Answer:
a)
b) imagen adjunta
Explanation:
a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.
Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:
Queremos encontrar la posición hasta detenerse, osea vf = 0.
b) Para este caso el gráfico se encuentra adjunto.
Espero que te sirva de ayuda!
Answer:
28,400 N
Explanation:
Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:
On the lower part of the hatch, there is a pressure equal to
So, the net pressure acting on the hatch is
which acts from above.
The area of the hatch is given by:
So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:
Answer:
the electric field at point A is
E = 5.5 ×10¹³N/C(-x direction)
Explanation:
given
electrostatics constant k = 9.0×10⁹
charge q = 31.7mC= 31.7×10⁻³C
distance r = 2.80mm
distance from midpoint to point A = 2.00mm
attached is the diagram of the solution, describing the position of the charge
note x = r/2, where x is the distance from midpoint of r to the particle
using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m
the electric field at point A is given as
vector <em>E </em>= 2E×cos θ( -x direction)
recall E =kq/x²
where k is the electrostatics constant = 1/4πε₀
where ε₀ is permittivity of free space
therefore using E =2{kq/x²}cosθ
∴cosθ = adjacent/hypotenuse
cosθ=1.40/2.44
E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)
E=2{4.79×10¹³}×(0.574)(-x)
E = 2×2.75 ×10¹³N/C(-x direction)
Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)
