Oxygen has Atomic number 8 so all isotopes have 8 protons and 8 electrons.
So the number of neutrons in Oxygen-18 = 18 - 8 = 10.
Option B is the correct one.
<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us. So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y =
t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
D. Budgeting time, avoiding stress, and prioritizing.
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s