The air temperature would drop quickly compared to the soil because the difference between solid and gas. But the ground will stay cooler longer than the air because it contains it better than the air.
<span>The ball's gravitational potential energy at its peak height was 0.4 joules, and its .... Ignore air resistance and determine (a) the kinetic energy at 28.1 m, (b) the .... At position B, just before it lands, it is falling at 15m/s. a) if the blocks potential ...</span>
Answer:
A)
B) ![d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}](https://tex.z-dn.net/?f=%20d%20%3D%204181.49%20kg%2Fm%5E%7B3%7D%20%3D%204.18%20g%2Fcm%5E%7B3%7D%20)
Explanation:
A) Using the Archimedes' force we can find the weight of water displaced:
![W_{d} = W_{a} - W_{w}](https://tex.z-dn.net/?f=%20W_%7Bd%7D%20%3D%20W_%7Ba%7D%20-%20W_%7Bw%7D%20)
Where:
: is the weight of the block in the air = 20.1 N
: is the weight of the block in the water = 15.3 N
![W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N](https://tex.z-dn.net/?f=%20W_%7Bd%7D%20%3D%20W_%7Ba%7D%20-%20W_%7Bw%7D%20%3D%2020.1%20N%20-%2015.3%20N%20%3D%204.8%20N%20)
Now, the mass of the water displaced is:
![m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7BW_%7Bd%7D%7D%7Bg%7D%20%3D%20%5Cfrac%7B4.8%20N%7D%7B9.81%20m%2Fs%5E%7B2%7D%7D%20%3D%200.49%20kg%20)
The volume of the block can be found using the mass of water displaced and the density of the water:
![V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bm%7D%7Bd%7D%20%3D%20%5Cfrac%7B0.49%20kg%7D%7B997%20kg%2Fm%5E%7B3%7D%7D%20%3D%204.92%20%5Ccdot%2010%5E%7B-4%7D%20m%5E%7B3%7D%20%3D%20492%20cm%5E%7B3%7D)
B) The density of the block can be found as follows:
![d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}](https://tex.z-dn.net/?f=%20d%20%3D%20%5Cfrac%7BW_%7Ba%7D%7D%7Bg%2AV%7D%20%3D%20%5Cfrac%7B20.1%20N%7D%7B9.81%20m%2Fs%5E%7B2%7D%2A4.92%20%5Ccdot%2010%5E%7B-4%7D%20m%5E%7B3%7D%7D%20%3D%204181.49%20kg%2Fm%5E%7B3%7D%20%3D%204.18%20g%2Fcm%5E%7B3%7D%20)
I hope it helps you!
Answer:
The minimum gauge pressure is 0.4969 atm.
Explanation:
Given that,
Density = 1040 kg/m³
Height = 4.94 cm
We need to calculate the pressure
Using formula of pressure
![P_{g}=\rho g h](https://tex.z-dn.net/?f=P_%7Bg%7D%3D%5Crho%20g%20h)
Where,
=density
h = height
Put the value into the formula
![P_{g}=1040\times9.8\times4.94](https://tex.z-dn.net/?f=P_%7Bg%7D%3D1040%5Ctimes9.8%5Ctimes4.94)
![P_{g}=50348.48\ Pa](https://tex.z-dn.net/?f=P_%7Bg%7D%3D50348.48%5C%20Pa)
Pressure in atmospheres
![P_{g}=\dfrac{50348.48}{101325}](https://tex.z-dn.net/?f=P_%7Bg%7D%3D%5Cdfrac%7B50348.48%7D%7B101325%7D)
![P_{g}=0.4969\ atm](https://tex.z-dn.net/?f=P_%7Bg%7D%3D0.4969%5C%20atm)
Hence, The minimum gauge pressure is 0.4969 atm.