Fundamental quantity : quantities which are independent on other physical quantity .
ex: length,mass,time, current, amount of substance, luminous intensity, thermodynamic temperature,
Derived quantity : quantities which are depend on fundamental quantities.
ex: Area, volume, density, speed, acceleration, force, velocity etc.
Answer:
t = 0.24 s
Explanation:
As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:
Translation: ΣF = ma
Rotation: ΣM = Iα ; where α = angular acceleration
Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:
ΣM = I(a/R)
Now we are going to resolve and combine these equations.
For translation: Fx - Ffr = ma
We know that Fx = mgSin27°, so we substitute:
(1) mgSin27° - Ffr = ma
For rotation: (Ffr)(R) = (2/3mR²)(a/R)
The radius cancel each other:
(2) Ffr = 2/3 ma
We substitute equation (2) in equation (1):
mgSin27° - 2/3 ma = ma
mgSin27° = ma + 2/3 ma
The mass gets cancelled:
gSin27° = 5/3 a
a = (3/5)(gSin27°)
a = (3/5)(9.8 m/s²(Sin27°))
a = 2.67 m/s²
If we assume that the acceleration is a constant we can use the next equation to find the velocity:
V = √2ad; where d = 0.327m
V = √2(2.67 m/s²)(0.327m)
V = 1.32 m/s
Because V = d/t
t = d/V
t = 0.327m/1.32 m/s
t = 0.24 s
A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2
.:. The acceleration is 28.27 m/s^2
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
Hope this Helps!!!
