<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)
cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0
therefore, wmax=1.2rad/s
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)
adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)
adn=r*w^2=200*1.2^2=288
ad= square root of adt^2+adn^2 = 288mm/s^2</span>
The volume of the room is the product of its dimensions:

Now, from the equation

where d is the density, m is the mass and V is the volume, we deduce

So, multiply the density and the volume to get the mass of air in the room.
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.
![(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]](https://tex.z-dn.net/?f=%280.355%2A0.095%29%2B%280.710%2A0.045%29%3D%280.355%2A0.035%29%2B%280.710%2Av_%7BB%27%7D%29%5C%5Cv_%7BB%27%7D%3D0.075%5Bm%2Fs%5D)
<span>the formula q = 375 g * 25 C * 4.186 J / (g*C) = 39,243.75 J q represents the heat in Joules , m the mass in grams, difference of temperature in Celsius degree, and 4.186 J/(g*C) is the specific heat of water( I assume the water is in liquid from and will remain liquid). Approximately 39.24 kJ once you round and transform to kJ..1 kJ=1000J</span>