The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a height of
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Substituting h0=8*a, t=6 and V0=-4*a into it,
8*a = (1/2)*g*36 - 4*a*6
Solving for a
a = 5.52 m/s^2
Answer:
Soory
Explanation:
I really dont know but i will send you wait
Answer:
False
Explanation:
it is very rare to get hyposmia
Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer:
![E = -6 \ N/C](https://tex.z-dn.net/?f=E%20%20%3D%20-6%20%5C%20%20N%2FC)
Generally given that the electric field is negative it mean that its direction is opposite to that of the force
Explanation:
From the question we are told that
The charge on the small object is ![Q = -4.00 \ nC = -4.00 *10^{-9} \ C](https://tex.z-dn.net/?f=Q%20%3D%20-4.00%20%5C%20nC%20%3D%20%20-4.00%20%2A10%5E%7B-9%7D%20%5C%20%20C)
The force is ![F = 24 \ nN = 24 *10^{-9} \ N](https://tex.z-dn.net/?f=F%20%20%3D%20%2024%20%20%5C%20nN%20%20%3D%20%2024%20%2A10%5E%7B-9%7D%20%5C%20%20N)
Generally the magnitude of the electric field is mathematically represented as
![E = \frac{F}{Q}](https://tex.z-dn.net/?f=E%20%20%3D%20%20%5Cfrac%7BF%7D%7BQ%7D)
=> ![E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}](https://tex.z-dn.net/?f=E%20%20%3D%20%20%5Cfrac%7B%2024%20%2A10%5E%7B-9%7D%7D%20%7B-4%20%2A10%5E%7B-9%20%7D%7D)
=> ![E = -6 \ N/C](https://tex.z-dn.net/?f=E%20%20%3D%20-6%20%5C%20%20N%2FC)
Generally given that the electric field is negative it mean that its direction is opposite to that of the force