Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So 

k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be


Answer:
a) I = 13.38 kg m / s, b) F = 1,373 10³ N
Explanation:
The impulse is given by the relation
I = ∫ F dt = Δp
I = p_f -p₀
I = m (v_f - v₀)
take the ball's exit direction as positive, whereby the ball velocities
v₀ = -90mph, the final velocity v_f = + 54 m / s
Let's reduce the units to
I = 0.142 [54- (-40.23) ]
the SI system
v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s
m = 142 g (1kg / 1000) = 0.142 kg
we calculate
I = 0.142 [54- (-40) ]
I = 13.38 kg m / s
b) let's use the definition of momentum
I = ∫ F .dt
I = F ∫ dt
F = I / t
F = 13.38 / 0.008
F = 1,373 10³ N
Answer:
A
Explanation:
I'm pretty sure it's A because when you look up the properties of alkali metal it states stuff about its melting points and what to expect. (correct me if I'm wrong but I had this same question on a test and I'm sure this is the answer lol)
<span>Static frictional force = 126.91 N
1st worker force = 110 N
2nd worker = 126.91 – 110 = 16.91 N</span>electron1 <span>· 7 years ago.</span>