Answer:

Explanation:
<u>Given:</u>
Force = f = 60 N
Mass = m = 12 kg
<u>Required:</u>
Acceleration = a = ?
<u>Formula:</u>
F = ma
<u>Solution:</u>
Rearranging formula
a = F / m
a = 60 / 12
a = 5 ms⁻²
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3><h3>Peace!</h3>
Explanation:
avg spd = total distance/total time
=80/8=10m/s
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:

The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
Answer:
The answer to your question is 636.6 ft
Explanation:
Data
base = 425 ft
angle = 39°
See the picture below
1.- Divide the triangle to get two right triangles.
Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft
2.- Use the trigonometric function sine to find the hypotenuse
sin 19.5 = 212.5/hyp
solve for hyp
hyp = 212.5 / sin 19.5
Result
hyp = 212.5/ 0.333
hyp = 636.6 ft
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.
4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>