what is the pressure of a tank of uniform cross sectional area 4.0m2 when the tank is filled with water a depth of 6m when given that 1 atm=1.013 x 10^5pa density of water=1000kgm-3 g=9.8m/s2​

Answers

Answer 1

The pressure of a tank of uniform cross-sectional area 4.0m2 when the tank is filled with water at a depth of 6m is 58800 Pa.

Pressure calculation

To find the pressure in the tank, we can use the formula for pressure:

Pressure = density x gravity x height

Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 6 m

Thus:

Pressure = 1000 kg/m³ x 9.8 m/s² x 6 m

Pressure = 58800 kg/(m·s²)

Since the unit of pressure is Pascal (Pa), which is equivalent to kg/(m·s²), the pressure in the tank is:

Pressure = 58800 Pa

Therefore, the pressure in the tank when it is filled with water to a depth of 6 m is 58800 Pascal.

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Related Questions

A 21.1-N force is applied to a cord wrapped around a pulley of mass M = 4.49-kg and radius R = 25.0-cm The pulley accelerates uniformly from rest to an angular speed of 25.5 rad/s in 2.31-s. If there is a frictional torque \tau = 2.80-mN at the axle,

(a) determine the moment of inertia of the pulley,

(b) determine the rough estimate of the moment of inertia.

(The pulley rotates about its center)

What is the difference be (a) and (b)?

Answers

Answer:

The difference between (a) and (b) is the deviation caused by the actual pulley not being a perfect solid disk. In (a), we took into account the additional frictional torque and calculated the more accurate moment of inertia. In (b), we made a rough estimate assuming the pulley to be a solid disk, which disregards factors such as the mass distribution and the presence of the axle. The difference between the two values is the deviation caused by these factors.

The diagram shows the electric field due to point charge Q. The negative charge, A, is within the field. Charge Q has vectors radially inward starting perpendicular from the surface. The farther you get from the charge, the shorter the vectors. All vectors point towards the charge. A point labeled A is just to the right of the charged object. Which statements are correct? Check all that apply. Charge Q is positive. Charge Q is negative. The electric field is uniform. The electric field is nonuniform. If charge A is negative, it moves away from charge Q. If charge A is positive, it moves away from charge Q.

Answers

The correct statements are:Charge Q is positive, the electric field is nonuniform and if charge A is negative, it moves away from charge Q.

Based on the given information, we can make the following conclusions:

Charge Q is positive: The diagram shows that the electric field vectors point radially inward towards charge Q. Since like charges repel each other, for the vectors to point towards charge Q, it must be positive.

The electric field is nonuniform: The statement mentions that "the farther you get from the charge, the shorter the vectors." This implies that the magnitude of the electric field decreases with distance from charge Q. Therefore, the electric field is nonuniform.If charge A is negative, it moves away from charge Q: In the diagram, charge A is within the electric field of charge Q. Since opposite charges attract each other, if charge A is negative, it will experience a force that pulls it towards charge Q. Therefore, it will move towards charge Q, not away from it.

If charge A is positive, it moves away from charge Q: This statement is incorrect. According to the previous conclusion, if charge A is positive, it will experience a force that attracts it towards charge Q. Therefore, it will move towards charge Q, not away from it.

The provided information does not specify the behavior of charge A when it is positive. It is possible that charge A could move towards charge Q, or it could experience other forces depending on its position and the magnitude of the charges involved.

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A student is standing on a skateboard that is not moving. The total mass of the student and the skateboard is 50 kilograms. The student throws a ball with a mass of 2 kilograms forward at 5 m/s. Assuming the skateboard wheels are frictionless, how will the student and the skateboard move?

Answers

Therefore,  the student and the skateboard will move backward by 5m/s to counterbalance the forward momentum.

Momentum explained.

According to to law of conservation of momentum, the total momentum before the ball is thrown is equal to the final momentum after the ball is thrown.

Momentum is mass × velocity.

The initial momentum is

mass of student + mass of skate ball * velocity.

50kg * 0 = 0kh m/s

Final momentum

mass of student + mass of skate ball * velocity.

The velocity is 5m/s

According to the question, the student and the skateboard move backward which counter balance the forward movement.

mass * -v

momentum = 50kg * -v

Therefore,  the student and the skateboard will move backward by 5m/s to counterbalance the forward momentum.

