It is wires or the sun from outside the sun as heat which light thing
Answer:
First Car Avg: Velocity=0.11 m/s
Second Car Avg: Velocity=0.28 m/s
Explanation:
Data
D1 (First Car Displacement)=0.25m
T1 (First Car total time taken)=2.23s
D2 (Second Car Displacement)=0.25m
T2(Second Car time taken)=3.13-2.23=0.9s
Avg: Velocity1=?
Avg: Velocity2=?
Solution
Avg: Velocity1=D1/T1
Avg: Velocity1=0.25/2.23
Avg: Velocity1=0.11 m/s
Avg: Velocity2=D2/T2
Avg: Velocity2=0.25/0.9
Avg: Velocity2=0.28 m/s
Answer:
x = 41.2 m
Explanation:
The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.
In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract
F = F₁₃ - F₂₃
let's write the expression for each force, let's set a reference frame on the charge q1
F₁₃ =
F₂₃ = ![k \frac{q_2 q_3}{(d-x)^2}](https://tex.z-dn.net/?f=k%20%5Cfrac%7Bq_2%20q_3%7D%7B%28d-x%29%5E2%7D)
they ask us that the net force be zero
F = 0
0 = F₁₃ - F₂₃
F₁₃ = F₂₃
k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}
q1 / x2 = q2 / (d-x) 2
(d-x)² =
x²
we substitute
(100 - x)² = 2/1 x²
100- x = √2 x
100 = 2.41 x
x = 41.2 m
<h2>The pressure will become double </h2>
Explanation:
The gas pressure is directly proportional to the mean root square velocity of the constituent molecules of gas .
P ∝
I
Here C₁ , C₂ ------------ Cₙ is the velocities of molecules .
By making these velocities double
The pressure P₀ ∝ 2
II
By dividing II by I
P₀ = 2 P
Thus pressure will become double than its previous value