The density of mercury is 13.6 g/ml. what is its density in lbs/L
1 answer:
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Answer:
a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft
Explanation:
For this exercise let's use the projectile launch relationships.
The initial height is I = 5 ft and the final height y = 4 ft
y = y₀ + t - ½ g t²
The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft
x = v₀ₓ t
t = x / v₀ₓ
We replace
y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²
v_{oy} = v₀ sin θ
v₀ₓ = vo cos θ
y –y₀ = x tan θ - ½ g x² / v₀² cos² θ
5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)
1 = 120 tan θ - 0.0213 sec² θ
Let's use the trigonometry relationship
Sec² θ = 1 - tan² θ
1 = 120 tan θ - 0.0213 (1 –tan²θ)
0.0213 tan²θ + 120 tanθ -1.0213 = 0
We change variables
u = tan θ
u² + 5633.8 u - 48.03 = 0
We solve the second degree equation
u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2
u = [- 5633.8 ± 5633.82] / 2
u₁ = 0.0085
u₂= -5633.81
u = tan θ
θ = tan⁻¹ u
For u₁
θ₁ = tan⁻¹ 0.0085
θ₁ = 0.487º
For u₂
θ₂ = -89.99º
The launch angle must be 0.487º
b) let's look for the time it takes for the arrow to arrive
x = v₀ₓ t
t = x / v₀ cos θ
t = 120 / (300 cos 0.487)
t = 0.400 s
The deer must be at a distance of
v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s
x = v t
x = 29.33 0.4
x = 11.73 ft
Answer:
B. QC > 0; QH < 0
Explanation:
Given that there are two reservoir of energy.
Sign convention for heat and work :
1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
From hot reservoir heat is going out that is why it is taken as negative
From cold reservoir heat is coming inside the reservoir that is why it is taken as positive
That is why the answer will be
,