Answer:
Explanation:
Initial speed, v = 10 x 10^3 m/s
Mass of the earth, M = 6 x 10^24 kg
Radius of the earth, R = 6.4 x 10^6 m
Maximum from the surface of earth, h = ?
Let m = Mass of the projectile
Solution:
Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface
![-G M m / ( R + h )=- G M m / R + (1/2) m v^2](https://tex.z-dn.net/?f=-G%20M%20m%20%2F%20%28%20R%20%2B%20h%20%29%3D-%20G%20M%20m%20%2F%20R%20%2B%20%281%2F2%29%20m%20v%5E2)
![- G M / ( R + h ) = - G M / R + (1/2) v^2](https://tex.z-dn.net/?f=-%20G%20M%20%2F%20%28%20R%20%2B%20h%20%29%20%3D%20-%20G%20M%20%2F%20R%20%2B%20%281%2F2%29%20v%5E2)
![-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2](https://tex.z-dn.net/?f=-2%5Ctimes%20G%20M%20%2F%20%28%20R%20%2B%20h%20%29%20%3D%20%28%20-%202%20G%20M%20%2F%20R%20%29%20%2B%20v%5E2)
=![( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2](https://tex.z-dn.net/?f=%28%20%28-%202%5Ctimes%206.67%5Ctimes10%5E%7B-11%7D%5Ctimes6%5Ctimes10%5E%7B24%7D%29%20%2F%286.4%5Ctimes10%5E6%29%7D%20%2B10000%5E2)
=![-2.50625\times10^7 J](https://tex.z-dn.net/?f=-2.50625%5Ctimes10%5E7%20J)
![=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7](https://tex.z-dn.net/?f=%3D-%208%5Ctimes10%5E%7B14%7D%20%2F%20%28%20R%20%2B%20h%20%29%3D-2.50625%5Ctimes%2010%5E7)
![R+h=31.92\times10^{6}](https://tex.z-dn.net/?f=R%2Bh%3D31.92%5Ctimes10%5E%7B6%7D)
![h=31.92\times10^{6}-6.4\times10^6](https://tex.z-dn.net/?f=h%3D31.92%5Ctimes10%5E%7B6%7D-6.4%5Ctimes10%5E6)
Newton's second law, which states that the force F acting on a body is equal to the mass m of the body multiplied by the acceleration a of its centre of mass, F = ma, is the basic equation of motion in classical mechanics.
Answer:
The volume of the bubble near the surface will be 9.47 m³
Explanation:
Given that,
Depth = 52.0 m
Volume = 1.50 m³
Temperature at bottom = 5.5°C
Temperature at the top = 18.5°C
We need to calculate the pressure at the depth 52.0 m
The pressure is
![P_{1}=P_{2}+\rho gh](https://tex.z-dn.net/?f=P_%7B1%7D%3DP_%7B2%7D%2B%5Crho%20gh)
Where,
= Pressure at the surface
= Pressure at the depth
Put the value into the formula
![P_{1}=101325+(1000\times9.8\times52.0)](https://tex.z-dn.net/?f=P_%7B1%7D%3D101325%2B%281000%5Ctimes9.8%5Ctimes52.0%29)
![P_{1}=610925\ N/m^2](https://tex.z-dn.net/?f=P_%7B1%7D%3D610925%5C%20N%2Fm%5E2)
We need to calculate the volume of the bubble just before it reaches the surface
Using equation of ideal gas
![PV=RT](https://tex.z-dn.net/?f=PV%3DRT)
![\dfrac{PV}{T}=constant](https://tex.z-dn.net/?f=%5Cdfrac%7BPV%7D%7BT%7D%3Dconstant)
Now, The equation of at bottom and top
![\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7B1%7DV_%7B1%7D%7D%7BT_%7B1%7D%7D%3D%5Cdfrac%7BP_%7B2%7DV_%7B2%7D%7D%7BT_%7B2%7D%7D)
![V_{2}=\dfrac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}](https://tex.z-dn.net/?f=V_%7B2%7D%3D%5Cdfrac%7BP_%7B1%7DV_%7B1%7DT_%7B2%7D%7D%7BP_%7B2%7DT_%7B1%7D%7D)
Put the value into the formula
![V_{2}=\dfrac{610925\times1.50\times(18.5+273)}{101325\times(5.5+273)}](https://tex.z-dn.net/?f=V_%7B2%7D%3D%5Cdfrac%7B610925%5Ctimes1.50%5Ctimes%2818.5%2B273%29%7D%7B101325%5Ctimes%285.5%2B273%29%7D)
![V=9.47\ m^3](https://tex.z-dn.net/?f=V%3D9.47%5C%20m%5E3)
Hence, The volume of the bubble near the surface will be 9.47 m³
Answer: The answer is in the figure attached.
Explanation:
The Work
done by a Force
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
It should be noted that it is a scalar magnitude, and its unit in the <u>International System of Units</u> is the Joule (J). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:
However, in the British Engineering and Gravitational Systems its unit is Foot-pound (ft-lb). Where ![1ft-lb=1.355818J](https://tex.z-dn.net/?f=1ft-lb%3D1.355818J)
Now, when the applied force
is constant and the direction of the force and the direction of the movement (traveled distance)
are parallel, the equation to calculate it is:
![W=(F)(d)](https://tex.z-dn.net/?f=W%3D%28F%29%28d%29)
So, taking into account the explanation above, the attached table shows the Work done for each situation.