Answer:
2 /s north
Explanation:
Given that,
Velocity due North is 8 m/s and due south is 6 m/s
We need to find the magnitude and the direction of the resulting velocity.
Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

It is positive. So, it is in North direction.
Answer:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s
According to a periodic table, Krypton was created during the fission of Uranium.
<h3>What is the atomic number?</h3>
<em>Atomic</em> number is a characteristic associated with an element and indicates its number of protons, when a fision occurs, the total number protons is conserved.
Thus, the fission of uranium is led by two elements with <em>atomic</em> numbers 56 and 36. According to a periodic table, those <em>atomic</em> numbers are associated to elements Barium (Ba) and Krypton (Kr), respectively.
According to a periodic table, Krypton was created during the fission of Uranium. 
To learn more on fission, we kindly invite to check this verified question: brainly.com/question/6572079
Answer:
If you apply a force to separate 2 opposite poles, the potential energy of the system increases.
Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ -
= 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N