Answer:
the acceleration due to gravity, g of the cylinder is 9.81m/s2
Explanation:
since the cylinder is rolling down from a hill, the acceleration due to gravity of any object coming down from the atmosphere is always constant which is g=9.8m/s2 or approximately 10m/s2
Intensity of electromagnetic wave is given as
![I = 2[\frac{B_{rms}^2}{2\mu_0}\times c]](https://tex.z-dn.net/?f=I%20%3D%202%5B%5Cfrac%7BB_%7Brms%7D%5E2%7D%7B2%5Cmu_0%7D%5Ctimes%20c%5D)
given that


here we know that


now we have


now we will have


frequency of wave is given as


now the induced EMF is given as



Answer:
C. N/m (newtons/meter)
Explanation:
Since the equation for potential energy for a spring is PE = 1/2kx², after you know <em>k </em>and <em>x</em>, you will get an answer in Joules.
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Where is the picture cant6 see it
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m