Answer:
Explanation:
Given
Object is thrown with a velocity of 
Acceleration due to gravity is -g (i.e. acting downward)
Vertical distance traveled by object is given by
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
at maximum height final velocity is zero


time taken to reach maximum height
using
v=u+at
0=9-gt

Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
A
Explanation:
The officer would have had permission regardless of anything else, kind of like letting someone into your house.
Answer:
1,373.4 N
Explanation:
The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.
Mass of books will be 10kg+20kg+30kg=60 kg
Total mass of table and books will be 500kg+60kg=560 kg
This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg
Force is product of mass and acceleration due to gravity hence F=gm
Taking g as 9.81 m/s2 then
F=140*9.81=1,373.4 N
Therefore, rhe normal force is equivalent to 1,373.4 N
<span>the ratio of the velocity of light in the medium over the velocity of light in a vacuum</span>