Newton's Third Law states that for every action there is an opposite and equal reaction:
If the gravitational force of the Earth on the Moon is F then the gravitational force of the Moon on the Earth is also F
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
Answer:
4.7 x 10³ rad / s
Explanation:
During the time light goes and comes back , one slot is replaced by next slot while rotating before the light source
Time taken by light to travel a distance of 2 x 500 m is
= (2 x 500) / 3 x 10⁸
= 3.333 x 10⁻⁶ s .
In this time period, two consecutive slots come before the source of light one after another by rotation. There are 400 slots so time taken to make one rotation
= 3.333 x 10⁻⁶ x 400
= 13.33 x 10⁻⁴ s
This is the time period so
T = 13.33 X 10⁻⁴
Angular speed
= 2π / T
= 
4.7 x 10³ rad / s
B. Decreasing surface area of a solid reactant. The more surface area showing, the quicker the reaction rate.
