Well, the lines of workings are cut off in the middle, and there's no 2nd image.
But I think it must have to do with the "cos" terms at the right end of the picture.
I'm guessing now, because the part I'm interested in would be just past the edge, where we can't see it.
I <em>think</em> that the first line says "cos(some angle)", and at the same place in the second line it says "cos(180 - the same angle)".
If that's what it says, then that's your answer, because cos(anything) is equal to the <em>negative</em> of cos(180 - the same thing).
That's the best I can do for you just now. Honestly, I don't see the connection yet between the question Dave is working on and the two lines I see in the picture.
Answer:
B: Greatest, A: Second greatest, C: Second geratest.
Explanation:
We can calculate the heat of fusion of a material (ΔHfus), which is the heat required to melt 1 kg of material, using the following expression.
ΔHfus = Q/m
where,
Material A
ΔHfus = 200 J/4 kg = 50 J/kg
Material B
ΔHfus = 300 J/5 kg = 60 J/kg
Material C
ΔHfus = 300 J/6 kg = 50 J/kg
Then, considering the heats of fusion,
B: Greatest, A: Second greatest, C: Second greatest.
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°
Answer:
220.7
Explanation:
distance traveled is 1.5*5=7.5m
PE gain is MGH=3*9.81*7.5=220.7
Answer:
The starting height of the ball is approximately 0.604 m
Explanation:
The given parameters are;
The mass of the the ball = 0.050
The speed with which it travels through the top loop = 2 m/s
The given height at which the ball moves at 2 m/s = 0.40 m
Therefore, we have;
1/2·m·v² = m·g·h
1/2·v² = g·h
h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204
The additional height = h = 0.204 m
Therefore;
The starting height of the ball ≈ The given height at which the ball moves at 2 m/s + h
The starting height of the ball ≈ 0.40 + 0.204 = 0.604 m
The starting height of the ball ≈ 0.604 m.