The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
<h3>
Time of motion of the girl</h3>
The time of motion of the girl is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- t is time of motion
- g is acceleration due to gravity
Substitute the given parameters and solve for time of motion;
50.8 = 0 + ¹/₂(9.8)t²
2(50.8) = 9.8t²
101.6 = 9.8t²
t² = 101.6/9.8
t² = 10.367
t = √10.367
t = 3.22 seconds
<h3>Final vertical velocity of the skydiver</h3>
vf = vi + gt
where;
vi is the initial vertical velocity = 0
vf = 0 + 9.8(3.22)
vf = 31.56 m/s
Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
Learn more about vertical velocity here: brainly.com/question/24949996
#SPJ1
Explanation:
As the given spheres are connected by a thin wire so, the potential on the spheres are the same.
......... (1)
Hence, total charge will be as follows.
= Q = -95.5 nC .......... (2)
Using the above two equations, the final equation will be as follows.
and, 
Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.
= 
= 82.714 nC
Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.
At the top:
Potential Energy = (mass) x (gravity) x (height)
= (30 kg) x (9.8 m/s²) x (3 meters)
= 882 joules
At the bottom:
Kinetic Energy = (1/2) x (mass) x (speed)²
= (1/2) x (30 kg) x (3 m/s)²
= (15 kg) x (9 m²/s²)
= 135 joules .
He had 882 joules of potential energy at the top,
but only 135 joules of kinetic energy at the bottom.
Friction stole (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
Answer:
The work done is 205 kJ.
Explanation:
Hi there!
Work can be calculated using the following equation:
W = F · Δx
Where:
W = work
F = applied force
Δx = displacement
In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:
W = ∫ F · dx
F = 3.6 N/m³ · x³ - 76 N
W = ∫ (3.6 x³ - 76)dx
W = 0.9 x⁴ - 76x
Evaluating from xi to xf:
W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m
W = 205 kJ
Answer:
The gravitational acceleration of a planet of mass M and radius R
a = G*M/R^2.
In this case we have:
G = 6.67 x 10^-11 N (m/kg)^2
R = 2.32 x 10^7 m
M = 6.35 x 10^30 kg
Now we can compute:
a = (6.67*6.35/2.32^2)x10^(-11 + 30 - 2*7) m/s^2 = 786,907.32 m/s^2
The acceleration does not depend on the mass of the object.
