Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
Answer : 0.814 newton
Explanation:
force (magnetic) acting on the wire is given by
F= ? , I=2.2amp , B = 0.37 T
F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N
Answer:
Explanation:
In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.
The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.
Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.
To find the weight of the book we simply multiply the mass of the book by gravity.
W = m*g
W = 1.3[kg] * 9.81[m/s^2]
W = 12.75 [N]
Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 =
× 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
I think it’s c because the other ones are just options not facts