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
The magnitude<span> of a </span>velocity<span> vector is </span>called<span> speed. Supposethat a wind is blowing in from the direction at a speed of 50 km/h. (This meansthat the direction from which the wind blows is west of the northerly direction.) Apilot is steering a plane in the direction at an airspeed (speed in still air) of250 km/h
</span>
The speed of the spaceship relative to the galaxy is 0.99999995c.
A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).
The Milky way galaxy is 100,000 light years in diameter.
The galaxy's diameter is a mere 1. 0 ly.
We know that ;

L = 1 light year
L₀ = 100,000 light year




Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.
Learn more about a light year here:
brainly.com/question/17423632
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Answer:
The motion of an object is accelerated when its speed increases.