Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π ×
................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Using the given equation you get:
E = 1.99x10^-25 / 9.0x10^-6
Divide 1.99 by 9.0: 1.99/9.0 = 0.22
For the scientific notation, when dividing subtract the two exponents:
25 -6 = 19
So you now have 0.22 x 10^-19
Now you need to change the 0.22 to be in scientific notation form:
2.2 x 10^-20
The answer is 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:
0 m/s²
Explanation:
The velocity is constant, so there is no acceleration.