The answer for this would be B!!
Answer:
the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.
Explanation:
When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.
Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.
The whole process must be analyzed using conservation of the moment.
p₀ = m v₀
celestines que clash case
p_f = (m + M) v
po = pf
m v₀ = (n + M) v
v = 
calculemos
v= 
v= 0.09 m/s
elastic shock case
p₀ = m v₀
p_f = m v₁ +M v₂
p₀ = p_f
m v₀ = m v₁ + m v₂
Answer:
Re=160ohm
Explanation:
Step#1
Rt=R1+R2 ( because both are in series)
Rt=(100+220 ) ohm
Rt=320 ohm
Step#2
Rt and R3 are parallel so,
Re= (Rt× R3) ÷ (Rt+R3)
Re= (320×320)÷( 320+320)
Re = 102,400÷ 640
Re=160ohm
The wave speed to this question is 400 meters
Answer:
a. the core will spin faster.
Explanation:
By law of conservation of angular momentum
(mvR)i= (mvR)f
m= mass of star
v= speed of star
R= radius of star
i= initial
f= final
since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.
Hence, a. the core will spin faster.