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what type of path do the hikers observe that the pack follows?

Answers

The group of hikers is easily following a well-worn trail, which they watch intently. This path elegantly meanders through the complex pattern created by nature as it traverses the challenging terrain.

A sinuous course engraved by discovery and adventure, it exposes itself as a tribute to the innumerable footprints that have come before. The route displays the magnificence of nature as it passes through lush woods, peaceful meadows, and bubbling streams, occasionally climbing towering hills and sinking into isolated valleys.

With each bend, spectacular views and legends of long-ago expeditions are revealed. The pack travels this route with unshakable resolve, matching the hikers' wonder as they see the pack's peaceful coexistence with the natural environment.

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when the trigger is pulled on a cordless drill it takes 0.36s for the drill bit to reach 5200rpm. If the drill spins counterclockwise then, what is the angular acceleration of the drill bit?

Answers

The angular acceleration of the drill is 1512.5 rad/s².

Time taken for the drill, t = 0.36 s

Angular velocity of the drill, ω = 5200 rpm = 544.5 rad/s

The change in angular velocity that a spinning object experiences per unit of time is expressed quantitatively as angular acceleration, also known as its rotational acceleration.

It is a vector quantity that has two distinct directions or senses as well as a component of magnitude. The unit of angular acceleration is rad/s².

So,

The expression for the angular acceleration is given by,

α = ω/t

α = 544.5/0.36

α = 1512.5 rad/s²

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A spring stretches 0.145-m when a 0.233-kg mass is gently suspended from it as in Fig. 11–3b. The spring is then set up horizontally with the 0.456-kg mass resting on a frictionless table as in Fig. 11–5. The mass is pulled so that the spring is stretched 0.192-m from the equilibrium point, and released from rest.

Determine:

(c) the magnitude of the maximum velocity vmax.

Answers

The maximum velocity of oscillation of the spring is 1.57m/s.

Displacement of the spring, x = 0.145 m

Mass of the object suspended from the spring, m = 0.233 kg

The spring constant of the spring is given by,

k = mg/x

k = 0.233 × 9.8/0.145

k = 15.74 N/m²

The angular frequency of the oscillation of the spring is given by,

ω = √(k/m)

ω = √(15.74/0.233)

ω = 8.21 rad/s

Amplitude of the horizontal oscillation of the spring, A = 0.192 m

Therefore, the maximum velocity of oscillation of the spring is given by,

v(max) = Aω

v(max) = 0.192 x 8.21

v(max) = 1.57 m/s

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what is the pressure of a tank of uniform cross sectional area 4.0m2 when the tank is filled with water a depth of 6m when given that 1 atm=1.013 x 10^5pa density of water=1000kgm-3 g=9.8m/s2​

Answers

The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)

Pressure (P) = Density (ρ) × g × Depth (h)

Given: Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Depth (h) = 6 m

Using the given values, one can calculate the pressure:

Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure

= 58800 kg·m⁻¹·s⁻²

Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x [tex]10^5[/tex] Pa:

Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x[tex]10^5[/tex] Pa)

Pressure = 580.124 atm

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What is a benefit of using active solar energy over utility-scale solar energy for a home?

Installation costs are less with active solar systems.
Homeowner is not responsible for installation costs.
Energy comes from the active system, not a grid.
Homeowners will see less cost savings over time.

Answers

Using active solar energy for a home offers benefits such as lower installation costs, homeowner control over the system, reduced reliance on the grid, and potential cost savings over time. Option A

A) Installation costs are less with active solar systems: Active solar energy systems, such as solar panels or solar water heaters, can be installed directly on the home or property, eliminating the need for extensive infrastructure development associated with utility-scale solar energy projects.

B) Homeowner is not responsible for installation costs: While utility-scale solar energy projects may require homeowners to bear the costs of installation and infrastructure development, active solar systems for homes typically allow homeowners to directly invest in their own renewable energy solutions.

This means that homeowners have control over the installation process and can choose the system that best fits their budget and energy needs.

C) Energy comes from the active system, not a grid: Active solar systems for homes generate energy on-site using sunlight, allowing homeowners to reduce their reliance on the traditional power grid.

