The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
Answer:
Explanation:
From newton's equation of motion of uniform acceleration
v = u + at
where v is final velocity , u is initial velocity , a is acceleration and time is t .
putting the values
v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )
= 90 m /s
So , final velocity is 90 m /s .
Answer:
t = 2.58*10^-6 s
Explanation:
For a nonconducting sphere you have that the value of the electric field, depends of the region:

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the sphere = 10.0/2 = 5.0cm=0.005m
In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

with this values of a you can use the following formula:

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s
Answer:
B) 0.3Hz
Explanation:
I just took the test i hope i helped and i hope you pass the test
I'm not good with math but I think it is 23.4