The force on each balloon is 2×10^−3 N.
Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

So,

A force is an influence that can change the motion of an object. A force can cause an object with mass to change its velocity (e.g. moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newton (N).
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The correct answer to the problemis velocity. I did the test.
Answer:
va = 4.79 m/s
vb = 1.29 m/s
Explanation:
Momentum is conserved:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂
22.75 = 3v₁ + 6.5v₂
For an elastic collision, kinetic energy is conserved.
½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²
79.625 = 3v₁² + 6.5v₂²
Two equations, two variables. Solve with substitution:
22.75 = 3v₁ + 6.5v₂
22.75 − 3v₁ = 6.5v₂
v₂ = (22.75 − 3v₁) / 6.5
79.625 = 3v₁² + 6.5v₂²
79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²
79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5
517.5625 = 19.5v₁² + (22.75 − 3v₁)²
517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²
0 = 28.5v₁² − 136.5v₁
0 = v₁ (28.5v₁ − 136.5)
v₁ = 0 or 4.79
We know v₁ isn't 0, so v₁ = 4.79 m/s.
Solving for v₂:
v₂ = (22.75 − 3v₁) / 6.5
v₂ = 1.29 m/s
Answer:
molecular formula =
Explanation:
Given data
c = 92.25%
H = 7.75%
molar mass = 104 g/mol
to find out
the empirical and molecular formula for styrene
solution
we know that
styrene 1 g contain = 0.9225 g C and 0.0775 g H
so
C = 104 × 0.9225 g / 12 g/mol
C = 7.995 mol = approx 8 mol
and
H = 104 × 0.0775 g / 1 g/mol
H = 8.06 mol = approx 8 mol
so we say that 1 mole of styrene have 8 mole of C and H
so
molecular formula =
Answer:
a) 10.54 sec
b) 284.58 m
c) 29.406 m/s
d) 39.92 m/s
Explanation:
Given data:
velocity of spacecraft = 27.0 m/s
rate of free fall acceleration is 2.79 m/s^2
distance of moving aircraft from mooon surface is 155 m
a. from kinematic eqaution of motion we have

where y = 155 m
Vi = 0 as this relation is for vertical motion, so the 27.0 m/s is not included
and a = 2.79 m/s^2.
Solving for t we get
t = 10.54 sec
b.
we know that 


c. from the kinematic formula
v = u + at

v = 29.4066 m/a
d. 
v = 39.92 m/s