The relevant equation we can use in this problem is:
h = v0 t + 0.5 g t^2
where h is height, v0 is initial velocity, t is time, g is
gravity
Since it was stated that the rock was drop, so it was free
fall and v0 = 0, therefore:
h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2
<span>h = 117.77 m</span>
Letra a obrigadoaijahahavqvqgqgqg
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of

from the center of the pattern. In the formula, m is the order of the minimum,

the wavelenght,

the distance of the screen from the slit and

the width of the slit.
In our problem, the distance of the first-order band (m=1) is

. The distance of the screen is D=86 cm while the wavelength is

. Using these data and re-arranging the formula, we can find a, the width of the slit:
Answer:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .
A will be the fastest and c the slowest because of the dip it has a is a straight line fastest way to get from a to b is a straight line b is the second fastest and d is last