Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
![v=gt](https://tex.z-dn.net/?f=v%3Dgt)
And the distance traveled downwards is:
![\displaystyle y=\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7Bgt%5E2%7D%7B2%7D)
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
![\displaystyle t=\sqrt{\frac{2y}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Csqrt%7B%5Cfrac%7B2y%7D%7Bg%7D%7D)
Replacing into the first equation:
![\displaystyle v=g\sqrt{\frac{2y}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3Dg%5Csqrt%7B%5Cfrac%7B2y%7D%7Bg%7D%7D)
Rationalizing:
![\displaystyle v=\sqrt{2gy}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Csqrt%7B2gy%7D)
Let's call v1 the final speed of the package dropped from a height H. Thus:
![\displaystyle v_1=\sqrt{2gH}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_1%3D%5Csqrt%7B2gH%7D)
Let v2 be the final speed of the package dropped from a height 4H. Thus:
![\displaystyle v_2=\sqrt{2g(4H)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D%5Csqrt%7B2g%284H%29%7D)
Taking out the square root of 4:
![\displaystyle v_2=2\sqrt{2gH}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D2%5Csqrt%7B2gH%7D)
Dividing v2/v1 we can compare the final speeds:
![\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%2Fv_1%3D%5Cfrac%7B2%5Csqrt%7B2gH%7D%7D%7B%5Csqrt%7B2gH%7D%7D)
Simplifying:
![\displaystyle v_2/v_1=2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%2Fv_1%3D2)
The final speed of the second package is twice as much as the final speed of the first package.
The gravity on Neptune is 11.15 m/s²
the gravity on earth is 9.81 m/s²
divide the Neptune and earth gravity we get 1.13
which means object on neptune is 1.13 heavier than earth
yield, weigh of the object on neptune is 1.13×15=17.04kg
Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
Answer:v nxfgdjngdnmgndjfnncnfndngndsnbxzmnfn
Explanation: