Answer:
Forces come in pairs, so the force of gravity (9.8 N) with the mans weight (794N) on the earth is counteracted with the normal force ( the force of the earth back on the man) which is the same
Explanation:
Mechanics is the study of motion and force on physical objects and their surroundings
From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM
M= mass of the Earth = 6×10²⁴ Kg.
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T² = 1653339719
∴ T = 40661.28 seconds.
Acceleration = (change in speed) / (time for the change)
-- during the first second, the object increases its speed to
(5 m/s²) · (1 s) = 5 m/s .
-- During the next 2 seconds, the object increases its speed by
(2 m/s²) · ( s) = 4 m/s
So at the end of the whole 3 seconds, its speed is (5 m/s) + (4 m/s) = 9 m/s
-- Over the whole time, its speed has changed from zero to 9 m/s.
Acceleration = (change in speed) / (time for the change)
Acceleration = (9 m/s) / (3 sec)
<em>Acceleration = 3 m/s²</em>
The important thing to note here is the direction of motion of the test rocket. Since it mentions that the rocket travels vertically upwards, then this motion can be applied to rectilinear equations that are derived from Newton's Laws of Motions.These useful equations are:
y = v₁t + 1/2 at²
a = (v₂-v₁)/t
where
y is the vertical distance travelled
v₁ is the initial velocity
v₂ is the final velocity
t is the time
a is the acceleration
When a test rocket is launched, there is an initial velocity in order to launch it to the sky. However, it would gradually reach terminal velocity in the solar system. At this point, the final velocity is equal to 0. So, v₂ = 0. Let's solve the second equation first.
a = (v₂-v₁)/t
a = (0-30)/t
a = -30/t
Let's substitute a to the first equation:
y = v₁t + 1/2 at²
49 = 30t + 1/2 (-30/t)t²
49 = 30t -15t
49 = 15 t
t = 49/15
t = 3.27 seconds
