For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Energy because only the energy moves ahead.
We use conservation of energy. Note that at the beginning of the drop, the pig has only potential energy, and at the bottom, it has only kinetic energy. Then:
Now, we multiply by
to get:
Then, we have that:
Now, we substitute.
while
. Therefore,
.
Answer ;
<em>Note : Satellite completes one orbit in the same time the earth rotates one about its axis</em>
circumference of the orbit = 2 ×π × 4.23×10⁷ (similar to the earth orbital speed
The time of earth/orbit traveled = 23.93 hrs ,
Speed of the orbit = (distance) ÷ (time)
= circumference ÷ time
= (2 ×π × 4.23×10⁷) ÷ 23.93 (1 hr = 3600 sec.)
= 3085.14 m/s
<em>The orbital velocity is 3085.14 m/s</em>
Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J