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 answer to that will be the Troposphere.
Velocity of an object is its rate of change of the object's position per interval of time. Velocity is a vector quantity which means that it consists of a magnitude and a direction. Magnitude is represented by the speed and the direction is represented by the angle. To determine the velocity components, we use trigonometric functions to determine the angle of the components. For the north component we, use the sine function while, for the west component, we use the cosine function. We calculate as follows:
north velocity component = (16.8 m/s) (sin 54°) = 16.4 m/s
<span>west velocity component = (16.8 m/s) (cos 54°) = 3.49 m/s</span>
The total flux through the cylinder is zero.
In fact, the electric flux through a surface (for a uniform electric field) is given by:
where
E is the intensity of the electric field
A is the surface
is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.
We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because 
and 
).
On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be
where r is the radius, and where we have taken 
since the perpendicular to the surface is parallel to the direction of the electric field, so 
. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so 
and 
, therefore the flux is
And the net flux through the cylinder is
