Un montacargas hidráulico es utilizado para levantar cajas con un peso de 1500 newtons sobre una superficie de 20 cm al cuadrado
1 answer:
Answer:
150 N
Explanation:
The question says that, "A hydraulic forklift is used to lift boxes with a weight of 1500 newtons over a surface of 20 cm squared. What is the force applied to a base of 2 cm squared?"
Given that,
Force applied, F₁ = 1500 N
Area, A₁ = 20 cm²
We need to find the force applied to a base of 2 cm². Using Pascal's law to find it such that,
So, the required weight is 150 N.
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Newton's second law allows us to find the force of the block on the table is 126 N
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
Σ F = m a
Where the bold letters indicate vectors, m is the mass and the acceleration of the body
A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.
Let's start by finding the acceleration of the elevator with kinematics
v = v₀ + a t
a =
Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time
a =
a = 3.5 m / s²
Let's write Newton's second law for each body
The book
N₂ - W₂ + N₁ = m a
Table
N₁ - W₁ - W₂ = M a
W₁ = Mg
W₂ = mg
N₁ = (M + m) g + M a
N₁ = (10 + 2) 9.8 + 10 3.5
N₁ = 152.6 N
This is the reaction of the earth to the support of the block and the table
N₂ = ma + m g - N₁
N₂ = m ( a +g) - N₁
N₂ = 2 (3.5 + 9.8) - 152.6
N₂ = 26.6 - 152.6
N₂ = -126 N
The negative sign indicates that the direction is opposite to the one assigned, this is the action of the block on the table.
In conclusion using Newton's second law we can find the forces of the block on the table is 126 N
Learn more here: brainly.com/question/19860811
Answer:
The change in volume is
Solution:
As per the question:
Coefficient of linear expansion of Copper,
Initial Temperature, T = = 273 K
Final Temperature, T' = = 273 + 100 = 373 K
Now,
Initial Volume of the block, V =
where
V' = Final volume
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
