe narrows to 2 centimeters in another section, determine the pressure in the narrow portion of the pipe., assuming that the water behaves like an ideal fluid and the density of water is 1000 kg/m3.

The horizontal distance from the point directly below C to where the skier lands is 54.2 m.

Applying conservation of energy and conservation of momentum principles.

First, let's find the initial potential energy of the skier at point A:

PE1 = mgh1

= (80 kg)(9.81 m/s^2)(h1)

Next, let's find the final kinetic energy of the skier at point C:

KE2 = (1/2)mv2^2

= (1/2)(80 kg)(42 m/s)^2

Since there is no friction,

PE1 = KE2 + PE3

where PE3 is the potential energy of the skier at point C:

PE3 = mgh3 = (80 kg)(9.81 m/s^2)(10 m)

Substituting the values, we get:

(80 kg)(9.81 m/s^2)(h1) = (1/2)(80 kg)(42 m/s)^2 + (80 kg)(9.81 m/s^2)(10 m)

to determin h1,

h1 = (1/2)(42 m/s)^2/9.81 m/s^2 + 10 m

h1 = 144.8 m

Therefore, the height of the hill is 144.8 m.

Since there is no external force acting on the skier in the horizontal direction, the horizontal momentum is conserved:

mvi = mvx

vx = v2cos(25°)

Substituting the values, we get:

vi = (42 m/s)cos(25°)

vi = 37.9 m/s

Therefore, the initial horizontal velocity of the skier at point C is 37.9 m/s.

to determine time of flight:

h = (1/2)gt^2

t = √(2h/g)

t = √(2(10 m)/(9.81 m/s^2))

t = 1.43 s

applying the horizontal velocity and the time of flight to find the horizontal distance traveled:

d = vixt

d = (37.9 m/s)(1.43 s)

d = 54.2 m

The horizontal distance from the point directly below C to where the skier lands is 54.2 m.

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Answer:

To solve this problem, we can use the conservation of energy and the conservation of momentum. We can assume that there is no friction or air resistance, so the total mechanical energy of the skier is conserved throughout the motion.

Let's denote the initial height of the hill as h1, the height of point C above the ground as h2, the horizontal distance from point A to the point directly below C as x, and the angle of the skier's velocity vector with respect to the horizontal as θ.

First, we can calculate the speed of the skier at the bottom of the hill, point B, using conservation of energy:

mgh1 = (1/2)mvB^2 where m is the mass of the skier and vB is the speed of the skier at point B.

Solving for vB, we get:

vB = sqrt(2gh1)

Next, we can calculate the velocity of the skier at point C using conservation of energy:

mgh1 = (1/2)mvB^2 + (1/2)mvC^2

where vC is the speed of the skier at point C.

Solving for vC, we get:

vC = sqrt(2gh1 + vB^2)

We can also express the velocity vector at point C in terms of its x and y components:

vCx = vCcos(θ)

vCy = vCsin(θ)

Using conservation of momentum, we can find the horizontal distance x from the point directly below C to where the skier lands:

mvCx(h2/(-vCy)) = mvCx(t) + (1/2)gt^2

where t is the time taken for the skier to reach the ground and we have used the fact that the vertical displacement from point C to the ground is h2. We can solve for t by substituting vCy = vC*sin(θ) and solving the quadratic equation:

(1/2)gt^2 + vC*sin(θ)*t - h2 = 0

Solving for t, we get:

t = (-vCsin(θ) + sqrt(vC^2sin(θ)^2 + 2gh2))/g

Finally, we can substitute this expression for t into the equation for x to get:

x = vCx*t

Substituting the expressions for vCx, t, and vC, we get:

x = (vC^2sin(θ)cos(θ) - sqrt(vC^4sin(θ)^2cos(θ)^2 + 2gh2vC^2sin(θ)^2))/(g*sin(θ))

Plugging in the given values, we get:

x ≈ 184.5 meters

Therefore, the horizontal distance from the point directly below C to where the skier lands is approximately 184.5 meters.

