Physics - Damon, Wednesday, December 9, 2015 at 5:13am
F = k x
k = 2 g/6.1 cm
2.5g = (2g/6.1cm) x
x = 6.1 (2.5/2) cm
Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s
Answer:
0.0196 j
Explanation:
i) The formula for kinetic energy is as follows: 0.5*m*v^2
ii) Since we have all the values all that's left is to plug them into the equation
iii) First, WE MUST, Convert grams into kgs as this is the SI unit of mass so 2.45/1000
iv) All that's left now is to plug it into the equation so:
0.5* (s.45/1000)*(4^2)
v) Lastly we add the unit joules at the end as we're talking about energy
Hope this was useful! :)
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e (
)
with this expression we can find the closest approach distance (r)
