Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>
Answer:
The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.
Explanation:
Let us call the mass of the balloon 
and the mass of the basket 
, then according to newton's second law:
,
where 
is the upward acceleration, and 
is the net propelling force (counts the gravitational force).
Also, the tension 
in the rope is 79.8 N more than the basket's weight; therefore,
and this tension must equal
Combining equations (2) and (3) we get:
since , we have
Putting this into equation (1) and substituting the numerical values of 
and 
, we get:
Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.
Because they are caused by your exercise
There are 3 significant figures, if that answers the question.
Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
Answer:
k = 17043.5 N/m = 17.04 KN/m
Explanation:
First we need to find the force applied by safe pn the spring:
F = Weight of Safe
F = mg
where,
F = Force Applied by the safe on the spring = ?
m = mass of safe = 800 kg
g = 9.8 m/s²
Therefore,
F = (800 kg)(9.8 m/s²)
F = 7840 N
Now, using Hooke's Law:
F = kΔx
where,
K = Spring Constant = ?
Δx = compression = 46 cm = 0.46 m
Therefore,
7840 N = k (0.46 m)
k = 7840 N/0.46 m
<u>k = 17043.5 N/m = 17.04 KN/m</u>