When crystals are pressed along their axis of symmetry, an electric charge is generated. This phenomenon is known as the piezoelectric effect.  

The piezoelectric effect was discovered by Pierre and Jacques Curie, who were French physicists. Piezoelectricity is the property of some materials to generate an electric charge in response to applied mechanical stress.

The charge generated by the piezoelectric effect is proportional to the amount of pressure applied to the crystal, as well as the strength of the crystal. The piezoelectric effect is used in a variety of applications, including microphones, speakers, and sensors.

In microphones and speakers, piezoelectric crystals convert sound waves into an electrical signal or vice versa. In sensors, the piezoelectric effect is used to measure pressure, acceleration, and strain. The piezoelectric effect is also used in medical applications, such as ultrasound technology.

Ultrasound imaging works by transmitting sound waves through the body and measuring the echoes that bounce back. The piezoelectric effect is used to generate and detect these sound waves.

In conclusion, the piezoelectric effect is a fascinating phenomenon that has a wide range of practical applications. It was discovered by Pierre and Jacques Curie in the late 19th century and has since been used in a variety of fields.

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If the voltage source or EMF is removed from a conductor, the electron flow within the conductor will eventually stop.

This is because the voltage source creates an electric field that causes the free electrons in the conductor to move, creating an electric current. When the voltage source is removed, the electric field disappears, and the free electrons in the conductor will no longer have a driving force to move them. As a result, the electrons will begin to lose energy and eventually come to a stop, leading to the cessation of the electric current.

It's worth noting that the time it takes for the electron flow to stop depends on various factors such as the resistance of the conductor, the capacitance of the circuit, and the amount of charge stored in any capacitors present in the circuit.

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The area required to produce a pressure of 4800 Pa with a force of 24 N is 0.005 m2.

Pressure is calculated as force per unit area, so we can rearrange the equation to solve for area.
Area = force/pressure
Substituting the given values, we get:
Area = 24 N / 4800 Pa
Simplifying,
Area = 0.005 m2
Therefore, the area required to produce a pressure of 4800 Pa with a force of 24 N is 0.005 m2.

Pressure equals force divided by area (P=FA P = F A ). The equation shows that pressure is directly proportional to force, but inversely proportional to area. At a constant area, pressure increases as the magnitude of the force applied also increases.

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Answer:

F = 1100 Newtons

Explanation:

Answer:

-15

Explanation:

displacement = (-32) + (+17)

= -15

note : displacement can be positive, negative as well as zero.

On an axis in which moving from right to left is positive, the displacement of a student who walks 32m to the right and then 17m to the left is -15m, and the distance is 49m.

To find the displacement and distance of a student who walks 32m to the right and then 17m to the left on an axis where moving from right to left is positive, you should follow these steps:

1. Assign a positive direction to moving from right to left (and a negative direction for left to right movement).
2. The student first moves 32m to the right, which is negative in this axis. So, this movement is -32m.
3. Next, the student moves 17m to the left, which is positive in this axis. This movement is +17m.
4. Calculate the displacement: Displacement is the overall change in position, so add the two movements together: -32m + 17m = -15m. The negative sign indicates that the student's final position is 15m to the right of the starting point.
5. Calculate the distance: Distance is the total length of the path traveled, regardless of direction. So, add the absolute values of the two movements: |-32m| + |17m| = 32m + 17m = 49m.

On an axis in which moving from right to left is positive, the displacement of a student who walks 32m to the right and then 17m to the left is -15m, and the distance is 49m.

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Answer:

2 lol

Explanation:

When a gas expands adiabatically, its internal energy decreases and work is done on the gas.

Adiabatic expansion occurs when a gas expands without any heat exchange with the surroundings.

As a result, the internal energy of the gas decreases because the gas does work on the surroundings.

This means that option A is correct, and the internal (thermal) energy of the gas decreases.

Option B is incorrect since the gas is doing work on the surroundings and therefore loses internal energy.

Option C is incorrect because work is done on the gas by the surroundings.

Option D is correct since work is done on the gas.

