A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
Read more on Brainly.com -
brainly.com/question/18743384#readmore
Answer:
3.75 MeV
Explanation:
The energy of the photon can be given in terms of frequency as:
E = h * f
Where h = Planck's constant
The frequency of the photon is 6 * 10^20 Hz.
The energy (in Joules) is:
E = 6.63 x10^(-34) * 6 * 10^(20)
E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J
We are given that:
1 eV = 1.06 * 10^(-19) Joules
This means that 1 Joule will be:
1 J = 1 / (1.06 * 10^(-19)
1 J = 9.434 * 10^(18) eV
=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV
This is the same as 3.75 MeV.
The correct answer is not in the options, but the closest to it is option C.
Yes. It means that the acceleration increases at a constant rate, for example 3 mph every second.
