Newton's three laws of motion can be used to describe the motion of the ice skating.
<h3>Newton's first law of motion</h3>
Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.
- Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.
<h3>Newton's second law of motion</h3>
Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.
- Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.
<h3>Newton's third law of motion</h3>
This law states that action and reaction are equal and opposite.
- Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.
Learn more about Newton's law here: brainly.com/question/3999427
I don't know this 1 I'm sorry I can't help you
Answer: The five major fields of environmental science are social sciences, geosciences, environmental chemistry, ecology, and atmospheric sciences.
Explanation:
<h3>Answer</h3>
m/s^2 (meter per sec square)
Explanation:
acc = change in velocity/time
= distance/time
----------------
time
= m/s
------
s
=m/s^2
Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law
