Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at
1 answer:
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Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:
Our values are given by,
As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:
Skate 2:
Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:
Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s