Answer:
Check the explanation
Explanation:
This is the step by step explanation to the above question:
![v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]](https://tex.z-dn.net/?f=v_i%20%3D%20v%20%5B%20f_L%20%2A%28v%20-%20v_b%29%20-%20f_s%2A%28v%20%2B%20v_b%29%5D%20%2F%20%5Bf_L%20%2A%20%28v%20-%20v_b%29%20%2B%20f_s%2A%28v%20%2Bv_b%29%5D)
= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]
v = 344 m/s
vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))
= 0.74 m/s
Answer:
What are we supposed to find, if it is kinetic energy then this is the solution.
K.E=1/2mv^2
K.E= kinetic energy
M=mass
V=velocity
K.E =0.5*55*0.6^2
K.E=9.9J
Explanation:
Answer:
If we’re talking about objects on the Earth, the gravitational potential energy is given by:
Explanation:
PEg=mgh
so the energy is proportional to the mass ( m ), but also to the strength of the gravitational field ( g ), and the height ( h ) to which the mass is lifted.
Let say for every 5 s of time interval the speed will remain constant
so it is given as
v(mi/h) 16 21 23 26 33 30 28
now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s
so here we will have
v(ft/s) 23.5 30.8 33.73 38.13 48.4 44 41.1
now for each interval of 5 s we will have to find the distance cover for above interval of time



so here it will cover 1298.1 ft distance in 30 s interval of time
multiply grav pull by mass of astro maybe with a calculator