This independence from the grid provides benefits such as energy self-sufficiency, reduced vulnerability to power outages, and potential savings on utility bills. It also allows homeowners to have a direct and tangible impact on reducing their carbon footprint.

D) Homeowners will see less cost savings over time: This statement is incorrect. Over time, homeowners who invest in active solar energy systems can potentially experience significant cost savings. By generating their own renewable energy, homeowners can reduce their reliance on electricity provided by the utility company, which often comes with rising costs.

As utility rates increase, the savings from generating solar energy can become more substantial, allowing homeowners to recoup their initial investment and potentially even earn credits through net metering programs.

In summary, using active solar energy for a home offers benefits such as lower installation costs, homeowner control over the system, reduced reliance on the grid, and potential cost savings over time. These advantages make it an attractive option for homeowners seeking to embrace renewable energy and reduce their environmental impact. Option A

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t is the relationship between the ping pong ball's release height and its bounce height, in this
timent? In your answer make sure to:
Restate the question and then Answer it by identifying a relationship shown in the data.
Cite three pieces of numerical evidence to fully show the relationship in the data.
Explain how each piece of evidence supports your claim. Be as specific as possible
Describe how the relationship in the data connects to the following concept:
"Potential energy can be converted into kinetic energy. Kinetic energy can also be
converted back into potential energy."

Answers

Answer: The relationship between the ping pong ball's release height and its bounce height in this experiment is directly proportional.

Numerical evidence:

1. When the ping pong ball is released from a height of 30 cm, it bounces to a height of 15 cm.
2. When the ping pong ball is released from a height of 45 cm, it bounces to a height of 22.5 cm.
3. When the ping pong ball is released from a height of 60 cm, it bounces to a height of 30 cm.
Explanation:
Each piece of evidence supports the claim of a direct proportional relationship between the release height and the bounce height. In each case, the bounce height is exactly half of the release height. For example, when the ball is released from 30 cm, it bounces to 15 cm, which is half of the release height. This pattern is consistent in all three pieces of evidence, indicating a direct proportional relationship.

The relationship in the data connects to the concept of potential energy being converted into kinetic energy and vice versa. When the ping pong ball is released from a certain height, it possesses potential energy due to its position above the ground. As the ball falls, this potential energy is converted into kinetic energy, which is the energy of motion. When the ball hits the ground and bounces back, this kinetic energy is converted back into potential energy as the ball gains height again. The direct proportional relationship observed in the data confirms this concept, as the ball's bounce height is directly related to the amount of potential energy it had at the release height.

I hope this helps! :)

HELPLP NEED FOR AN ASSIGNENT ! In the string pull illustration above, a shark on the button string results in the bottom string breaking, does this occur because of the balls weight of its mass?

Answers

Answer:

Yes, the shark's weight or mass is what causes the bottom string to break. The weight of the shark creates tension on the bottom string, which can cause it to snap if the tension becomes too great.

How much effort will be required on the small piston having cross section area zam to lift a lead of 4000N on a large piton having cross sectional area 1m². also calculate pressure exerted on the small piston.​

Answers

Answer:

4000 Nm^-2

Explanation:

Dude that "zam" drove me away, anyway:

Given:

Force on the large piston (F1) = 4000 N

Cross-sectional area of the large piston (A1) = 1 m²

Cross-sectional area of the small piston (A2) = zam (let's assume zam represents the area in square meters)

According to Pascal's law, the pressure exerted on the large piston (P1) is equal to the pressure exerted on the small piston (P2):

P1 = P2

Pressure is defined as force divided by area:

P1 = F1 / A1

P2 = F2 / A2

Since P1 = P2, we can equate the two expressions:

F1 / A1 = F2 / A2

Rearranging the equation to solve for F2, the force on the small piston:

F2 = (F1 / A1) * A2

Substituting the given values:

F2 = (4000 N / 1 m²) * zam

Now, to calculate the pressure exerted on the small piston (P2), we can divide the force by the area:

P2 = F2 / A2

Substituting the values we obtained:

P2 = [(4000 N / 1 m²) * zam] / zam

The area "zam" cancels out in the equation, leaving us with:

P2 = 4000 N/m²

Therefore, the pressure exerted on the small piston is 4000 N/m².

To determine the effort required on the small piston, we need to know the area of the small piston. Once we have that information, we can substitute it into the equation for F2 to calculate the effort required

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