There is one major difference between electromagnetic waves and mechanical waves. Electromagnetic waves can travel through space and medium. Mechanical waves needs matter to transfer.
The rate of energy absorption from the outdoor air is 4400 KJ/h
<u>Explanation:</u>
Coefficient of performance (COP), is an expression of the efficiency of a heat pump. When calculating the COP for a heat pump, the heat output from the condenser (Q) is compared to the power supplied to the compressor (W).
The coefficient of performance, ![C O P=\frac{Q_{H}}{W_{i n}}](https://tex.z-dn.net/?f=C%20O%20P%3D%5Cfrac%7BQ_%7BH%7D%7D%7BW_%7Bi%20n%7D%7D)
We replace the value in the equation with given data,
![C O P=\frac{(8000 K J / h)}{1 K W}=8000 K J / h](https://tex.z-dn.net/?f=C%20O%20P%3D%5Cfrac%7B%288000%20K%20J%20%2F%20h%29%7D%7B1%20K%20W%7D%3D8000%20K%20J%20%2F%20h)
By making a conversion, we get
![C O P=\frac{(8000 \mathrm{KJ} / \mathrm{h})}{1 \mathrm{KW}} \times \frac{(1 \mathrm{KW})}{3600}=2.22](https://tex.z-dn.net/?f=C%20O%20P%3D%5Cfrac%7B%288000%20%5Cmathrm%7BKJ%7D%20%2F%20%5Cmathrm%7Bh%7D%29%7D%7B1%20%5Cmathrm%7BKW%7D%7D%20%5Ctimes%20%5Cfrac%7B%281%20%5Cmathrm%7BKW%7D%29%7D%7B3600%7D%3D2.22)
The energy balance, ![Q_{L}=Q_{H}-W_{i n}](https://tex.z-dn.net/?f=Q_%7BL%7D%3DQ_%7BH%7D-W_%7Bi%20n%7D)
Now, replace the values in the equation, we get
![Q_{L}=(8000 \mathrm{KJ} / \mathrm{h})-(3600 \mathrm{KJ} / \mathrm{h})=4400 \mathrm{KJ} / \mathrm{h}](https://tex.z-dn.net/?f=Q_%7BL%7D%3D%288000%20%5Cmathrm%7BKJ%7D%20%2F%20%5Cmathrm%7Bh%7D%29-%283600%20%5Cmathrm%7BKJ%7D%20%2F%20%5Cmathrm%7Bh%7D%29%3D4400%20%5Cmathrm%7BKJ%7D%20%2F%20%5Cmathrm%7Bh%7D)
Remember your kinematic equations for constant acceleration. One of the equations is
![x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}](https://tex.z-dn.net/?f=%20x_%7Bf%7D%20%3D%20%20x_%7Bi%7D%20%2B%20%20v_%7Bi%7D%28t%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E%7B2%7D%20)
, where
![x_{f}](https://tex.z-dn.net/?f=x_%7Bf%7D%20)
= final position,
![x_{i}](https://tex.z-dn.net/?f=x_%7Bi%7D%20)
= initial position,
![v_{i}](https://tex.z-dn.net/?f=v_%7Bi%7D)
= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means
![x_{i}](https://tex.z-dn.net/?f=x_%7Bi%7D%20)
= 100m. You final position is where you ended up after t seconds passed, so
![x_{f}](https://tex.z-dn.net/?f=x_%7Bf%7D%20)
= 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was
![v_{i}](https://tex.z-dn.net/?f=v_%7Bi%7D)
= 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
Answer:
![0.42 m/s^2](https://tex.z-dn.net/?f=0.42%20m%2Fs%5E2)
Explanation:
There are two forces acting on the person as he's standing on the scale in the elevator:
- Its weight,
, downward, with m = 45.0 kg being his mass and g = 9.8 m/s^2 being the acceleration of gravity
- The normal reaction exerted by the scale on the person, N, upward, which is equal to the value read on the scale, N = 460 N
Using Newton's second law, we can write
![N-mg = ma](https://tex.z-dn.net/?f=N-mg%20%3D%20ma)
Where a is the acceleration of the person (and of the elevator). Solving for a,
![a=\frac{N-mg}{m}=\frac{460-(45)(9.8)}{45}=0.42 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BN-mg%7D%7Bm%7D%3D%5Cfrac%7B460-%2845%29%289.8%29%7D%7B45%7D%3D0.42%20m%2Fs%5E2)
And since we chose positive as the upward direction, the acceleration is upward.
Maybe holding a tray in the cafeteria line