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:
27J
Explanation:
From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.
Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;
U = Q + (F×d);
U- change in internal energy
Q is the energy received by the system and is positive when energy is received by the system.
Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.
U=12+(3×5)= 27J
The potential energy= mass times gravity times height. However, we are missing height. Gravity is a constant that is 9.8 on Earth. We then solve for height by dividing 350 by 10 and then 9.8 to get about 3.5.
TLDR: 3.5
Answer:
its 1/2 the mass of the object times by its velocity ^ 2
Answer:
Only the goalie is allowed inside the goal crease. The only exception when another player is allowed in the goal area is when they take off from outside the goal area, and shoots or passes the ball before landing. To avoid interference with other players, the player must then exit the goal area as soon as possible.
Explanation: