Answer:
m₁ / m₂ = 1.3
Explanation:
We can work this problem with the moment, the system is formed by the two particles
The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero
p₀o = m₁ v₁ + m₂ v₂
pf = 0
m₁ v₁ + m₂ v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂= - (-6.2) / 4.7
m₁ / m₂ = 1.3
Another way to solve this exercise is to use the mass center relationship
Xcm = 1/M (m₁ x₁ + m₂ x₂)
We derive from time
Vcm = 1/M (m₁ v₁ + m₂v₂)
As they say the velocity of the center of zero masses
0 = 1/M (m₁ v₁ + m₂v₂)
m₁ v₁ + m₂v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂ = 1.3
Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
Answer: Option <em>a.</em>
Explanation:
Kepler's 2nd law of planetary motion states:
<em>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</em>
It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.
A change in momentum is called impulse.
