F=ma, so 100=m×10. Solve for m by dividing by 10. The mass is 10 kg.
Does this count as a response? Or do you want me to answer a question?
Answer:
change in relative vorticity 0.0590
Explanation:
Given data
pressure = 1000 hPa
temperature lapse rate q1 = 3.1◦C per 50 hPa
pressure = 850 hPa
temperature lapse rate q2= -0.61◦C per 50 hPa
to find out
change in relative vorticity
solution
we will apply here formula that is
N = (g / potential temperature ) × (potential vertical temperature) × exp^1/2 ............................1
here we know g = 9.8 m/s
and q1 = potential temperature=3.3 degree celsius
potential vertical temperature gradient = 3.1 - 0.61 / 1000 -850
potential vertical temperature gradient = 0.0166 degree celsius/hpa
so
N = 9.8 / 2.75 × 0.0166 × exp^1/2
N = 0.0590

In order to find it's square root, we could make it into two square roots.

Let us find the square roots of both radicals seprately.

Each pair of a number inside square root gives a number out .



Therefore,


The net force on the box parallel to the plane is
∑ F[para] = mg sin(24°) = ma
where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.
Solve for a :
g sin(24°) = a ≈ 3.99 m/s²
The box starts at rest, so after 7.0 s it attains a speed of
(3.99 m/s²) (7.0 s) ≈ 28 m/s