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:
0.5 m/s².
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 10 m/s
Time (t) = 20 s
Acceleration (a) =?
Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:
a = (v – u) / t
Where:
a is the acceleration.
v is the final velocity.
u is the initial velocity.
t is the time.
With the above formula, we can obtain the acceleration of the car as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 10 m/s
Time (t) = 20 s
Acceleration (a) =?
a = (v – u) / t
a = (10 – 0) / 20
a = 10/20
a = 0.5 m/s²
Therefore, the acceleration of the car is 0.5 m/s².
Vector 1 has components
and vector 2 has
Add these vectors to get the resultant, which has components
The magnitude of the resultant is
with direction 
such that
or about 50º N of E.
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
Answer:
lift per meter of span = 702 N/m
Explanation:
See attached pictures.
