Answer:
23: Acceleration 24:1m/s^2
Explanation:
Answer
given,
range of the projectile = 4.3 m
time of flight = T = 0.829 s
v = 5.19 m/s
vertical component of velocity of projectile
v_y = gt'
a) Launch angle
θ = 38°
b) initial speed of projectile
v = 6.59 m/s
c) maximum height reached by the projectile
Answer:
Explanation:
The x component is the adjacent side making up the given angle (39.4)
The vector is the hypotenuse.
The definition of the cos (x) is adjacent / hypotenuse.
cos(39.4) = adjacent / 47.3 Multiply both sides by 47.3
47.3 * cos(39.4) = adjacent Cos(39.4) = 0.7727
adjacent = 36.55
Would love to help you but there are no options for me to choose from
Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2
)
Answer:
108 N
Explanation:
From the question,
Applying
F' = mgμ................ Equation 1
Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.
From the question,
Given: m = 18 kg, μ = 0.6, g = 10 m/s²
Substitute these values into equation 1
F' = 18×0.6×10
F' = 108 N
