Select the correct answer.

A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy

goblinko [34]

Answer:

Air resistance

Explanation:

Despite the law of conservation of energy stating that energy can neither be created nor destroyed but can only be transformed from one state to another, some energy is usually lost in the process of transformation and its majorly attributed to frictional loss. Friction opposes normal movement hence in air, air resistance tends to reduce the original energy compared to the initial. That is why the final energy in this case is slightly less than the original energy.

3 0

4 years ago

Plzzz someone help me!!!!.

Vlad [161]

Answer:

He could re read his answear.

Explanation:

4 0

3 years ago

You are walking from your math class to your science class. You are carrying books

kolezko [41]

Answer:

1800J

Explanation:

Given parameters:

Weight of the book = 20N

Total distance covered = 45m + 15m + 30m = 90m

Unknown:

Total work performed on the books = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

Work done = Force x distance

Work done = 20 x 90 = 1800J

8 0

3 years ago

If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Tasya [4]

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

7 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0

3 years ago