Answer:
Explanation:
Given:
<u>Using the equation of heat:</u>
<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>
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:
123 J transfer into the gas
Explanation:
Here we know that 123 J work is done by the gas on its surrounding
So here gas is doing work against external forces
Now for cyclic process we know that
so from 1st law of thermodynamics we have
so work done is same as the heat supplied to the system
So correct answer is
123 J transfer into the gas