Answer:

Explanation:
First of all, we need to find the volume of the room, which is given by

Now we can find the mass of the air by using

where
is the density of the air
is the volume of the room
Substituting,

Answer:
6 m/s is the missing final velocity
Explanation:
From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).
Object X had a mass of 300 kg, while object Y had a mass of 100 kg.
Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.
We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.
In numbers, and calling
the initial momentum of object X and
the initial momentum of object Y, we can derive the total initial momentum of the system: 
Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):
Final momentum of the system: 
We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s
Answer:
the tension in the string is 5.59 N
Explanation:
Here ,
m_1 = 0.385 Kg
m_2 = 0.710 Kg
Using second law of motion ,
a = F_net / effective mass
a = (0.710- 0.385)×9.8/(0.710 + 0.385 + 0.0125/0.15^2)
a = 1.93 m/s^2
Now , let tension be T ,
then,
mg-T=ma
0.710×g - T = 0.710×1.93
T = 5.59 N
the tension in the string is 5.59 N
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