Given :
The mass of the balloon was 1890 kg and had a volume of 11,430 m3 .
The balloon floats at a constant height of 6.25m above the ground.
To Find :
The density of the hot air in the balloon.
Solution :
We know,
Volume × ( Density of surrounding air - Density of hot air ) = mass
Putting given values in above equation, we get :
Therefore, the density of hot air in the balloon is 1.125 kg m³.
Answer:
f1/f2 =W1/W2 = 1/3
.0 f2 = 3f1
As ,
1/F= 1/f1 +1/f2
...1/40 = 1/f1 - 1/3f1
f1=> 80/3 cm
... f2 = 2f1 = 3 x 80/3 = 80 cm
Answer:
W = M g weight of ball
T cos θ = W balancing vertical forces
T sin θ = F balancing horizontal forces
tan θ = F / W dividing equations
F = W tan θ when θ equals zero F equals zero
Answer:
5 feet
Explanation:
First lets imagine that the values are in meters and kg, to make it easier for me.
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The force created by the teacher is 120 × 5 = 600N.
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The force created by the 50 kg student is 50 × 8 = 400N
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The force that needs to be created by the 40 kg student is 600N - 400N = 200N.
To create that force that student needs to sit 200 ÷ 40 = 5m.
So the 40 lbs student is sitting 5 feet from the fulcrum.
