THat would be humus. This answer apparently wasn't long enough so I'm just adding junk.
Answer:
Explanation:
We shall apply law of conservation of momentum .
m₁ v₁ = m₂ v₂
m₁ = 80 kg
v₁ = ?
m₂ = .43 kg
v₂ = 15 m /s
80 v₁ = .43 x 15
v₁ = .43 x 15 / 80
= .08 m /s
8 cm /s
option A is correct.
Law of conservation of momentum is applied only where no external force acts . Here during the period when ball was thrown , apart from internal force , external force like gravitational force was also acting but we still applied this law because the time duration of action of throw was very small
so we neglected the effect of external force . So it was just an approximation.
Answer:
(a) Position Vectors V₁= -2î km, V₂=5î km
(b) Displacement Δx=7 km
Explanation:
Given data
Distance=2 km west at t=0
Distance=5 km east at t=6 min
Positive x is the east direction
To find
(a)Car position vector at given times
(b)Displacement between 0 to 6.0 min
Solution
For Part (a) car position vector at given times
At t=0 the distance=2 km west so conclude that x₁=-2 because it is in negative side So vector V₁
V₁= -2î km
At t=6.0 the distance=5 km east so conclude that x₂=5 because it is in positive side So vector V₂
V₂=5î km
For (b) displacement between 0 to 6.0 min
According to following mathematical law we can conclude that
Δx=x₂-x₁
Δx=5-(-2)km
Δx=7 km
To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
<span>a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. </span>
<span>b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. </span>
<span>f = (ma), = .226 x 105.88, = 23.92N. </span>
I think the answer to your question is twenty-four percent.
