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1 year ago

A large tank is filled with water to a depth of 38 m. An opening is located 14 m

1 answer:

5 0

When a large tank is filled with water to a depth of 38 m. An opening is located 14 m above the bottom of the tank then the water would emerge from the opening at a velocity of 21.70 m/s

<h3>What is mechanical energy?</h3>

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.

As total mechanical energy ias s the sum of all the kinetic as well potential energy stored in the system.

ME = KE + PE

The total kinetic energy

KE = 1/2*m*v²

Similarly, the total potential energy

PE = m*g*h

As given in the problem When a large tank is filled with water to a depth of 38 m. An opening is located 14 m above the bottom of the tank

height of the water above the opening

h= 38-14

h= 24 m

The potential energy of water above the opening is getting converted into the kinetic energy

m*g*h = 1/2*m*v²

v = √(2*g*h)

By substituting the respective values of the height and gravity

v = √(2*9.81*24)

v = 21.70 m/s

Thus, when water is poured into a sizable tank until it reaches a 38 m depth. If a hole were to be made at a height of 14 m above the tank's bottom, water would rush out of it at a speed of 21.70 m/s.

Learn more about mechanical energy from here

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Answer:

Part 1

Stationary

Part 2

20 Newtons

Part 3

Force

Part 4

4.0 m/s²

Part 5

Normal

Part 6

Cart m

Part 7

The gravitational force is less than magnetic force

Explanation:

Part 1

The position time graph of the object is an horizontal straight line passing across the top of the position 3 boxes vertically up from the origin

As the time increases by the units of number boxes to the left, the position of the object does not change and remains at the 3 boxes up above the origin, therefore, the object is stationary

Part 2

By Newton's third law of motion, the action action obtained from a force is equal to the reaction given to the force, therefore, we have;

The force exerted by the student on the scale = The force exerted by the scale on the student = 20 N

Part 3

A force is a the directional push on an object or pull from the object as a form of interaction with another object which tends to alter or maintain the motion of the object

Part 4

The given parameters are'

The mass of block A = 1.0 kg

The mass of block B = 2.0 kg

Both blocks, "A" and "B" are initially at rest

The applied horizontal force, F = 12-N

The nature of the surface over which the blocks move = Smooth surface

Force, F = Mass, m × Acceleration, a

F = m × a

The blocks two blocks experience a common acceleration, a

The combined mass of the two blocks, m = 1.0 kg + 2.0 kg = 3.0 kg

m = 3.0 kg

Therefore, a = F/m = 12-N/(3.0 kg) = 4 m/s²

Part 5

A normal force is a force acting perpendicularly to a surface that supports the weight of an object

Part 6

The given parameters are;

M = 2 kg

The mass of the left cart attached to the spring = M = 2 kg

The mass of the right cart attached to the spring = 2·M = 2 × 2 kg = 4 kg

Therefore, given that the force exerted by one cart on the other after the spring is removed, we have;

Force, F = M × a₁ = 2·M × a₂

Where;

a₁ = The average acceleration of the cart with mass, M

a₂ = The average acceleration of the cart with mass, 2·M

M × a₁ = 2·M × a₂

∴ a₁ = 2·a₂

The acceleration, a₁, of car M = 2 × The acceleration, a₂, of car 2·M

The acceleration of cart M is two times the acceleration of cart 2·M

Cart M will experience a greater average acceleration

Part 7

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