Answer:
F₄ = 29.819 N
Explanation:
Given
F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N
F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N
F₃ = (0 i + 0 j + 4 k) N
Then we have
F₁ + F₂ + F₃ + F₄ = 0
⇒ F₄ = - (F₁ + F₂ + F₃)
⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N
The magnitude of the force will be
F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N
Answer:
4. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight.
Explanation:
When you push on anything, Newtons Third Law comes into effect: "when two objects interact, they apply forces to each other of equal magnitude and opposite direction." So pretty much, the table you are putting some of your weight on is taking some of the downward force and in return the scale isn't going to push up with as much force.
I'm not the best at explaining, but im in physics right now and thats somewhat how my teacher explained it.
Answer:
10.85 m/s
Explanation:
m = 1200 kg, h1 = 19 m, h2 = 13 m
Let v be the velocity
Use the conservation of energy
Potential energy at 19 m = Potential energy at 13 m + kinetic energy
m x g x h1 = m x g x h2 + 1/2 mv^2
m x g (h1 - h2) = 0.5 x m v^2
g (h1 - h2) = 0.5 v^2
9.8 (19 - 13) = 0.5 x v^2
v = 10.85 m/s
Answer:
a = 50 [m/s²]
Explanation:
This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force =25 [N] (units of Newtons)
m = mass = 0.5 [kg]
a = acceleration [m/s²]