D = 110 m, t = 5 s
v o = 110 cs : 5 m = 22 m/s
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v = v o - a t
v = 0 m/s, v o = 22 m/s, t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g.
Answer:
x' = 1.01 m
Explanation:
given,
mass suspended on the spring, m = 0.40 Kg
stretches to distance, x = 10 cm = 0. 1 m
now,
we know
m g = k x
where k is spring constant
0.4 x 9.8 = k x 0.1
k = 39.2 N/m
now, when second mass is attached to the spring work is equal to 20 J
work done by the spring is equal to


x'² = 1.0204
x' = 1.01 m
hence, the spring is stretched to 1.01 m from the second mass.
Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
No friction present means: Ek = Ep
So Ek = mgh = 10 * 9.8 * 2 = 196 J