A skydiver weighing 588 N reaches a velocity of 45 m/s [down] before opening her parachute. After falling an additional 30 m in
4 seconds, her velocity has decreased to 25 m/s.
What is the acceleration of the skydiver over the 30 m displacement?
What was the upward force exerted on the parachute during that time?
What is the acceleration of the skydiver over the 30 m displacement?
What was the upward force exerted on the parachute during that time?
1 answer:
Answer:
The acceleration of the skydiver assuming that it is constant due to the parachute is -5.
The upwards force exerted during that time is 888 N.
Explanation:
The skydiver weighs 588 N , this means that mg=588 N, where m is the mass of the skydiver.
Initial Velocity(u)=45 m/s
Final Velocity(v)=25 m/s
Time elapsed=4 s
a=??
Applying first equation of motion,
25=45 + 4a
4a=-20
a=-5
The acceleration of the skydiver assuming that it is constant due to the parachute is -5.
From the FBD attached we get that
F-mg=m x a
F-588=300
F=888 N
The upwards force exerted during that time is 888 N.
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Answer:
A) x _electron = 0.66 10² m , B) x _Eart = 1.13 10² m , C) d_sphere = 1.37 10⁻² mm
Explanation:
A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of
Let's use proportions rule
x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5
x _electron = 0.66 10⁵ mm = 0.66 10² m
B) the radii of the Earth and the sun are
= 6.37 10⁶ m
tex]R_{Sum}[/tex] = 6.96 10⁸ m
Distance = 1.5 10¹¹ m
x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5
x _Eart = 1.13 10² m
C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions
d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶
d_sphere = 1.37 10⁻² mm