Answer:
C) one-half as great
Explanation:
We can calculate the acceleration of gravity in that planet, using the following kinematic equation:
In this case, the sphere starts from rest, so 
. Replacing the given values and solving for g':
The acceleration due to gravity near Earth's surface is 
. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer:
The strength of the magnetic field that the line produces is 
.
Explanation:
From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:
(1)
Where 
is the permiability constant, I is the current and r is the distance from the wire.
Notice that it is necessary to express the current, I, from kiloampere to ampere.
⇒ 
Finally, equation 1 can be used:
Hence, the strength of the magnetic field that the line produces is .
Answer: 
Explanation:
Centripetal acceleration 
is calculated by the following equation:
Where:
is the Earth's orbital speed
is the orbital radius
