Answer:
D
Explanation:
Energy is never created nor destroyed.
Answer:
50 s
Explanation:
Given:
Δx = 0.005 m
v₀ = 0 m/s
t = 0.50 s
Find: a
Δx = v₀ t + ½ at²
0.005 m = (0 m/s) (0.50 s) + ½ a (0.50 s)²
a = 0.04 m/s²
Given:
Δx = 50 m
v₀ = 0 m/s
a = 0.04 m/s²
Find: t
Δx = v₀ t + ½ at²
50 m = (0 m/s) t + ½ (0.04 m/s²) t²
t = 50 s
First find the time it takes for the ball to reach the ground using the vertical component of its position vector:
Meanwhile, the horizontal component of the ball's position vector is
After about 4.52 s, the ball has traveled a horizontal distance of
which you would round to 200 m, so the answer is B.
Answer:
93.125 × 10^(19)
Explanation:
We are told the asteroid has acquired a net negative charge of 149 C.
Thus;
Q = -149 C
charge on electron has a value of:
e = -1.6 × 10^(-19) C
Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.
Thus;
n = Q/e
n = -149/(-1.6 × 10^(-19))
n = 93.125 × 10^(19)
Thus, it has 93.125 × 10^(19) more electrons than protons
Answer:
θ₁ = 5.4°
θ₂ = 10.86°
Explanation:
The angle ca be found by using grating equation:
mλ = d Sinθ
where,
m = order of diffraction
λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m
d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line
θ = angle = ?
FOR m = 1:
(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁
Sin θ₁ = 0.09425
θ₁ = Sin⁻¹(0.09425)
<u>θ₁ = 5.4°</u>
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FOR m = 2:
(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁
Sin θ₂ = 0.1885
θ₂ = Sin⁻¹(0.1885)
<u>θ₂ = 10.86°</u>
