Answer: 8 meters per second
Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s
The answer is D. Isotopes.
Hope that helped.
Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.
Explanation:
Ok, let's write the definitions:
Displacement: The displacement is equal to the difference between the final position and the initial position.
Distance traveled: Total distance that you moved.
So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:
The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.
But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.
So no, we can have a displacement equal to zero, but a distance traveled different than zero.
She ran for 3s
Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.
Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible
