The force applied to small piston = 2.2 x 10³ N
<h3>Further explanation</h3>
Given
a radius of 5 cm and 15 cm
weight 20000 N
Required
Force applied
Solution
Pascal Law :
F₁/A₁=F₂/A₂
A₁ = π.5²
A₂ = π.15²
F₁/ π.5² cm² = 20000/π.15² cm²
F₁ = 2222.22 N⇒2.2 x 10³ N
Answer:
Ex(P) = -3.602 x 10^6 N/C
Explanation:
- q2 = 2.6 μC = the net charge on the conducting shell
- inner radius of conducting shell = a = 2.2 cm =0.022m
- outer radius of conducting shell = b = 4.5 cm = 0.045m
1) To get Ex(P), the value of the x-component of the electric field at point P, located a distance 8.8 cm along the x-axis from q1 ;
Ex(P) = k(q1+q2)/r^2
= 9 x 10^9 (-5.7 + 2.6) x 10^-6 /0.088^2
Ex(P) = -27.9 x 10^3/ 0.007744
Ex(P) = -3.602 x 10^6 N/C
Answer:
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Explanation:
I: x = 1/2at²
II: v = at
I+II: x = 1/2a(v²/a²) = 1/2 v²/a
a = 1/2 v²/x
v = 30
x = 49
(a) Since the lake has a circular shape, the distance they traveled is exactly 3/4 of a circumference. The radius of the lake is 2.51 km, and the circumference is given by

, therefore the distance covered is

(b) We can consider the lake to be on a xy-plane with the origin of the axes being at the center of the lake. In this system of coordinates, the starting point of the motion is at the west side of the lake, so at coordinates (-2.51 km,0). The final point is after 3/4 of circumference, therefore at the north side, at coordinates (0, 2.51 km).
So we can calculate the magnitude of the displacement as


(c) Considering only the initial and final point of the motion, the couple moved 2.51 km north (on the x-axis) and 2.51 km east (on the y-axis). Therefore, we can calculate the angle of the displacement with respect to the east direction:

from which