Answer:
10N
Explanation:
1. Every Action has an equal and opposite reAction.
2. If 10N of force is acted upon an wrench, then the wrench will react with an equal amount of force, but in the opposite direction.
V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
According to the position vs time graph, the <em>average</em> <em>velocity</em> of the motorcycle is the change in position divided by the change in time. Also, note that the slope is linear and positive throughout the 5 hours, it doesn't change direction.
Therefore, we have
Avg velocity = change in direction/change in time
Avg velocity = (150km - 30km)/(5h - 0h)
Avg velocity = 24km/hr south.
Given,
A player kicks a soccer hits at an angle of 30° at a speed of 26 m/s
We can resolute the trajectory of soccer into horizontal and vertical components.(Please see the attached file)
We can have,
Horizontal velocity component of ball= 26cos(30°) = 26×(√3÷2) = 22.51 m/s
And vertical velocity component of ball = 26sin(26°) = 26×(1÷2) = 13 m/s
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
