Answer:
Discovered different planets.
Explanation:
<h2>
Answer:g=9.79
,A object of mass
![m](https://tex.z-dn.net/?f=m)
at the surface of earth experiences a force
![mg](https://tex.z-dn.net/?f=mg)
</h2>
Explanation:
Let
be the mass of earth.
Let
be the radius of earth.
Let
be the universal gravitational constant.
Given,
![M=5.96\times 10^{24}Kg](https://tex.z-dn.net/?f=M%3D5.96%5Ctimes%2010%5E%7B24%7DKg)
![R=6.37\times 10^{6}m](https://tex.z-dn.net/?f=R%3D6.37%5Ctimes%2010%5E%7B6%7Dm)
![G=6.67259 \times 10^{-11}](https://tex.z-dn.net/?f=G%3D6.67259%20%5Ctimes%2010%5E%7B-11%7D)
![Nm^{2}Kg^{-2}](https://tex.z-dn.net/?f=Nm%5E%7B2%7DKg%5E%7B-2%7D)
Let
be the acceleration due to gravity.
Then,![g=\dfrac{GM}{R^{2}}](https://tex.z-dn.net/?f=g%3D%5Cdfrac%7BGM%7D%7BR%5E%7B2%7D%7D)
![g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}](https://tex.z-dn.net/?f=g%3D%5Cfrac%7B6.67259%20%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.96%5Ctimes%2010%5E%7B24%7D%7D%7B%286.37%5Ctimes%2010%5E%7B6%7D%29%5E%7B2%7D%7D)
![g=9.79ms^{-2}](https://tex.z-dn.net/?f=g%3D9.79ms%5E%7B-2%7D)
A object of mass
at the surface of earth experiences a force ![mg](https://tex.z-dn.net/?f=mg)
Answer:
A) E = 3.70*10^{4} N/C
B) E = 2.281*10^3 N/C
Explanation:
given data:
charge density ![\lambda = 130*10^{-9} C/m](https://tex.z-dn.net/?f=%20%5Clambda%20%3D%20130%2A10%5E%7B-9%7D%20C%2Fm)
length of wire = 9.50 cm
a) at x = 4.5 m above midpoint, electric field is calculated as
![E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B%202%5Cpi%20%5Cepsilon%7D%20%2A%20%5Cfrac%7B%20%5Clambda%7D%7Bx%5Csqrt%7B%28x%5E2%2Fa%5E2%29%2B1%7D%7D)
x = 4.5 cm
midpoint a = 4.5 cm = 0.0475 m
![E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}](https://tex.z-dn.net/?f=E%20%3D2%7B%5Cfrac%7B1%7D%7B%208.99%2A10%5E9%7D%20%2A%20%5Cfrac%7B130%2A10%5E%7B-9%7D%20%7D%7B0.045%5Csqrt%7B%284.5%5E2%2F4.75%5E2%29%2B1%7D%7D)
E = 3.70*10^{4} N/C
B) when wire is in circle form
![Q = \lambda * L](https://tex.z-dn.net/?f=Q%20%3D%20%5Clambda%20%2A%20L)
![= 130*10^{-9} *9.5*10^{-2}](https://tex.z-dn.net/?f=%3D%20130%2A10%5E%7B-9%7D%20%2A9.5%2A10%5E%7B-2%7D)
= 1.235*10^{-8} C
Radius of circle
![r = \frac{L}{2\pi}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7BL%7D%7B2%5Cpi%7D)
![r = \frac{9.5*10^{-2}}{2\pi}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B9.5%2A10%5E%7B-2%7D%7D%7B2%5Cpi%7D)
r = 1.511*10^{-2} m
![E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B%202%5Cpi%20%5Cepsilon%7D%20%2A%20%5Cfrac%7BQx%7D%7B%28x%5E2%2Br%5E2%29%5E%7B3%2F2%7D%7D)
![E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}](https://tex.z-dn.net/?f=E%20%3D8.99%2A10%5E%7B9%7D%20%2A%20%5Cfrac%7B1.23%2A10%5E%7B-8%7D%2A4.5%2A10%5E%7B-2%7D%7D%7B%28%284.5%2A10%5E%7B-2%7D%29%5E2%2B%281.511%2A10%5E%7B-2%7D%29%5E2%29%5E%7B3%2F2%7D%7D)
E = 2.281*10^3 N/C
Answer:
In kinematics questions we need to separate the question into different parts if the acceleration changes. Here, there are three time intervals where acceleration is different.
1) a(t) = 96t. We can find the velocity function of the rocket by integrating the acceleration function. Then we can integrate again to find the position function.
![v(t) = \int{a(t)} \, t = \int {96t} \, dt = 48t^2 + C](https://tex.z-dn.net/?f=v%28t%29%20%3D%20%5Cint%7Ba%28t%29%7D%20%5C%2C%20t%20%3D%20%5Cint%20%7B96t%7D%20%5C%2C%20dt%20%3D%2048t%5E2%20%2B%20C)
'C' is the integration constant. We can find this constant by investigating the initial conditions.
![v(t = 0) = 48(0)^2 + C = 0\\C = 0](https://tex.z-dn.net/?f=v%28t%20%3D%200%29%20%3D%2048%280%29%5E2%20%2B%20C%20%3D%200%5C%5CC%20%3D%200)
We know that the rocket is initially at rest, so 'C' should be zero.
![s(t) = \int {v(t)} \, dt = \int {48t^2} \, dt = 16t^3 + C](https://tex.z-dn.net/?f=s%28t%29%20%3D%20%5Cint%20%7Bv%28t%29%7D%20%5C%2C%20dt%20%20%3D%20%5Cint%20%7B48t%5E2%7D%20%5C%2C%20dt%20%3D%2016t%5E3%20%2B%20C)
Again, the rocket started from ground zero, so C = 0.
