Answer:
200 Watts.
Explanation:
Power is defines as the amount of work expended per unit time. Mathematically, it is expressed as Power = Workdone/Time
Given parameters
Energy used up 100Joules
Distance moved by brick = 1 meters
Time taken by the machine = 0.5 secs
Power can also be written as Energy/Time
Required
We need to calculate the amount of power used up.
Power = 100J/0.5s
Power = 100/(1/2)
Power = 100 * 2/1
Power = 200Watts.
This shows that the machine would expend 200Watts of power
Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F = 
F = 
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹
θ = tan⁻¹
0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East
Answer:
Explanation:
Cutting a string in half because
b is irreversible
c is a cheical and d is also a chemical change
Answer:The correct options are:
1. A system is a group of objects analyzed as one unit.
2. Energy that moves across system boundaries is conserved.
Explanation:
A system is defined as group of interrelated or interacting items existing as a single unit or a whole to achieve a specific objective.Energy lost by the system is equal to the energy gained by the surroundings.
Two statements are true about a system:
- A system is a group of objects analyzed as one unit.
- Energy that moves across system boundaries is conserved.