Energy can neither be created nor be destroyed.
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.
![90\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} \\= 25 \frac{m}{s}](https://tex.z-dn.net/?f=90%5Cfrac%7Bkm%7D%7Bh%7D%2A%5Cfrac%7B1000m%7D%7B1km%7D%2A%5Cfrac%7B1h%7D%7B3600s%7D%20%20%20%5C%5C%3D%2025%20%5Cfrac%7Bm%7D%7Bs%7D)
![v_{f} =v_{i} + (a*t)](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2B%20%28a%2At%29)
where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:
![\theta =37.01^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D37.01%5E%7B%5Ccirc%7D)
We can apply the first Newton's law in x and y-direction.
If we do a free body diagram of the system we will have:
x-direction
All the forces acting in this direction are:
(1)
Where:
- T(1) is the tension due to the rope 1
- T(2) is the tension due to the rope 2
Here we just conclude that T(1) = T(2)
y-direction
The forces in this direction are:
(2)
Here W is the weight of the steel beam.
We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.
Knowing that T(1) = T(2) and W = mg, we have:
![T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0](https://tex.z-dn.net/?f=T_%7B1%7Dcos%28%5Ctheta%29%2BT_%7B1%7Dcos%28%5Ctheta%29-m_%7Bsteel%7Dg%3D0)
![2T_{1}cos(\theta)-m_{steel}g=0](https://tex.z-dn.net/?f=2T_%7B1%7Dcos%28%5Ctheta%29-m_%7Bsteel%7Dg%3D0)
![2T_{1}cos(\theta)=m_{steel}g](https://tex.z-dn.net/?f=2T_%7B1%7Dcos%28%5Ctheta%29%3Dm_%7Bsteel%7Dg)
T(1) must be equal to 5479 N, so we have:
![cos(\theta)=\frac{m_{steel}g}{2T_{1}}](https://tex.z-dn.net/?f=cos%28%5Ctheta%29%3D%5Cfrac%7Bm_%7Bsteel%7Dg%7D%7B2T_%7B1%7D%7D)
![cos(\theta)=\frac{892*9.81}{2*5479}](https://tex.z-dn.net/?f=cos%28%5Ctheta%29%3D%5Cfrac%7B892%2A9.81%7D%7B2%2A5479%7D)
Therefore, the maximum angle allowed is θ = 37.01°.
You can learn more about tension here:
brainly.com/question/12797227
I hope it helps you!
Answer:
![s=6.86m/s^2](https://tex.z-dn.net/?f=s%3D6.86m%2Fs%5E2)
Explanation:
Hello,
In this case, considering that the acceleration is computed as follows:
![a=\frac{v_{final}-v_{initial}}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_%7Bfinal%7D-v_%7Binitial%7D%7D%7Bt%7D)
Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:
![a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B28.82m%2Fs-0m%2Fs%7D%7B4.2s%7D%5C%5C%20%5C%5Cs%3D6.86m%2Fs%5E2)
Regards.