Finally, option E is incorrect because the temperature of the gas decreases as its internal energy decreases due to the expansion.

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1)The strength of the magnetic field for particle 1 will be 2.8 T.

2)The magnitude of the magnetic force exerted on the particle will be 0.12 N.

It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.

The given data in the problem is;

q₁ has a charge = 2.7 μc

v₁ is the velocity of particle 1 = 1 773 m/s.

F is magnetic force = 5.75 × 10–3 n,

q₂ has a charge of 42.0 μc

v₂ is the velocity of 1.21 × 103 m/s.

\(\rm F_{B2}\)   is the magnitude of the magnetic force exerted on particle 2=?

The megnetic force for case 1 is found as;

\(\rm F_{B1}= qvB SIN \alpha_1 \\\\ 5.75 \times 10^{-3} =2.7 \times 10^{-6}\times 773 \times B sin 90^0 \\\\ B=\frac{5.75 \times 10^{-3}}{2.7 \times 10^{-6} \timesd 773 \times sin 90^0} \\\\ B=2.8 \ T\)

The megnetic force for case 2 is found as;

\(\rm F_{B2} = q_2v_2bsin \alpha_2 \\\\ 42\times 10^{-6} \times 1.21 \times 10^3 \times 2.8 \times sin 55^0 \\\\ F_{B2}=0.12 T\)

Hence the value of the megnetic force exerted on particle 2 will be 0.00122 T.

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Answer:

last one and second one

Explanation:

Particle q1 has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10–3 N, what is the strength of the magnetic field?

2.8

T

In the same magnetic field, particle q2 has a charge of 42.0 μC and a velocity of 1.21 × 103 m/s. What is the magnitude of the magnetic force exerted on particle 2?

0.12

N

Answer:68

Explanation:

6.8x100

We have that for the Question "A block weighing (Fg) 100 N is resting on a steel table (us = 0.68) ." it can be said that

The minimum force to start this block moving is \(F_{min}=37N\)

From the question we are told

A block weighing (Fg) 100 N is resting on a steel table (us = 0.68)

The minimum force to start this block moving is. N.

Generally the equation for the Maximum force  is mathematically given as

\(F_{max} =0.74*50\\\\F_{max} =37N\)

Therefore

The Minimum force = max Frictional force

Therefore

\(F_{min}=37N\)

Hence

The minimum force to start this block moving is \(F_{min}=37N\)

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We do not expect to see tidal disruption of sun-like stars by black holes larger than about 10^8 M☉ due to their weaker gravitational tidal forces.

Tidal disruption occurs when a star gets too close to a black hole, and the gravitational forces from the black hole pull on the star more strongly than the internal forces holding it together. This causes the star to be torn apart and accreted onto the black hole. The tidal disruption radius, which is the distance from the black hole at which this happens, depends on the mass and size of the star as well as the mass of the black hole. For a sun-like star, the tidal disruption radius is proportional to the black hole mass. However, once the black hole mass exceeds about 10^8 M☉, the tidal disruption radius becomes larger than the size of the star, making tidal disruption less likely to occur. Therefore, we do not expect to see tidal disruption of sun-like stars by black holes larger than about 10^8 M☉.

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Answer:

10,000 N

Explanation:

Given that a rocket pushes exhaust 10, 000 N. Without any other forces acting on the rocket, how much force does the rocket go forward?

Solution

Since there is no any other additional forces, according to Newton's 3rd law of motion which state that:

In every action there will be equal and opposite reaction.

The action expelled by the rocket through its exhaust system is of 10000N.

