Answer:
Explanation:
The question relates to time of flight of a projectile .
Time of flight = 2 u sinθ / g
u is speed of projectile , θ is angle of projectile
= 2 x 48.5 sin42 / 9.8
= 6.6 seconds .
Maximum height attained
= u² sin²θ / g
= 48.5² sin²42 / 9.8
= 107.47 m .
Answer:
![\phi=4.52 rad](https://tex.z-dn.net/?f=%5Cphi%3D4.52%20rad)
Explanation:
From the question we are told that
Distance b/e antenna's ![d=9.00m](https://tex.z-dn.net/?f=d%3D9.00m)
Frequency of antenna Radiation![F_r=120 MHz \approx 120*10^6Hz](https://tex.z-dn.net/?f=F_r%3D120%20MHz%20%5Capprox%20120%2A10%5E6Hz)
Distance from receiver ![d_r=150m](https://tex.z-dn.net/?f=d_r%3D150m)
Intensity of Receiver ![i= 10](https://tex.z-dn.net/?f=i%3D%2010)
Distance difference of the receiver b/w antenna's ![(r^2-r^1)=1.8m](https://tex.z-dn.net/?f=%28r%5E2-r%5E1%29%3D1.8m)
Generally the equation for Phase difference
is mathematically given by
![\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)](https://tex.z-dn.net/?f=%5Cphi%3D%5Cfrac%7B2%5Cpi%7D%7B%5Cfrac%7Bc%7D%7Bf_r%7D%7D%20%2A%28r%5E2-r%5E1%29)
![\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8](https://tex.z-dn.net/?f=%5Cphi%3D%5Cfrac%7B2%2A%5Cpi%7D%7B%5Cfrac%7B3%2A10%5E%7B8%7D%7D%7B120%2A10%5E6%7D%7D%20%2A1.8)
![\phi=\frac{4\pi}{5} *1.8](https://tex.z-dn.net/?f=%5Cphi%3D%5Cfrac%7B4%5Cpi%7D%7B5%7D%20%20%2A1.8)
<h3>
![\phi=4.52 rad](https://tex.z-dn.net/?f=%5Cphi%3D4.52%20rad)
</h3>
Therefore phase difference f between the two radio waves produced by this path difference is given as
![\phi=4.52 rad](https://tex.z-dn.net/?f=%5Cphi%3D4.52%20rad)
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a
7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd
Answer:the rate changes during the position of the object
Explanation:so there is no object that has the same rate but unless it is a specific one like a care but it changes during the position of the object