Answer:

Explanation:
Given that,
The length of a string, l = 0.87 m
Speed of the ball, v = 3.36 m/s
We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

So, the acceleration of the ball is
.
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)
S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m
Distance double 720m*2=1440m
V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
Answer:
0.25m/s
Explanation:
Given parameters
m₁ = 5kg
v₁ = 1.0m/s
m₂ = 15kg
v₂ = 0m/s
Unknown:
velocity after collision = ?
Solution:
Momentum before collision and after collision will be the same. For inelastic collision;
m₁v₁ + m₂v₂ = v(m₁ + m₂)
Insert parameters and solve for v;
5 x 1 + 15 x 0 = v (5 + 15 )
5 = 20v
v =
= 0.25m/s
Answer:
Inter Quartile Range
Explanation:
Quartile is a positional statistical average, which divided the data into 4 equal halves.
Q1 (Lower Quartile) has 25% data below it, 75% above it. Q3 (Upper Quartile) has 75% data below it, 25% above it.
Interquartile range is the measure used to calculate how far the lower & upper quartiles are.