Explanation:
The given data is as follows.
m = 5000 kg, h = 800 km = ![0.8 \times 10^{6} m](https://tex.z-dn.net/?f=0.8%20%5Ctimes%2010%5E%7B6%7D%20m)
, r = R + h = ![7.17 \times 10^{6} m](https://tex.z-dn.net/?f=7.17%20%5Ctimes%2010%5E%7B6%7D%20m)
kg, G = ![6.67 \times 10^{-11} Nm^{2}/kg^{2}](https://tex.z-dn.net/?f=6.67%20%5Ctimes%2010%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D)
As we know that,
![\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D%20%3D%20%5Cfrac%7BGmM_%7Be%7D%7D%7Br%5E%7B2%7D%7D)
v = ![\sqrt{\frac{GM_{e}}{r^{2}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7Br%5E%7B2%7D%7D%7D)
And, it is known that formula to calculate angular velocity is as follows.
![\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7Br%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7Br%5E%7B3%7D%7D%7D)
v = ![\sqrt{\frac{GM_{e}}{r^{3}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7Br%5E%7B3%7D%7D%7D)
= ![\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D%20%5Ctimes%205.98%20%5Ctimes%2010%5E%7B-24%7D%20kg%5E%7B-2%7D%7D%7B%287.17%20%5Ctimes%2010%5E%7B6%7D%20m%29%5E%7B3%7D%7D%7D)
= ![1.0402 \times 10^{-3} rad/s](https://tex.z-dn.net/?f=1.0402%20%5Ctimes%2010%5E%7B-3%7D%20rad%2Fs)
Thus, we can conclude that speed of the satellite is
.
Answer:11
Explanation: because is the number that repeted 2 times
Answer:
![x' = 10 x](https://tex.z-dn.net/?f=x%27%20%3D%2010%20x)
Explanation:
By energy conservation we know that spring energy is converted into kinetic energy of the block
so we will have
![\frac{1}{2}kx^2 = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
so we will have
![v = \sqrt{\frac{k}{m}}(x)](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%28x%29)
now we will have same thing for another mass 4m which moves out with speed 5v
so we have
![5v = \sqrt{\frac{k}{4m}}(x')](https://tex.z-dn.net/?f=5v%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7B4m%7D%7D%28x%27%29)
now from above two equations we have
![\frac{5v}{v} = \frac{x'}{2x}](https://tex.z-dn.net/?f=%5Cfrac%7B5v%7D%7Bv%7D%20%3D%20%5Cfrac%7Bx%27%7D%7B2x%7D)
so we have
![x' = 10 x](https://tex.z-dn.net/?f=x%27%20%3D%2010%20x)
Answer:
![a=2.36\ m/s^2](https://tex.z-dn.net/?f=a%3D2.36%5C%20m%2Fs%5E2)
T=157.06 N
Explanation:
Given that
Mass of first block = 21.1 kg
Mass of second block = 12.9 kg
First mass is heavier than first that is why mass second first will go downward and mass second will go upward.
Given that pulley and string is mass less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.
So now from force equation
![m_1g-m_2g=(m_1+m_2)a](https://tex.z-dn.net/?f=m_1g-m_2g%3D%28m_1%2Bm_2%29a)
21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a
![a=2.36\ m/s^2](https://tex.z-dn.net/?f=a%3D2.36%5C%20m%2Fs%5E2)
Lets tension in string is T
![m_1g-T=m_1a](https://tex.z-dn.net/?f=m_1g-T%3Dm_1a)
![T=m_1(g-a)](https://tex.z-dn.net/?f=T%3Dm_1%28g-a%29)
T=21.1(9.81-2.36) N
T=157.06 N
Answer:
![F = 2.6 \times 10^3 N](https://tex.z-dn.net/?f=F%20%3D%202.6%20%5Ctimes%2010%5E3%20N)
Explanation:
Maximum height reached by the ball after being popped by the bat is given as
![y = 40 m](https://tex.z-dn.net/?f=y%20%3D%2040%20m)
now we know by energy conservation final speed of the ball after being hit by the bat is given as
![\frac{1}{2}mv^2 = mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20mgh)
![v = \sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%7D)
![v = \sqrt{2(9.81)(40)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2%289.81%29%2840%29%7D)
![v = 28 m/s](https://tex.z-dn.net/?f=v%20%3D%2028%20m%2Fs)
now the change in momentum of the ball is given as
![\Delta P = m(v_f - v_i)](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20m%28v_f%20-%20v_i%29)
![\Delta P = 0.145(28\hat j + 35 \hat i)](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%200.145%2828%5Chat%20j%20%2B%2035%20%5Chat%20i%29)
now force is given as rate of change in momentum
![F = \frac{\Delta P}{\Delta t}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%5CDelta%20P%7D%7B%5CDelta%20t%7D)
![F = \frac{0.145(28\hat j + 35 \hat i)}{2.5 \times 10^{-3}}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B0.145%2828%5Chat%20j%20%2B%2035%20%5Chat%20i%29%7D%7B2.5%20%5Ctimes%2010%5E%7B-3%7D%7D)
![F = (1.62 \hat j + 2.03 \hat i)\times 10^3 N](https://tex.z-dn.net/?f=F%20%3D%20%281.62%20%5Chat%20j%20%2B%202.03%20%5Chat%20i%29%5Ctimes%2010%5E3%20N)
so magnitude of the force is given as
![F = \sqrt{1.62^2 + 2.03^2} \times 10^3 N](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7B1.62%5E2%20%2B%202.03%5E2%7D%20%5Ctimes%2010%5E3%20N)
![F = 2.6 \times 10^3 N](https://tex.z-dn.net/?f=F%20%3D%202.6%20%5Ctimes%2010%5E3%20N)