The reactive force will be equal and opposite which will be same -10,000 N

Therefore, the rocket will go forward with a force of magnitude 10,000 N

Ey of the electric field at the origin = (-Qk/R²α) × [(α/2 + α)/2]sinφ + kQ/RC²Ey = (-Qk/R²α) × [3α/4]sinφ + kQ/RC²

To compute the value of Ey of the electric field at the origin, using the same four steps we used in class for the rod and the ring, we have:Step 1The value of the electric field created by a small piece of the thin plastic rod at the origin is:dE=kdq/r²where:dq = -Qdθ / α   is the charge of a small element of the rod at an angle θ.α is the angle between the two ends of the rod.The minus sign in dq indicates that the rod is negatively charged.k is Coulomb's constant, k=9×10^9 N·m²/C².r is the distance between a small element of the rod and the origin and is given by:r= RsinθThe electric field at the origin produced by a small element of the rod is then:dE=kdq/R²sin²θ= -Qdθ/α × k/R²sin²θdE= -Qdθ/α × k/R²(1-sin²θ) = -Qdθ/α × k/R²cos²θThe x-component of the electric field produced by a small element of the rod is given by:Ex= dEcosθ = -Qdθ/α × k/R²cos³θStep 2We need to integrate this expression over the whole rod. Since the rod is uniformly charged, the angle element is:dq = -Qdθ/αTherefore, the electric field at the origin due to the entire rod is:Edue to the rod = ∫dE = ∫ (-Qdθ/α × k/R²cos²θ) from θ = -α/2 to θ = α/2Edue to the rod = (-Qk/R²α) × ∫cos²θdθ from θ = -α/2 to θ = α/2

We can use the trigonometric identity:cos²θ= (1+cos2θ)/2to evaluate this integral.Edue to the rod = (-Qk/R²α) × ∫(1+cos2θ)/2 dθ from θ = -α/2 to θ = α/2Edue to the rod = (-Qk/R²α) × [θ/2 + (sin2θ)/4] from θ = -α/2 to θ = α/2Edue to the rod = (-Qk/R²α) × [(α/2 + sinα)/2]The electric field due to the rod at the origin is:Edue to the rod = (-Qk/R²α) × [(α/2 + sinα)/2]Step 3The value of the electric field at the origin produced by the ring is:Ering= kQ/RC²where Q is the charge of the ring, R is its radius, and C is the distance between the ring and the origin.Step 4The total electric field at the origin is:Etotal = Edue to the rod + EringTherefore,Ey = EtotalsinφWhere Ey is the y-component of the electric field, and φ is the angle between the x-axis and the line connecting the origin and the center of the ring.Ey = (Edue to the rod + Ering)sinφ = (-Qk/R²α) × [(α/2 + sinα)/2]sinφ + kQ/RC²sin(π/2)Ey = (-Qk/R²α) × [(α/2 + sinα)/2]sinφ + kQ/RC²For a small value of α, sinα ≈ α.

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The initial velocity of the car when the brake was initially applied is 6.3 m/s.

option D is the correct answer.

The velocity of the car when the brake was initially applied is calculated as follows;

Mathematically, kinetic energy formula is given

K.E = ¹/₂mv²

where;

mv² = 2 K.E

v² = ( 2 K.E ) / ( m)

v = √ [  ( 2 K.E ) / ( m) ]

Substitute the given parameters and solve of the initial velocity of the car when the brake was applied.

v = √ [  ( 2 x 23,180 ) / ( 1168) ]

v = 6.3 m/s

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The complete question is below:

A 1,168-kg car comes to a stop without skidding. The car's brakes do -23,180 J of work

to stop the car.

Which of the following was the car's velocity when the brakes were initially applied?

A. 3.2 m/s

B. 1.7 m/s

C. 5.1 m/s

D. 6.3 m/s

The size of the habitable zone around a star of spectral class g is larger to the size of the habitable zone around a spectral class m star.

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Answer:

Weight

Explanation:

The second law of motion by Newton can be used to determine weight.

Answer:

Dry lubricants or solid lubricants are materials that, despite being in the solid phase, are able to reduce friction between two surfaces sliding against each other without the need for a liquid oil medium.

Explanation:

Answer:

Your answer is graphite :) Have a nice day

Explanation:

Answer:

It allows people in different places and different countries to use the same units, avoid mistakes and understand each other more easily. The common base 10 of all units makes it easier and has more accurate calculations that are made without cumbersome conversion factors.

