Answer:
Explanation:
The strength of the gravitational field at the surface of a planet is given by
(1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For the Earth:
For the unknown planet,
Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:
And substituting g = 9.8 m/s^2,
-- If the object is moving with speed of 10.954 meters per second, then
it has 300J of kinetic energy no matter where it may be located.
-- If the object is 6.118 meters above somewhere, then it has 300J of
gravitational potential energy relative to that place.
This is an example Newton's Third Law. All the kinectic energy from the moving car transferred the potential energy of the parked car. This potential is not much since the brakes are on (hopefully) and it's not in a non-moving position.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
