Answer:if youre looking for the weight of the thermas in genral it should be 500n
Explanation:using the formula w=mg
w=500x10
giving us 500 newtons which is the weight.
There are 3 bases before you reach home plate.
Answer:
28,400 N
Explanation:
Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:
![p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa](https://tex.z-dn.net/?f=p_%7Btop%7D%20%3D%20p_%7Batm%7D%20%2B%20%5Crho%20g%20h%3D1.013%5Ccdot%2010%5E5%20Pa%20%2B%20%281000%20kg%2Fm%5E3%29%289.8%20m%2Fs%5E2%29%2821.0%20m%29%3D3.071%20%5Ccdot%2010%5E5%20Pa)
On the lower part of the hatch, there is a pressure equal to
![p_{bot}=p_{atm}=1.013\cdot 10^5 Pa](https://tex.z-dn.net/?f=p_%7Bbot%7D%3Dp_%7Batm%7D%3D1.013%5Ccdot%2010%5E5%20Pa)
So, the net pressure acting on the hatch is
![p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa](https://tex.z-dn.net/?f=p%3Dp_%7Btop%7D-p_%7Bbot%7D%3D3.071%20%5Ccdot%2010%5E5%20Pa%20-%201.013%5Ccdot%2010%5E5%20Pa%3D2.058%20%5Ccdot%2010%5E5%20Pa)
which acts from above.
The area of the hatch is given by:
![A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%28%5Cfrac%7B0.420%20m%7D%7B2%7D%29%5E2%3D0.138%20m%5E2)
So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:
![F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N](https://tex.z-dn.net/?f=F%3DpA%3D%282.058%5Ccdot%2010%5E5%20Pa%29%280.138%20m%5E2%29%3D28%2C400%20N)
The relationship between wavelength
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
, frequency f and speed of light c for an electromagnetic wave is
![\lambda= \frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%20%5Cfrac%7Bc%7D%7Bf%7D%20)
Using the data of the problem, we find
I think it's a pulley and a lever.