We should conclude the first part by calculating the final position and final velocity of the rocket.
![s(t=3) = 16(3)^3 = 432ft\\v(t=3) = 48(3)^2 = 432ft/s](https://tex.z-dn.net/?f=s%28t%3D3%29%20%3D%2016%283%29%5E3%20%3D%20432ft%5C%5Cv%28t%3D3%29%20%3D%2048%283%29%5E2%20%3D%20432ft%2Fs)
2) For the second part, the rocket is in free fall, so
![a(t) = -32.2ft/s^2\\v(t) = -32.2t + C\\v(t=3) = -32.2*3 + C = 432\\C = 335.4\\v(t) = -32.2t + 335.4\\s(t) = -16.1t^2 + 335.4t + C\\s(t=3) = -16.1(3)^2 + 335.4*3 + C = 432\\C = 432 + 144.9 - 1006.2 = -429.3\\s(t) = -16.1t^2 + 335.4t - 429.3](https://tex.z-dn.net/?f=a%28t%29%20%3D%20-32.2ft%2Fs%5E2%5C%5Cv%28t%29%20%3D%20-32.2t%20%2B%20C%5C%5Cv%28t%3D3%29%20%3D%20-32.2%2A3%20%2B%20C%20%3D%20432%5C%5CC%20%3D%20335.4%5C%5Cv%28t%29%20%3D%20-32.2t%20%2B%20335.4%5C%5Cs%28t%29%20%3D%20-16.1t%5E2%20%2B%20335.4t%20%2B%20C%5C%5Cs%28t%3D3%29%20%3D%20-16.1%283%29%5E2%20%2B%20335.4%2A3%20%2B%20C%20%3D%20432%5C%5CC%20%3D%20432%20%2B%20144.9%20-%201006.2%20%3D%20-429.3%5C%5Cs%28t%29%20%3D%20-16.1t%5E2%20%2B%20335.4t%20-%20429.3)
The maximum height that the rocket reaches is when its velocity is zero.
So,
![v(t) = -32.2t + 335.4 = 0\\t = 10.4 s](https://tex.z-dn.net/?f=v%28t%29%20%3D%20-32.2t%20%2B%20335.4%20%3D%200%5C%5Ct%20%3D%2010.4%20s)
The maximum height is
![s(t=10.4) = -16.1t^2 + 335.4t - 429.3 = -16.1(10.4)^2 + 335.4*10.4 - 429.3 = -1741.3 + 3488.1 - 429.3 = 1318 ~ft](https://tex.z-dn.net/?f=s%28t%3D10.4%29%20%3D%20-16.1t%5E2%20%2B%20335.4t%20-%20429.3%20%3D%20-16.1%2810.4%29%5E2%20%2B%20335.4%2A10.4%20-%20429.3%20%3D%20-1741.3%20%2B%203488.1%20-%20429.3%20%3D%201318%20~ft)
The final positions for the part 2 is
![s(t=19) = -16.1(19)^2 + 335.4*19 - 429.3 = -5812.1 + 6372 - 429.3 = 131.2~ft\\v(t=19) = -32.2*19 + 335..4 = -276.4~ft/s](https://tex.z-dn.net/?f=s%28t%3D19%29%20%3D%20-16.1%2819%29%5E2%20%2B%20335.4%2A19%20-%20429.3%20%3D%20-5812.1%20%2B%206372%20-%20429.3%20%3D%20131.2~ft%5C%5Cv%28t%3D19%29%20%3D%20-32.2%2A19%20%2B%20335..4%20%3D%20-276.4~ft%2Fs)
3) With the parachute, the velocity is dropped from -276.4 to 16 in 5 s.
![a(t) = \frac{-16 - (-276.4))}{5} = 52ft/s^2\\v(t) = 52t + C\\v(t= 19) = 52*19 + C= -276.4\\988+ C = -276.4\\C = -1264.4\\v(t) = 52t - 1264.4](https://tex.z-dn.net/?f=a%28t%29%20%3D%20%5Cfrac%7B-16%20-%20%28-276.4%29%29%7D%7B5%7D%20%3D%2052ft%2Fs%5E2%5C%5Cv%28t%29%20%3D%2052t%20%2B%20C%5C%5Cv%28t%3D%2019%29%20%3D%2052%2A19%20%2B%20C%3D%20-276.4%5C%5C988%2B%20C%20%3D%20-276.4%5C%5CC%20%3D%20-1264.4%5C%5Cv%28t%29%20%3D%2052t%20-%201264.4)
![s(t) = 26t^2 - 1264.4t + C \\s(t=19) = 26(19)^2 - 1264.4*19 + C = 131.2\\9386 - 24023.6 + C = 131.2\\C = 14768.8\\s(t) = 26t^2 - 1264.4t + 14768.8](https://tex.z-dn.net/?f=s%28t%29%20%3D%2026t%5E2%20-%201264.4t%20%2B%20C%20%5C%5Cs%28t%3D19%29%20%3D%2026%2819%29%5E2%20-%201264.4%2A19%20%2B%20C%20%3D%20131.2%5C%5C9386%20-%2024023.6%20%2B%20C%20%3D%20131.2%5C%5CC%20%3D%2014768.8%5C%5Cs%28t%29%20%3D%2026t%5E2%20-%201264.4t%20%2B%2014768.8)
The rocket lands
![s(t) = 26t^2 - 1264.4t + 14768.8 = 0\\t = 29s.](https://tex.z-dn.net/?f=s%28t%29%20%3D%2026t%5E2%20-%201264.4t%20%2B%2014768.8%20%3D%200%5C%5Ct%20%3D%2029s.)
Answer: When a car is struck by lightning, the resulting electric field inside the car is zero.
Explanation: