Answer:
30 half of 30 m/ss
Explanation:
hit the thaks pls Thanks!
<em>d) ≈ 0.87 m/s²</em>
<em>Hi there ! </em>
- <em><u>Newton</u></em><em>'s second law</em>
<em>F = m×a => a = F/m</em>
<em>a = 59N/68kg</em>
<em>1 N = 1kg·m/s²</em>
<em>= (59kg·m/s²)/68kg</em>
<em>= 0.8676 m/s²</em>
<em>≈ 0.87 m/s²</em>
<em>Good luck !</em>
Answer:
1 m
Explanation:
Given that
Mass of the child, m = 15 kg
Distance of the pivot, d = 1.5 m
Force applied, F = 220 N
To solve this problem, we first need to find the torque around the pivot.
Torque, t = mgd, where
m is the mass of the child
g is the acceleration due to gravity and
d is the distance of the pivot.
Thus, we can say that the torque is
T = 15 * 9.8 * 1.5
T = 220.5 Nm
This torque we have gotten would be used to find the distance, using the inverse of the equation.
T = F * d
d = T / F
d = 220.5 / 220
d = 1 m
Therefore, the minimum distance on the other side of the pivot required is 1 m
Answer:
0. 3 s
Explanation:
The motion of the boat is up and down in simple harmonic with amplitude of 20 cm . Time period of oscillation T = 3.5 s and top extreme point is at level with the boarding dock .
The traveller has to wait until the boat reaches middle point ( 1o cm . ) that is half its equilibrium position of oscillation .
Time period is 3.5 s . During its oscillation from extreme top position to its half of the equilibrium position , that is 10 cm below the stationary dock , time taken by boat
t = T / 12 = 3.5 /12 = .3 s .
Man shall have to wait for a period of 0.3 s
Ans 0. 3 s .
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