<h2>
Distance traveled in 1 second after drop is 4.9 m</h2><h2>
Distance traveled in 4 seconds after drop is 78.4 m</h2>
Explanation:
We have s = ut + 0.5at²
For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²
Substituting
s = 0 x t + 0.5 x 9.8 x t²
s = 4.9t²
We need to find distance traveled in 1 s and 4 s
Distance traveled in 1 second
s = 4.9 x 1² = 4.9 m
Distance traveled in 4 seconds
s = 4.9 x 4² = 78.4 m
Distance traveled in 1 second after drop = 4.9 m
Distance traveled in 4 seconds after drop = 78.4 m
Answer:
Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
Answer:
3525.19 kg
Explanation:
The computation of the mass of the car is shown below:
As we know that
Fc = m × V^2 ÷ R
m = Fc × R ÷ V^2
Provided that:
Fc = 34.652 kN = 34652 N
R = Radius = 24.98 m
V = speed = 15.67 m/s
So,
m = 34652 × 24.98 ÷ 15.67^2
= 3525.19 kg
The force on the ship is more than a car
