Answer:
The answer to your question is:
Explanation:
Data
Duane Albert
d = 5 m ; v = 3 m/s v = 4.2 m/s
a) b)
Duane's Albert's
d = 5 + (3)t d = 4.2t
d = 5 + 3t
c) 5 + 3t = 4.2t
4.2t - 3t = 5
1.2t = 5
t = 4.17 s
d)
Duane's
d= 5 + 3(4.17)
d = 17.51 m
Alberts
d = 4.2(4.17)
d = 17.51 m
Answer:
0.0002°, 0.1691°, 0.338°
Explanation:
Difference between the two line = 5.97 * 10-⁸m
d = 1 / N
N = 5.0 * 10³
d = 2.0 * 10⁴m
nL = Nsin¤
For first order
588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤
Sin¤ = 2.944*10-³
¤ = sin-¹ 0.002944
¤ = 0.1687°
First order ¤ =
Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)
Sin-¹ (0.002947) = 0.1689°
Angular separation = 0.1689 - 0.1687 = 0.0002°
Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)
Second order ¤ = 0.3378°
Angular difference = 0.3378° - 0.1687° = 0.1691°
Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°
Angular difference = 0.5067° - 0.1687° = 0.338°
Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html
The half-life equation
in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>
<em>
</em>
<em>
</em>
Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.
<em>2 * 5700 = </em>11400 yr
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J