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Tasya [4]
2 years ago
15

Ahmed fills a basket with shopping weighing 10 kg. How much work is being done on the shopping basket when he lifts it verticall

y by 1.5 m? (g= 9.8 m/s²)
Physics
1 answer:
NISA [10]2 years ago
6 0

The work done by the shopping basket is 147 J.

<h3>When is work said to be done?</h3>

Work is said to be done whenever a force moves an object through a certain distance.

The amount of work done on the shopping basket can be calculated using the formula below.

Formula:

  • W = mgh

Where:

  • W = Amount of work done by the basket
  • m = mass of the shopping basket
  • h = height of the shopping basket
  • g = acceleration due to gravity.

Form the question,

Given:

  • m = 10 kg
  • h = 1.5 m
  • g = 9.8 m/s²

Substitute these values into equation 2

  • W = 10(1.5)(9.8)
  • W = 147 J.

Hence, The work done by the shopping basket is 147 J.

Learn more about work done here: brainly.com/question/18762601

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A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?
Alinara [238K]

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

since all the lides are of equal length you can just choose one side.

20squared is 400

20 x 20 = 400cm squared

Hope this helps :)

6 0
3 years ago
An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o
kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0
3 years ago
Which of the fundamental forces explains the structure of atoms and molecules?
spayn [35]
The electromagnetic force<span> holds atoms and molecules together.
like a magnet's pull on steel.</span>
7 0
3 years ago
Read 2 more answers
A person pushes a large 42.9 kg box at a constant velocity of 9 m/s across a horizontal floor for 3.8 s. Find the
Gwar [14]

Answer:

10 kJ

Explanation:

W = Fd

W = (μN)(vt)

W = μ(mg)vt

W = 0.7(42.9)(9.81)(9)(3.8)

W = 10,075.12506 J

W ≈ 10 kJ

6 0
3 years ago
Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far
nata0808 [166]

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

Mass of object 1  = 3.1 x 10⁵kg

Mass of object 2 = 6.5 x 10³kg

Gravitational force  = 65N

Unknown:

Distance between them  = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

    Fg  =    \frac{G mass 1 x mass 2}{distance ^{2} }  

   G = 6.67 x 10⁻¹¹

        65  = \frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3}   }{distance^{2} }    

   Distance  = 4.55 x 10⁹m

         

3 0
3 years ago
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