Answer:

Explanation:

As we know that the SI units are extended versions of the MSK system. SI system helps us to calculate the appropriate measurement of any quantity. SI system is convenient(easy) to use and it is also followed all over the world which can help in international trade as well. So, SI system been developed.

The distance between the first crest (or trough) and the following crest is known as a wave's wavelength (or trough). In this instance, the distance between each crest is specified as 1 m. The wave's wavelength is therefore 1 m.

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Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

\(h = ut+\dfrac{gt^2}{2}\)

Putting all given values, we get :

\(12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3\)

t = 2.56  and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

Answer: The charges must be like charges (both positive or both negative).

Explanation:

Answer:

When the rock is on top of the building, it does not move, so it only has potential energy.

The potential energy can be written as:

U = m*g*h

where m is the mass, g is the gravitational acceleration, h is the height.

Now, as the rock starts to fall down, the potential energy transforms into kinetic energy.

K = (m/2)*v^2

Where v is the velocity.

When the rock hits the ground, all the potential energy has ben converted into kinetic energy, then:

U = K

m*g*h = (m/2)*v^2

Here we can isolate v:

v = √(2*g*h)

and g = 9.8m/s^2

      h = 10.5m

v = √(2*10.5m*9.8m/s^2) = 14.34m/s

Now the second question:

"what is the Plot the position, velocity and acceleration vs. time"

I suppose that you need to select the correct plot for each thing, the images are not given, so let's analyze how each plot is:

The motion equations are:

Acceleration:

Here we have only the gravitational acceleration, so we can write:

a(t) = -g

This is a constant, the graph will be a horizontal line at y = -g.

Velocity:

We integrate the acceleration over time, the constant of integration is the initial velocity, that in this case is zero.

v(t) = -g*t

This is a linear equation with slope equal to -g, and y-intercept equal to zero.

Position.

We integrate again over time, this time the constant of integration will be the initial height of the rock = 10.5m

The equation is:

p(t) = -(g/2)*t^2 + 10.5m

This is a quadratic equation with a negative leading coefficient, so the arms go downwards.

Jupiter's volume is more than ten times as large as Saturn's volume. This statement is true. Jupiter is the largest planet in our solar system with a volume of about 1,431,281,810,739 km³ while Saturn is the second-largest planet with a volume of about 827,129,915,150 km³.

Jupiter is approximately 11 times larger than Saturn. The two planets belong to the gas giant category, and they share many similarities such as having a large number of moons. Jupiter is famous for its Great Red Spot and powerful magnetic field, while Saturn is well-known for its stunning ring system. Both planets have been the focus of scientific research and exploration, and they continue to fascinate scientists and stargazers alike. In conclusion, Jupiter's volume is more than ten times as large as Saturn's volume.

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When calculating the moment of inertia of the platter using its mass and dimensions, assumptions were made about the platter being a solid, uniform object with constant density, a rigid body that does not deform under external forces, rotating about a fixed axis, and no external torques acting on it.

When we used the mass and dimensions of the platter to calculate its moment of inertia, we made several assumptions.

Firstly, we assumed that the platter was a solid, uniform object with a constant density. This allowed us to use the formula for the moment of inertia of a uniform solid object, which is I = (1/2)mr², where m is the mass of the object and r is the radius of gyration.

Secondly, we assumed that the platter was a rigid body, meaning that its shape would not change under the influence of external forces. This allowed us to use the formula for the moment of inertia of a rigid body, which is I = ∑mr², where the summation is taken over all the particles in the body.

Thirdly, we assumed that the platter was rotating about a fixed axis of rotation. This allowed us to use the formula for the moment of inertia of a rotating object, which is I = mr², where r is the distance between the axis of rotation and the particle.

Finally, we assumed that there were no external torques acting on the platter, which means that the angular momentum of the platter was conserved.

This allowed us to use the conservation of angular momentum principle to solve for the angular velocity of the platter given its initial angular velocity and the moment of inertia calculated using the above assumptions.

In conclusion, by making these assumptions, we were able to calculate the moment of inertia of the platter using its mass and dimensions, and use this to predict its rotational motion under various conditions.

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